3.3 Transpose, symmetric, and skew-symmetric matrices

Motivation

Once matrix multiplication is available, the shape of a matrix matters as much as its numerical entries. A row of coefficients and a column of coordinates play different roles in a product, yet many arguments need to pass from one role to the other without changing the underlying data. Transpose performs exactly that conversion: rows become columns, columns become rows, and the dimensions change accordingly.

For square matrices, this index swap can be pictured as reflection across the main diagonal. Matrices fixed by that reflection are symmetric; matrices sent to their negatives are skew-symmetric. These conditions are more than visual patterns. They determine how products behave, explain why multiplication order matters, and lead to a canonical way of separating any square matrix into two structurally simpler parts.

Throughout this note, matrix entries are real numbers. All transpose identities remain valid over an arbitrary field, but the statements that divide by $2$ or deduce $a_{ii}=0$ from $a_{ii}=-a_{ii}$ require that $2$ be nonzero and invertible. Thus they also hold over any field of characteristic not equal to $2$. In characteristic $2$, symmetric and skew-symmetric conditions are no longer distinct, so those conclusions must be reformulated.

Transpose swaps rows and columns

Definition 3.3.1

(Transpose)

If $A = [a_{ij}]$ is an $m \times n$ matrix, then its transpose $A^T$ is the $n \times m$ matrix whose $(j,i)$ entry is $a_{ij}$.

Equivalently,

$$(A^T)_{ji} = a_{ij}.$$

So every row of $A$ becomes a column of $A^T$, and every column of $A$ becomes a row.

This operation does not alter the entries themselves; it changes where each entry is read. If $A$ is $m \times n$, then the row index of $A$ ranges from $1$ to $m$ and its column index ranges from $1$ to $n$. After transposition, those index ranges exchange places. This dimension check should be made before using any transpose identity, because an equality of matrices is meaningful only when both sides have the same size.

The sequence below sets out the geometric picture before the examples: transpose is an index swap, and for square matrices that index swap becomes reflection across the main diagonal.

Guided visual comparison

The live comparison below lets you switch among several examples. Compare $A$ with $A^T$, then look at what happens to entries on opposite sides of the main diagonal.

Worked example 3.3.2

(Compute a transpose)

Let

$$A = \begin{bmatrix} 1 & 4 & -2 \\ 0 & 3 & 5 \end{bmatrix}.$$

Then

$$A^T = \begin{bmatrix} 1 & 0 \\ 4 & 3 \\ -2 & 5 \end{bmatrix}.$$

The $2 \times 3$ matrix becomes a $3 \times 2$ matrix because rows and columns swap roles.

Three basic identities should be treated as part of the definition-level vocabulary of the subject:

$$(A^T)^T = A, \qquad (A + B)^T = A^T + B^T, \qquad (cA)^T = cA^T.$$

When products are defined, transpose reverses the order:

$$(AB)^T = B^T A^T.$$

That reversal is not a cosmetic detail. It reflects the fact that matrix multiplication is built from row-column pairings, and transposition switches the role of rows and columns.

Worked example 3.3.3

(Transpose identities in action)

Take

$$A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 3 & 0 \\ 4 & 5 \end{bmatrix}.$$

Then

$$A + B = \begin{bmatrix} 4 & 2 \\ 4 & 6 \end{bmatrix}, \qquad (A + B)^T = \begin{bmatrix} 4 & 4 \\ 2 & 6 \end{bmatrix}.$$

On the other hand,

$$A^T + B^T = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} + \begin{bmatrix} 3 & 4 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 2 & 6 \end{bmatrix}.$$

The product rule is similar, but the order matters:

$$AB = \begin{bmatrix} 11 & 10 \\ 4 & 5 \end{bmatrix}, \qquad (AB)^T = \begin{bmatrix} 11 & 4 \\ 10 & 5 \end{bmatrix},$$

while

$$B^T A^T = \begin{bmatrix} 3 & 4 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 4 \\ 10 & 5 \end{bmatrix}.$$

Theorem 3.3.4

(Basic properties of transpose)

Let $A$ and $B$ have the same size when they are added, let $c$ be a scalar, and suppose $A$ is $m \times n$ and $B$ is $n \times p$ when they are multiplied. Then:

1. $(A^T)^T = A$
2. $(A + B)^T = A^T + B^T$
3. $(cA)^T = cA^T$
4. $(AB)^T = B^T A^T$

The first three rules follow immediately by comparing entries. The product rule deserves a complete calculation because it explains why the order reverses.

Proof

(Entrywise proof of the product rule)

Let $A=[a_{ij}]$ be $m \times n$ and $B=[b_{ij}]$ be $n \times p$. Then $AB$ is $m \times p$, so both $(AB)^T$ and $B^T A^T$ are $p \times m$. For $1 \le i \le p$ and $1 \le j \le m$, the $(i,j)$ entry of the first matrix is

$$\bigl((AB)^T\bigr)_{ij} =(AB)_{ji} =\sum_{k=1}^{n} a_{jk}b_{ki}.$$

On the other hand, the $(i,j)$ entry of $B^T A^T$ is

$$\begin{aligned} (B^T A^T)_{ij} &=\sum_{k=1}^{n}(B^T)_{ik}(A^T)_{kj} \\ &=\sum_{k=1}^{n}b_{ki}a_{jk} \\ &=\sum_{k=1}^{n}a_{jk}b_{ki}. \end{aligned}$$

The last equality uses commutativity of scalar multiplication, not commutativity of matrix multiplication. Corresponding entries therefore agree, so $(AB)^T=B^T A^T$. The dimensions also explain the reversed order: $B^T$ is $p \times n$ and $A^T$ is $n \times m$, whereas $A^T B^T$ would generally not even be defined.

Why transpose matters beyond rearranging entries

For real column vectors $u,v$ of the same size, the scalar $u^Tv=\sum\limits_i u_iv_i$ is called their dot product. This formula is all that is needed here; its geometric properties are developed in Inner products, norms, and angles.

Suppose $A$ is $m \times n$, $x$ is a column in $\mathbb{R}^n$, and $y$ is a column in $\mathbb{R}^m$. The product $Ax$ lies in $\mathbb{R}^m$, so its dot product with $y$ is defined. Written as matrix multiplication, the same scalar can be regrouped as

$$(Ax)^T y=x^T A^T y=x^T(A^T y).$$

The first equality uses the product-transpose rule; the second only changes parentheses, using associativity. Thus an application of $A$ to $x$ can be transferred to an application of $A^T$ to $y$ when the two sides occur inside a dot product. Later courses describe $A^T$ as the matrix that moves a linear map from one side of the Euclidean inner product to the other.

The dimensions clarify the roles. The matrix $A$ sends an $n$-component input to an $m$-component output, whereas $A^T$ sends an $m$-component coefficient vector back to $n$ components. This does not say that $A^T$ undoes $A$; that would be a statement about an inverse. It says instead that transpose records how the same coefficients are read after rows and columns exchange roles. This distinction becomes essential in orthogonality, projections, and least-squares problems, where a rectangular matrix cannot have an ordinary two-sided inverse but always has a transpose.

This identity also provides a reliable dimension audit: if either side is not a scalar, then at least one matrix or vector has been placed in the wrong order.

Symmetric and skew-symmetric matrices

Definition 3.3.5

(Symmetric and skew-symmetric matrices)

Let $A$ be a square matrix.

• $A$ is symmetric if $A^T = A$.
• $A$ is skew-symmetric if $A^T = -A$.

Notice the word square. If $A$ is not square, then $A$ and $A^T$ do not even have the same size, so the equations $A^T = A$ and $A^T = -A$ cannot be true.

Symmetry says the matrix matches its reflection across the main diagonal. Skew-symmetry says the reflected matrix is the negative of the original. Entrywise, these statements are

$$A\text{ symmetric} \iff a_{ij}=a_{ji}, \qquad A\text{ skew-symmetric} \iff a_{ij}=-a_{ji}$$

for every pair of indices $i,j$. The diagonal deserves special attention. If $A$ is skew-symmetric, then $a_{ii}=-a_{ii}$, so $2a_{ii}=0$. Over the real numbers this forces $a_{ii}=0$. This short argument is exactly where the characteristic-not-$2$ assumption enters.

Worked example 3.3.6

(Recognize symmetry)

The matrix

$$\begin{bmatrix} 2 & -1 & 4 \\ -1 & 3 & 0 \\ 4 & 0 & 5 \end{bmatrix}$$

is symmetric, because the $(i,j)$ and $(j,i)$ entries agree.

The matrix

$$\begin{bmatrix} 0 & 2 & -1 \\ -2 & 0 & 4 \\ 1 & -4 & 0 \end{bmatrix}$$

is skew-symmetric, because the transpose changes every off-diagonal entry's sign and leaves the diagonal as $0$.

Three quick consequences are especially useful:

• every diagonal entry of a real skew-symmetric matrix is $0$
• the zero matrix is both symmetric and skew-symmetric
• the identity matrix is symmetric

The third item is a special case of the fact that diagonal matrices are fixed by transpose.

Theorem 3.3.7

(Useful transpose-based identities)

For every square matrix $A$,

$$(A + A^T)^T = A + A^T, \qquad (A - A^T)^T = -(A - A^T).$$

So $A + A^T$ is symmetric and $A - A^T$ is skew-symmetric.

Proof

(Why the identities hold)

Apply the basic transpose rules:

$$(A + A^T)^T = A^T + (A^T)^T = A^T + A = A + A^T.$$

For the difference,

$$(A - A^T)^T = A^T - (A^T)^T = A^T - A = -(A - A^T).$$

The previous result is the starting point for the most important structural fact in this section.

Theorem 3.3.8

(Decomposition into symmetric and skew-symmetric parts)

Every real square matrix $A$ can be written uniquely as

$$A = S + K,$$

where $S$ is symmetric and $K$ is skew-symmetric. In fact,

$$S = \frac{1}{2}(A + A^T), \qquad K = \frac{1}{2}(A - A^T).$$

The conclusion is both an existence statement and a uniqueness statement. A formula alone is not enough unless both parts are checked.

Proof

(Existence and uniqueness of the decomposition)

For existence, define

$$S=\frac12(A+A^T), \qquad K=\frac12(A-A^T).$$

The preceding transpose identities give $S^T=S$ and $K^T=-K$, so $S$ is symmetric and $K$ is skew-symmetric. Moreover,

$$S+K =\frac12(A+A^T)+\frac12(A-A^T) =A.$$

Thus a decomposition exists. To prove uniqueness, suppose instead that some symmetric matrix $S$ and some skew-symmetric matrix $K$ satisfy $A=S+K$. Taking transposes and using the defining properties gives a second equation,

$$A^T=S^T+K^T=S-K.$$

Adding and subtracting the two equations yields

$$A+A^T=2S, \qquad A-A^T=2K.$$

Because $2$ is invertible over the real numbers,

$$S=\frac12(A+A^T), \qquad K=\frac12(A-A^T).$$

Any proposed decomposition must therefore use exactly these two matrices. This proves uniqueness as well as existence. The same proof works over every field of characteristic not equal to $2$.

Worked example 3.3.9

(Decompose a matrix)

Let

$$A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & -1 & 3 \\ 4 & 2 & 0 \end{bmatrix}.$$

Then

$$A^T = \begin{bmatrix} 1 & 2 & 4 \\ 0 & -1 & 2 \\ 1 & 3 & 0 \end{bmatrix}.$$

So

$$\frac{1}{2}(A + A^T) = \begin{bmatrix} 1 & 1 & \frac{5}{2} \\ 1 & -1 & \frac{5}{2} \\ \frac{5}{2} & \frac{5}{2} & 0 \end{bmatrix},$$

and

$$\frac{1}{2}(A - A^T) = \begin{bmatrix} 0 & -1 & -\frac{3}{2} \\ 1 & 0 & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} & 0 \end{bmatrix}.$$

The first matrix is symmetric, the second is skew-symmetric, and their sum is $A$.

There is also a useful symmetric pattern whenever a matrix is multiplied by its transpose. It converts a possibly rectangular matrix into square matrices that encode row or column interactions.

Theorem 3.3.10

(Products with a transpose are symmetric)

If the product is defined, then both $A^T A$ and $A A^T$ are symmetric.

Proof

(Proof by transposition)

If $A$ is $m \times n$, then $A^T A$ is $n \times n$ and $AA^T$ is $m \times m$, so symmetry is dimensionally possible in both cases. For the first product,

$$(A^T A)^T = A^T (A^T)^T = A^T A.$$

For the second,

$$(AA^T)^T=(A^T)^T A^T=AA^T.$$

Each product equals its transpose, which is precisely the definition of a symmetric matrix.

This identity is one of the main reasons transpose shows up again in later topics such as orthogonality, projections, and least-squares problems. For example, when $x$ is a compatible real column vector,

$$x^T A^T A x=(Ax)^T(Ax)=\lVert Ax\rVert^2\ge 0.$$

Here $\lVert z\rVert^2=\sum\limits_i z_i^2$ for a real column vector $z$, so the nonnegativity follows directly from squares.

Thus $A^T A$ carries information about the lengths of vectors after applying $A$. This observation will later connect transpose to geometry, not merely to entry rearrangement.

Commuting and non-commuting matrices

Definition 3.3.11

(Commuting matrices)

Two square matrices $A$ and $B$ of the same order commute if

$$AB = BA.$$

For addition, commutativity is automatic. For multiplication, it is exceptional. This is one of the most important differences between scalar algebra and matrix algebra.

The zero matrix and the identity matrix commute with every square matrix of the same size. Diagonal matrices of the same size also commute with one another, because their products remain diagonal and the diagonal entries multiply in the ordinary commutative way.

Transpose gives a precise test for when products of structured matrices retain structure. If $A$ and $B$ are symmetric, then $(AB)^T=BA$; consequently $AB$ is symmetric exactly when $AB=BA$. For two skew-symmetric matrices the same formula $(AB)^T=BA$ appears, because the two minus signs cancel. By contrast, if $S$ is symmetric and $K$ is skew-symmetric, then $(SK)^T=-KS$, so a commuting product $SK=KS$ is skew-symmetric. If the product is zero, it is also symmetric.

Worked example 3.3.12

(A non-commuting pair)

Take

$$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}.$$

Then

$$AB = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}, \qquad BA = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}.$$

So $AB \ne BA$.

That kind of example is not a curiosity. It is the reason matrix identities must always preserve the order of factors.

Theorem 3.3.13

(A useful skew-symmetric commuting test)

If $A$ and $B$ are skew-symmetric square matrices of the same order, then

$$AB \text{ is symmetric } \iff AB = BA.$$

Proof

(Why this is true)

Because $A^T = -A$ and $B^T = -B$,

$$(AB)^T = B^T A^T = (-B)(-A) = BA.$$

So $AB$ is symmetric exactly when $(AB)^T = AB$, and that happens exactly when $BA = AB$.

Worked example 3.3.14

(Symmetric times skew-symmetric)

Let

$$S = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}, \qquad K = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.$$

Here $S$ is symmetric and $K$ is skew-symmetric, but they do not commute:

$$SK = \begin{bmatrix} 0 & 2 \\ -3 & 0 \end{bmatrix}, \qquad KS = \begin{bmatrix} 0 & 3 \\ -2 & 0 \end{bmatrix}.$$

If two matrices of this kind do commute, then their product is skew-symmetric, because

$$(SK)^T = K^T S^T = (-K)S = -KS = -SK.$$

Common mistakes

Common mistake

(Transpose reverses products)

The correct identity is $(AB)^T = B^T A^T$, not $A^T B^T$.

Common mistake

(Symmetric does not mean commuting)

Symmetric is a property of one matrix. Commuting is a property of two matrices. They are different statements.

Common mistake

(A zero diagonal does not guarantee skew-symmetry)

If $A^T=-A$, then every diagonal entry is $0$, but the converse is false. For example, $\begin{bmatrix}0&1\\1&0\end{bmatrix}$ has a zero diagonal and is symmetric, not skew-symmetric. The off-diagonal conditions $a_{ij}=-a_{ji}$ must also be checked.

Common mistake

(Transpose is not the same as inverse)

Transpose always exists and only exchanges indices. An inverse exists only for certain square matrices and must satisfy $AA^{-1}=A^{-1}A=I$. The symbols $A^T$ and $A^{-1}$ therefore describe different operations, even though they coincide for special matrices studied later.

Summary

Transpose exchanges the two matrix indices, changes an $m \times n$ matrix into an $n \times m$ matrix, and reverses multiplication order. Symmetric and skew-symmetric matrices are square matrices characterized by $A^T=A$ and $A^T=-A$; over the real numbers, the latter condition forces a zero diagonal.

The formulas

$$\frac12(A+A^T) \quad\text{and}\quad \frac12(A-A^T)$$

are not arbitrary tricks. They are the uniquely determined symmetric and skew-symmetric parts of $A$, obtained by solving the two equations $A=S+K$ and $A^T=S-K$. Products such as $A^T A$ are automatically symmetric, while products of two structured square matrices may require commutativity to retain symmetry or skew-symmetry. These ideas prepare the algebra needed for orthogonality, quadratic forms, projections, and least-squares problems.

Quick checks

Quick check

(If $A$ is $3 \times 2$, what is the size of $A^T$?)

Swap rows and columns.

Solution · Answer

$A^T$ is $2 \times 3$.

Quick check

(What must every diagonal entry of a real skew-symmetric matrix be?)

Use $A^T = -A$ on the diagonal.

Solution · Answer

Every diagonal entry must be $0$.

Quick check

(Which identity is correct: $(AB)^T = A^T B^T$ or $(AB)^T = B^T A^T$?)

Compare an arbitrary entry on the two sides, or test a noncommuting pair of square matrices.

Solution · Answer

$(AB)^T = B^T A^T$.

Quick check

(If $S$ is symmetric and $K$ is skew-symmetric, what is $S^T$ and what is $K^T$?)

Read the definitions directly.

Solution · Answer

$S^T = S$ and $K^T = -K$.

Guided exercises

Quick check

(Find the symmetric and skew-symmetric parts of $A = \begin{bmatrix}2&1\\4&3\end{bmatrix}$.)

Use $S = 1/2 (A + A^T)$ and $K = 1/2 (A - A^T)$.

Solution · Guided solution

First compute

$$A^T = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}.$$

Then

$$S = \frac{1}{2} \begin{bmatrix} 4 & 5 \\ 5 & 6 \end{bmatrix} = \begin{bmatrix} 2 & \frac{5}{2} \\ \frac{5}{2} & 3 \end{bmatrix},$$

and

$$K = \frac{1}{2} \begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -\frac{3}{2} \\ \frac{3}{2} & 0 \end{bmatrix}.$$

Check that $S$ is symmetric, $K$ is skew-symmetric, and $A = S + K$.

Quick check

(Let $A$ and $B$ be symmetric matrices. If $AB = BA$, why is $AB$ symmetric?)

Use the transpose identity and the commuting assumption.

Solution · Guided solution

Because

$$(AB)^T = B^T A^T = BA = AB.$$

So $AB$ equals its transpose, which is exactly the definition of symmetric.

Quick check

(For a square matrix $A$, explain in words why $A + A^T$ is always symmetric.)

You do not need a full formal proof, but you must mention transpose.

Solution · Guided solution

If you transpose $A + A^T$, you get $(A + A^T)^T = A^T + (A^T)^T = A^T + A = A + A^T$, so the matrix equals its own transpose and is therefore symmetric.

Quick check

(For a square matrix $A$, explain in words why $A - A^T$ is always skew-symmetric.)

Again, focus on what happens after taking the transpose.

Solution · Guided solution

If you transpose $A - A^T$, you get $(A - A^T)^T = A^T - A = -(A - A^T)$, so the transpose is the negative of the original matrix.

Related notes

This note builds on 3.2 Matrix multiplication, identity matrices, and linear systems. Continue to 3.4 Special matrices to study diagonal, triangular, identity, and elementary matrix families.

Supplementary exercises with complete solutions

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 1

For the matrices A,B below: (a)i find A^T; ii find the first row of 2A^T+3B. (b) Given AB=I3, determine u,v,w with justification.

$$A=\begin{bmatrix}1&1&0&-1\\1&0&1&0\\-1&1&-1&0\end{bmatrix},\quad B=\begin{bmatrix}1&1&1\\p&u&x\\q&v&y\\r&w&z\end{bmatrix}.$$

Solution

Transpose exchanges the row and column indices. For (b), the second column of AB must be the second identity column. Its three scalar equations determine the requested entries; the other columns impose no additional condition on u,v,w.

$$(a)i\quad A^T=\begin{bmatrix}1&1&-1\\1&0&1\\0&1&-1\\-1&0&0\end{bmatrix}. \qquad(a)ii\quad 2(1,1,-1)+3(1,1,1)=(5,5,1).$$ $$(b)\quad A\begin{bmatrix}1\\u\\v\\w\end{bmatrix}=\begin{bmatrix}1+u-w\\1+v\\-1+u-v\end{bmatrix}=\begin{bmatrix}0\\1\\0\end{bmatrix} \Longrightarrow v=0,\ u=1,\ w=2.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 2

For A:m×n, B,C:n×p, D:p×q, prove (a) [A(B+C)]^T=B^TA^T+C^TA^T; (b) (ABD)^T=D^TB^TA^T.

Solution

Use the established product-transpose and sum-transpose rules. All products are defined: the final sizes are p×m in (a) and q×m in (b).

$$(a)\quad[A(B+C)]^T=(B+C)^TA^T=(B^T+C^T)A^T=B^TA^T+C^TA^T.$$ $$(b)\quad(ABD)^T=((AB)D)^T=D^T(AB)^T=D^TB^TA^T.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 3

For square A, prove (a) (A²)^T=(A^T)²; (b) by induction (A^n)^T=(A^T)^n for all positive integers n.

Solution

The product rule proves (a). For (b), start at n=1. Under the induction hypothesis at n=k, apply the same rule to A^k A. Powers of the single matrix A^T combine by their definition, without assuming that arbitrary matrices commute.

$$(a)\quad(A^2)^T=(AA)^T=A^TA^T=(A^T)^2.$$ $$(b)\quad(A^1)^T=A^T;\qquad (A^{k+1})^T=(A^kA)^T=A^T(A^k)^T=A^T(A^T)^k=(A^T)^{k+1}.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 4

For A1,A2 with the same m rows, verify directly that [A1 A2]^T is the vertical stack of A1^T,A2^T. Also explain the two general block transpose formulas for horizontal and vertical concatenations.

Solution

Let A1 have r columns and A2 have s columns. In the transposed matrix, rows 1,...,r come from A1 and rows r+1,...,r+s from A2, with the same entry indices as their own transposes. This proves equality entry by entry. Repeating this argument gives any finite number of blocks. Applying transpose again gives the vertical-stack formula.

$$([A_1\ A_2]^T)_{ji}=\begin{cases}(A_1)_{ij},&1\le j\le r,\\(A_2)_{i,j-r},&r<j\le r+s.\end{cases} \qquad [A_1\ A_2]^T=\begin{bmatrix}A_1^T\\A_2^T\end{bmatrix}.$$ $$[A_1\ \cdots\ A_p]^T=\begin{bmatrix}A_1^T\\\vdots\\A_p^T\end{bmatrix},\qquad \begin{bmatrix}B_1\\\vdots\\B_p\end{bmatrix}^T=[B_1^T\ \cdots\ B_p^T].$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 5

For the 7×7 incidence matrix below, write a_i for row i and (a) find each row self-product; (b) using the stipulated unique common 1, find products of distinct rows; (c) find alpha,beta in AA^T=alpha I7+beta J7. Adapt the argument to the 13×13 matrix B specified below.

$$A=\begin{bmatrix}1&1&0&1&0&0&0\\0&1&1&0&1&0&0\\0&0&1&1&0&1&0\\0&0&0&1&1&0&1\\1&0&0&0&1&1&0\\0&1&0&0&0&1&1\\1&0&1&0&0&0&1\end{bmatrix}.\quad J_7=(1)_{1\le i,j\le7}.\qquad B_{r+1,c+1}=\begin{cases}1&c-r\pmod {13}\in\{0,1,3,9\},\\0&\text{otherwise},\end{cases}\quad0\le r,c\le12.$$

Solution

Each A row has three ones, giving diagonal Gram entries 3; different rows have dot product 1 by the given observation. Thus alpha=2,beta=1. For the optional B, number columns and rows by residues modulo13. Row r has ones in r+S, $S=\{0,1,3,9\}$. The twelve nonzero ordered differences of S are distinct and exhaust all nonzero residues, so distinct translated rows intersect once. Each row has four ones.

$$(a)\quad a_i a_i^T=3\quad(1\le i\le7).\qquad(b)\quad a_i a_j^T=1\ (i\ne j).$$ $$(c)\quad (AA^T)_{ij}=\begin{cases}3&i=j,\\1&i\ne j,\end{cases}\qquad AA^T=2I_7+J_7.$$ $$\{s-t:s,t\in S,s\ne t\}=\{1,3,9,12,2,8,10,11,6,4,5,7\}\pmod {13}. \qquad BB^T=3I_{13}+J_{13}.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 6

Let real A=[[a,b],[c,d]], with AA^T=A^TA and a≠d. Prove A symmetric.

Solution

Compare just the off-diagonal entries. Their equality factors as (a-d)(c-b)=0. The nonzero first factor forces c=b, which is exactly the missing symmetry condition.

$$(AA^T)_{12}=ac+bd,\quad(A^TA)_{12}=ab+cd; \qquad ac+bd-ab-cd=(a-d)(c-b)=0. \qquad a\ne d\Longrightarrow c=b\Longrightarrow A^T=A.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 7

Prove the zero n×n real matrix is the only matrix both symmetric and skew-symmetric.

Solution

Under both conditions A=-A, so every entry satisfies 2aij=0 and is zero over the reals. Conversely the zero matrix satisfies both transpose equations. Both necessity and existence are therefore checked.

$$A^T=A,\ A^T=-A\Longrightarrow2A=O\Longrightarrow A=O. \qquad O^T=O=-O.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 8

Let A,B be symmetric square matrices of the same size. (a) Prove AB symmetric if AB=BA. (b) Determine and justify whether the converse holds.

Solution

Transpose gives a precise test for when products of structured matrices retain structure. If $A$ and $B$ are symmetric, then $(AB)^T=BA$; consequently $AB$ is symmetric exactly when $AB=BA$. For two skew-symmetric matrices the same formula $(AB)^T=BA$ appears, because the two minus signs cancel. By contrast, if $S$ is symmetric and $K$ is skew-symmetric, then $(SK)^T=-KS$, so a commuting product $SK=KS$ is skew-symmetric. If the product is zero, it is also symmetric.

Therefore (a) holds, and the converse in (b) is true.

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 9

Let real A=[aij] be skew-symmetric n×n. Directly prove akk=0 for each k=1,...,n.

Solution

The diagonal deserves special attention. If $A$ is skew-symmetric, then $a_{ii}=-a_{ii}$, so $2a_{ii}=0$. Over the real numbers this forces $a_{ii}=0$. This short argument is exactly where the characteristic-not-$2$ assumption enters.

Since the diagonal index is arbitrary, every diagonal entry is zero.

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 10

If real A:n×n is skew-symmetric and C:n×p, prove C^TAC is skew-symmetric p×p.

Solution

Transpose the product in reverse order and use A^T=-A. The resulting matrix is p×p, so its equality to the negative transpose proves the assertion.

$$(C^TAC)^T=C^TA^T(C^T)^T=C^T(-A)C=-C^TAC.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 11

For real column vectors u,v and A=uv^T-vu^T: (a) prove A skew-symmetric. (b) If u^Tu=v^Tv=1 and u^Tv=p, i find alpha,beta in A²=alpha(uv^T+vu^T)+beta(uu^T+vv^T); ii find gamma in A³=gamma A.

Solution

Part(a) follows by reversing each rank-one product. In (b), scalar products can be pulled out of the four expanded terms; v^Tu=p because it is the transpose of the scalar u^Tv. This gives alpha=p,beta=-1. To compute A³, first compute Au,Av and then A²u,A²v; multiply the two rank-one terms of A by A² on the left. No independence of u,v is assumed, so this also covers p=±1.

$$(a)\quad A^T=vu^T-uv^T=-A.$$ $$(b)i\quad A^2=u(v^Tu)v^T-u(v^Tv)u^T-v(u^Tu)v^T+v(u^Tv)u^T =p(uv^T+vu^T)-(uu^T+vv^T).$$ $$(b)ii\quad Au=pu-v,\quad Av=u-pv; \quad A^2u=p(pu-v)-(u-pv)=(p^2-1)u, \quad A^2v=(pu-v)-p(u-pv)=(p^2-1)v. \quad A^3=(A^2u)v^T-(A^2v)u^T=(p^2-1)A,\qquad\gamma=p^2-1.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 12

(a) Compute C² for C=[[0,a],[b,0]]. (b)i symmetric S implies S² symmetric; ii disprove converse. (c)i skew-symmetric Q implies Q² symmetric; ii disprove converse. (d) Disprove: P² symmetric implies P symmetric or skew-symmetric.

Solution

The same 2×2 counterexample P with a=1,b=2 refutes all three asserted converses: its square is diagonal while its two off-diagonal entries are neither equal nor negatives. For the two forward statements apply the transpose-product rule; the two minus signs cancel in the skew case.

$$(a)\quad C^2=\begin{bmatrix}ab&0\\0&ab\end{bmatrix}=abI_2.$$ $$(b)i\quad(S^2)^T=S^TS^T=S^2.\qquad(c)i\quad(Q^2)^T=(-Q)(-Q)=Q^2.$$ $$(b)ii,(c)ii,(d)\quad P=\begin{bmatrix}0&1\\2&0\end{bmatrix},\quad P^2=2I_2, \quad P^T\ne P,\quad P^T\ne-P.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 13

A is symmetric, B skew-symmetric, both n×n. Classify (a) AB+BA; (b) AB-BA; (c) A²B⁴A²; (d) A³B⁴A³; (e) A⁴B⁵A⁴; (f) A³B⁵A³, with proof.

Solution

Taking transpose directly gives (a) skew-symmetric and (b) symmetric. In every sandwich the outside factors are the same power of symmetric A, so reversing order leaves them in place and contributes only the parity sign from B. Thus (c),(d) are symmetric and (e),(f) skew-symmetric. These are guaranteed classifications; a zero resulting matrix belongs to both classes. No commutation assumption is needed.

$$(AB+BA)^T=-BA-AB=-(AB+BA), \qquad(AB-BA)^T=-BA+AB=AB-BA.$$ $$(A^rB^sA^r)^T=A^r(-B)^sA^r=(-1)^s A^rB^sA^r. \qquad(c),(d):s=4;\quad(e),(f):s=5.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 14

(a) Real A:m×n satisfies x^TAy=0 for every real x∈R^m,y∈R^n: prove A=0. (b) Real symmetric B:n×n satisfies u^TBu=0 for all real u: prove B=0. (c) Give a nonzero real C with u^TCu=0 for every real u, disproving the assertion without symmetry.

Solution

For (a) choose standard basis columns to select each entry. For (b), polarize the quadratic expression: symmetry makes the two mixed terms equal, hence all x^TBy vanish; apply (a). For (c) a 2×2 nonzero skew-symmetric matrix suffices; its quadratic expression cancels for every pair of real coordinates. This is a counterexample to the universally quantified statement (for fixed n≥2 it embeds as a diagonal block; n=1 admits no such nonzero example).

$$(a)\quad a_{ij}=e_i^TAe_j=0.$$ $$(b)\quad0=(x+y)^TB(x+y)-x^TBx-y^TBy=x^TBy+y^TBx=2x^TBy. \qquad x^TBy=0\quad\forall x,y\Longrightarrow B=O.$$ $$(c)\quad C=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\ne O, \quad (s,t)C\begin{bmatrix}s\\t\end{bmatrix}=st-ts=0.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 15

For the 3×3 matrices A,B below, determine whether they commute and justify by multiplication.

$$A=\begin{bmatrix}3&1&-2\\2&-2&0\\-1&1&2\end{bmatrix},\quad B=\begin{bmatrix}1&1&1\\1&-1&1\\0&1&2\end{bmatrix}.$$

Solution

Compute the nine row-column products in each order. Both yield four times the identity, so the matrices commute. The equality is specific to these parameters.

$$AB=\begin{bmatrix}3+1&3-1-2&3+1-4\\2-2&2+2&2-2\\-1+1&-1-1+2&-1+1+4\end{bmatrix}=4I_3.$$ $$BA=\begin{bmatrix}3+2-1&1-2+1&-2+2\\3-2-1&1+2+1&-2+2\\2-2&-2+2&4\end{bmatrix}=4I_3.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 16

For the exact upper triangular A and lower triangular B, each 4×4 and displayed below, decide whether AB=BA and justify.

$$A=\begin{bmatrix}1&1&1&1\\0&1&1&1\\0&0&1&1\\0&0&0&1\end{bmatrix},\quad B=\begin{bmatrix}1&0&0&0\\1&1&0&0\\1&1&1&0\\1&1&1&1\end{bmatrix}.$$

Solution

Multiplying in each order gives the following matrices. In particular their (1,1) entries are 4 and1, so they do not commute.

$$AB=\begin{bmatrix}4&3&2&1\\3&3&2&1\\2&2&2&1\\1&1&1&1\end{bmatrix},\qquad BA=\begin{bmatrix}1&1&1&1\\1&2&2&2\\1&2&3&3\\1&2&3&4\end{bmatrix}. \qquad AB\ne BA.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 17

Under AB=BA=0: (a)i verify (A+B)²=A²+B²; ii verify the cube formula. (b) Prove (A+B)^n=A^n+B^n for every integer n>1 by induction without the binomial theorem. (c) Verify the conclusion for the 3×3 matrices A,B below.

$$A=\begin{bmatrix}2&2&2\\-2&-2&-2\\1&1&1\end{bmatrix},\quad B=\begin{bmatrix}-3&-4&-2\\6&6&0\\-3&-2&2\end{bmatrix}\qquad\text{(c)}.$$

Solution

Distributivity gives the square. In any higher step the mixed terms vanish since A^kB=A^(k-1)(AB)=0 and B^kA=B^(k-1)(BA)=0. This proves the cube and then the induction from n=2. In (c), write A=ur with u=(2,-2,1)^T and r=(1,1,1); directly rB=0 and Bu=0, hence both products vanish. Apply (b).

$$(a)i\quad(A+B)^2=A^2+AB+BA+B^2=A^2+B^2.$$ $$(a)ii\quad(A+B)^3=(A^2+B^2)(A+B)=A^3+A^2B+B^2A+B^3=A^3+B^3.$$ $$(b)\quad(A+B)^k=A^k+B^k\Longrightarrow (A+B)^{k+1}=(A^k+B^k)(A+B)=A^{k+1}+B^{k+1}\quad(k\ge2).$$ $$(c)\quad rB=(-3+6-3,-4+6-2,-2+0+2)=(0,0,0), \quad Bu=\begin{bmatrix}-6+8-2\\12-12\\-6+4+2\end{bmatrix}=0. \quad AB=u(rB)=O,\ BA=(Bu)r=O\Longrightarrow(A+B)^n=A^n+B^n\ (n\ge2).$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 18

(a) Prove diagonal matrices of the same size commute. (b) If GAH and GBH are diagonal and HG=I_p, prove square A,B commute.

Solution

For (a) the off-diagonal entries of each product vanish and corresponding diagonal entries are scalar products in either order. For (b) use (a) on the two diagonal matrices, substitute HG=I between A and B, and multiply the resulting equality on the left by H and right by G. This uses the given one-sided inverse equation directly.

$$(a)\quad(DE)_{ij}=\delta_{ij}d_{ii}e_{ii}=\delta_{ij}e_{ii}d_{ii}=(ED)_{ij}.$$ $$(b)\quad GABH=(GAH)(GBH)=(GBH)(GAH)=GBAH. \quad H(GABH)G=H(GBAH)G\Longrightarrow AB=BA.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 19

For commuting p×p matrices A,B, (a) prove A^mB^n=B^nA^m for all positive integers m,n; (b) prove the full binomial expansion (A+B)^n=sum k=0..n binom(n,k) A^(n-k)B^k for all positive integers n.

Solution

For (a), first induct on n to move A past B^n, then induct on m. For (b), the base n=1 is immediate. Multiply the expansion at n on the right by A+B; (a) moves the added A left past B^k. Reindex the second sum and use Pascal’s identity. The two boundary terms retain coefficient1. Thus induction supplies every term of the claimed expansion; zero powers mean the identity.

$$(a)\quad AB^{n+1}=(AB^n)B=B^nAB=B^{n+1}A,\quad AB=BA; \quad A^{m+1}B^n=A(A^mB^n)=AB^nA^m=B^nA^{m+1}.$$ $$(b)\quad(A+B)^1=A+B. \quad(A+B)^{n+1}=\sum_{k=0}^{n}\binom nk A^{n+1-k}B^k+ \sum_{k=0}^{n}\binom nk A^{n-k}B^{k+1} =\sum_{j=0}^{n+1}\binom{n+1}j A^{n+1-j}B^j. \quad\binom nj+\binom n{j-1}=\binom{n+1}j\ (1\le j\le n).$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 20

For commuting square A,B and every positive integer n, (a) prove A^(n+1)-B^(n+1)=(A-B)sum k=0..n A^(n-k)B^k; (b) deduce I-B^(n+1)=(I-B)(I+B+...+B^n).

Solution

Distribute A-B across the finite sum. Commutation lets B pass the power of A, so adjacent terms cancel and only the first and last remain. This proves (a) directly for every positive n. In (b), substitute A=I, which commutes with B; no inverse of I-B is required.

$$(a)\quad(A-B)\sum_{k=0}^nA^{n-k}B^k =\sum_{k=0}^n A^{n+1-k}B^k-\sum_{k=0}^n A^{n-k}B^{k+1} =A^{n+1}-B^{n+1}.$$ $$(b)\quad(I-B)\sum_{k=0}^n B^k=I-B^{n+1}.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 21

If square B,C commute and f,g are scalar-coefficient polynomials: (a) prove f(B) commutes with C^m for each positive integer m; (b) prove f(B),g(C) commute.

Solution

Expand each polynomial as a finite scalar linear combination. Q19(a), or the permitted fact, moves every B^k past C^m for positive k; the constant term is a multiple of I and also commutes. This proves (a). Distribute once more over g(C), including its constant term, to obtain (b).

$$f(x)=\sum_{k=0}^r a_kx^k,\qquad g(x)=\sum_{j=0}^s b_jx^j.$$ $$(a)\quad f(B)C^m=\sum_{k=0}^r a_k B^kC^m =\sum_{k=0}^r a_k C^m B^k=C^mf(B).$$ $$(b)\quad f(B)g(C)=\sum_{j=0}^s b_j f(B)C^j =\sum_{j=0}^s b_j C^j f(B)=g(C)f(B).$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 22

For the 3×3 matrices J,K,L below, verify [J,K]=L, [K,L]=J, [L,J]=K.

$$J=\begin{bmatrix}0&1&0\\-1&0&0\\0&0&0\end{bmatrix},\quad K=\begin{bmatrix}0&0&-1\\0&0&0\\1&0&0\end{bmatrix},\quad L=\begin{bmatrix}0&0&0\\0&0&1\\0&-1&0\end{bmatrix}.$$

Solution

Write the matrices in standard matrix units Eij; Eij Ekl=delta(j,k) Eil follows from the product definition. Each of the six products below has only one surviving matrix unit. Subtract in the specified order to verify all three identities.

$$J=E_{12}-E_{21},\quad K=E_{31}-E_{13},\quad L=E_{23}-E_{32}. \qquad JK=E_{23},\quad KJ=E_{32};\qquad KL=E_{12},\quad LK=E_{21};\qquad LJ=E_{31},\quad JL=E_{13}. \qquad[J,K]=E_{23}-E_{32}=L,\quad[K,L]=E_{12}-E_{21}=J,\quad[L,J]=E_{31}-E_{13}=K.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 23

For same-size square matrices, prove (a) [A,beta B+gamma C]=beta[A,B]+gamma[A,C]; (b) [A,A]=0; (c) [A,B]=-[B,A]=[-B,A]=[B,-A]; (d) [alpha A+beta B,C]=alpha[A,C]+beta[B,C]; (e) if A commutes with [A,B], prove [A^(k+1),B]=(k+1)[A,B]A^k for all positive integers k.

Solution

Expand the definition XY-YX for (a)-(d), keeping matrix order. For (e), first expand the product rule for commutators. Let K=[A,B]; the hypothesis is AK=KA. Starting with [A,B]=K, induct on the positive power m. The product rule and commutation combine m identical contributions and produce [A^m,B]=mKA^(m-1). Take m=k+1.

$$(a)\quad A(\beta B+\gamma C)-(\beta B+\gamma C)A =\beta(AB-BA)+\gamma(AC-CA). \quad(b)\ [A,A]=A^2-A^2=O. \quad(c)\ AB-BA=-(BA-AB)=(-B)A-A(-B)=B(-A)-(-A)B. \quad(d)\ (\alpha A+\beta B)C-C(\alpha A+\beta B) =\alpha(AC-CA)+\beta(BC-CB).$$ $$(e)\quad[XY,B]=XYB-BXY=X(YB-BY)+(XB-BX)Y. \quad[A,B]=K;\quad[A^{m+1},B]=A^m[A,B]+[A^m,B]A =A^mK+mKA^m=(m+1)KA^m. \quad[A^{k+1},B]=(k+1)[A,B]A^k.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 24

Let A,B,C,D be real square matrices of the same size. (a) Prove [A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0. (b) If AB=BA, prove [A,[B,C]]=[B,[A,C]] for every C. (c) Determine whether the four-term identity below holds for all real square matrices A,B,C,D of the same size.

$$[A,[B,[C,D]]]+[B,[C,[D,A]]]+[C,[D,[A,B]]]+[D,[A,[B,C]]]=O\qquad\text{(c)}.$$

Solution

In (a) expand all twelve ordered triple products; each cancels a matching negative term. For (b), the third term in (a) is zero since [A,B]=0, and antisymmetry gives [C,A]=-[A,C]. Part(c) is false as printed, so no proof exists under its stated hypotheses. Use the explicit 2×2 real matrix-unit counterexample below. Its four terms are respectively 0,0,E11-E22,0. Their nonzero sum refutes the statement.

$$(a)\quad[A,[B,C]]=ABC-ACB-BCA+CBA, \quad[B,[C,A]]=BCA-BAC-CAB+ACB, \quad[C,[A,B]]=CAB-CBA-ABC+BAC. \quad[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=O.$$ $$(b)\quad[A,[B,C]]-[B,[A,C]]=-[C,[A,B]]=O.$$ $$(c)\quad A=E_{11},\ B=E_{12},\ C=E_{21},\ D=E_{22}. \quad[C,D]=-E_{21},\quad[B,[C,D]]=-(E_{11}-E_{22}),\quad[A,[B,[C,D]]]=O. \quad[D,A]=O\Longrightarrow[B,[C,[D,A]]]=O. \quad[A,B]=E_{12},\quad[D,[A,B]]=-E_{12},\quad[C,[D,[A,B]]]=E_{11}-E_{22}. \quad[B,C]=E_{11}-E_{22},\quad[A,[B,C]]=O\Longrightarrow[D,[A,[B,C]]]=O. \quad\text{sum}=E_{11}-E_{22}=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\ne O.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 25

Prove (a) [A,A^T] symmetric; (b) symmetric A,B give skew-symmetric [A,B]; (c) symmetric A and skew-symmetric B give symmetric [A,B]; (d) skew-symmetric A,B give skew-symmetric [A,B].

Solution

Transpose each difference, reverse products and substitute the respective symmetry assumptions. A positive transpose sign proves symmetry; a negative sign proves skew-symmetry. As always the zero result has both properties.

$$(a)\quad(AA^T-A^TA)^T=AA^T-A^TA.$$ $$(b)\quad(AB-BA)^T=BA-AB=-(AB-BA).$$ $$(c)\quad(AB-BA)^T=(-B)A-A(-B)=AB-BA.$$ $$(d)\quad(AB-BA)^T=(-B)(-A)-(-A)(-B)=BA-AB=-(AB-BA).$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 26

(a) Express (A+B)² and (A-B)(A+B) as alpha A²+beta B²+gamma AB+delta[A,B]. (b) Prove (A+B)²=A²+2AB+B² iff (A-B)(A+B)=A²-B².

Solution

From the definition, BA=AB-[A,B]. Substitute this in both distributive expansions. The coefficients are respectively (1,1,2,-1) and (1,-1,0,1). Each equality in (b) holds precisely when [A,B]=0, proving the equivalence in both directions without presupposing commutation.

$$(a)\quad(A+B)^2=A^2+AB+BA+B^2=A^2+B^2+2AB-[A,B], \quad(A-B)(A+B)=A^2+AB-BA-B^2=A^2-B^2+[A,B].$$ $$(b)\quad(A+B)^2=A^2+2AB+B^2\iff[A,B]=O \iff(A-B)(A+B)=A^2-B^2.$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 27

For C=sA+tB,D=uA+vB with scalar sv-tu≠0, prove [A,B]=0 iff [C,D]=0.

Solution

Expand both products in the commutator; the A² and B² terms cancel. The remaining difference is the nonzero scalar sv-tu times [A,B]. A scalar multiple by a nonzero real is zero exactly when its matrix factor is zero, which proves both implications.

$$[C,D]=suA^2+svAB+tuBA+tvB^2 -\bigl(usA^2+utAB+vsBA+vtB^2\bigr) =(sv-tu)(AB-BA)=(sv-tu)[A,B]. \quad sv-tu\ne0\Longrightarrow\bigl([C,D]=O\iff[A,B]=O\bigr).$$

Real scalars and compatible matrix sizes are assumed throughout; transpose is ordinary transpose, not conjugate transpose. [X,Y]=XY-YX; I is the identity; O is the zero matrix.

Question 28

(a) Prove diagonal C,D commute and CD is diagonal. (b) For diagonal D and orthogonal P prove PDP^T symmetric. (c) For diagonal C,D and orthogonal P, put G=PCP^T,H=PDP^T; prove GH=HG and GH symmetric. (d)i compute AB,BA for A=[[1,2],[2,3]], B=[[2,0],[0,3]]; ii use these to refute both “symmetric factors have symmetric product” and “symmetric factors commute”.

Solution

For (a), only matching diagonal indices survive multiplication. For (b), a real diagonal matrix equals its transpose; transpose the sandwich. In (c), the middle P^TP is I, so GH=P(CD)P^T and HG=P(DC)P^T, equal by (a). Their common value is symmetric by (b), with diagonal CD. In (d), both given factors are symmetric; the explicitly unequal off-diagonal entries of AB show it is not symmetric, and the two products differ. This single example disproves both universal claims.

$$(a)\quad(CD)_{ij}=\delta_{ij}c_{ii}d_{ii}=(DC)_{ij}.$$ $$(b)\quad(PDP^T)^T=PD^TP^T=PDP^T.$$ $$(c)\quad GH=PC(P^TP)DP^T=P(CD)P^T=P(DC)P^T=HG; \quad(GH)^T=(P(CD)P^T)^T=P(CD)^TP^T=P(CD)P^T=GH.$$ $$(d)i\quad AB=\begin{bmatrix}2&6\\4&9\end{bmatrix},\quad BA=\begin{bmatrix}2&4\\6&9\end{bmatrix}. \quad(d)ii\quad A^T=A,\quad B^T=B,\quad (AB)^T\ne AB,\quad AB\ne BA.$$