3.4 Special matrices

Not every matrix deserves a special name. The families in this note do because their visible patterns impose algebraic restrictions. Once those restrictions are translated into index notation, they let us predict products, transposes, and row operations without expanding every entry.

Throughout the note, entries come from a field $\mathbb F$, usually $\mathbb R$ in this course. Statements about diagonal and triangular matrices concern square matrices unless another size is stated explicitly.

What a zero pattern lets us predict

A general $n \times n$ matrix contains $n^2$ entries, and none of them is forced by shape alone. A special matrix carries extra information before any calculation begins. For example, knowing that a matrix is lower triangular immediately tells us that every entry $a_{ij}$ with $i\lt j$ is zero. That single condition controls many later computations.

The point of naming matrix families is therefore not classification for its own sake. A good name compresses a useful hypothesis:

• diagonal means that only the main diagonal can contain nonzero entries;
• upper or lower triangular means that an entire side of the main diagonal is already zero;
• identity means that multiplication leaves every compatible matrix unchanged;
• elementary means that one elementary row operation has been encoded as matrix multiplication.

These descriptions are related rather than disjoint. In particular, a diagonal matrix is simultaneously upper triangular and lower triangular. The identity matrix is diagonal, hence belongs to both triangular families as well. A matrix can therefore carry several useful labels at once.

The advantage becomes clearest in proofs. To show that a product of two lower triangular matrices is lower triangular, we do not need to calculate all $n^2$ entries. We only examine an arbitrary entry above the diagonal and show, by splitting its row-by-column sum into two index ranges, that every term must vanish. This is a model for how structural information replaces brute force.

Recognize the matrix families

Definition 3.4.1

(Diagonal, triangular, identity, and elementary matrices)

Let $A=(a_{ij})\in\mathbb F^{n\times n}$.

The matrix $A$ is diagonal when $a_{ij}=0$ whenever $i\ne j$. We write

$$A=\operatorname{diag}(a_{11},a_{22},\ldots,a_{nn}).$$

It is upper triangular when $a_{ij}=0$ for $i\gt j$, so every entry below the main diagonal is zero. It is lower triangular when $a_{ij}=0$ for $i\lt j$, so every entry above the main diagonal is zero. Consequently, a matrix is diagonal exactly when it is both upper triangular and lower triangular.

The identity matrix $I_n$ is the diagonal matrix $\operatorname{diag}(1,\ldots,1)$.

An elementary matrix of order $n$ is obtained from $I_n$ by exactly one elementary row operation: interchanging two rows, multiplying one row by a nonzero scalar, or adding a scalar multiple of one row to another.

The words “above” and “below” refer to indices, not to the signs of the entries: $(i,j)$ is above the diagonal when $i\lt j$, on it when $i=j$, and below it when $i\gt j$. Diagonal entries themselves are unrestricted in a triangular matrix; in particular, they may be zero.

Forced zero positions distinguish diagonal and triangular matrices; the identity and elementary matrices are also characterized by their multiplication rules.

The following widget concentrates on diagonal, identity, symmetric, and skew-symmetric recognition. Its labels are intentionally not an exhaustive list of matrix families: a matrix shown as “neither” in that narrower comparison may still be upper or lower triangular. Use the entries themselves, rather than a single label, when applying the definitions in this note.

Which structures survive an operation?

Theorem 3.4.2

(Transpose exchanges the two triangular families)

Let $A\in\mathbb F^{n\times n}$. Then $A$ is upper triangular if and only if $A^T$ is lower triangular. Likewise, $A$ is lower triangular if and only if $A^T$ is upper triangular. A diagonal matrix therefore remains diagonal after transposition.

Theorem 3.4.3

(Triangular matrices are closed under multiplication)

Let $A,B\in\mathbb F^{n\times n}$. If both matrices are lower triangular, then $AB$ is lower triangular. If both are upper triangular, then $AB$ is upper triangular.

As a special case, the product of two diagonal matrices of order $n$ is diagonal.

Theorem 3.4.4

(Elementary left multiplication performs a row operation)

Suppose $E\in\mathbb F^{n\times n}$ is obtained by applying one elementary row operation $\mathcal R$ to $I_n$. For every $A\in\mathbb F^{n\times p}$, the product $EA$ is defined and is exactly the matrix obtained by applying $\mathcal R$ to the rows of $A$.

The size conditions are part of these statements. Closure under multiplication requires matrices of the same order, while the elementary-matrix theorem only requires the number of rows of $A$ to equal the order of $E$.

Use the forced zeros in each proof

For the transpose theorem, suppose $A$ is upper triangular. To prove that $A^T$ is lower triangular, choose a position above its diagonal, so $i\lt j$. By the definition of transpose,

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

Since $j\gt i$, the entry $a_{ji}$ lies below the diagonal of the upper triangular matrix $A$, and is therefore zero. Hence every entry above the diagonal of $A^T$ vanishes. The reverse implication follows by applying the same argument to $A^T$ and using $(A^T)^T=A$. Interchanging “upper” and “lower” gives the second equivalence.

The multiplication theorem needs a little more care. Let $A$ and $B$ be lower triangular $n\times n$ matrices. Fix any position $(i,j)$ above the diagonal; thus $i\lt j$. The corresponding entry of the product is

$$(AB)_{ij}=\sum_{k=1}^{n}a_{ik}b_{kj}.$$

Split the possible values of $k$ into the two exhaustive cases $k\le i$ and $k\gt i$.

• If $k\le i$, then $k\le i\lt j$, so $k\lt j$. The entry $b_{kj}$ is above the diagonal of the lower triangular matrix $B$, hence $b_{kj}=0$.
• If $k\gt i$, then $a_{ik}$ is above the diagonal of the lower triangular matrix $A$, hence $a_{ik}=0$.

Thus every term $a_{ik}b_{kj}$ in the sum is zero, regardless of $k$. Therefore $(AB)_{ij}=0$ for every $i\lt j$, proving that $AB$ is lower triangular for arbitrary $n$, not merely in the $3\times3$ case. The upper triangular result follows either from the analogous index argument or from $(AB)^T=B^TA^T$ and the transpose theorem.

Because a diagonal matrix is both upper and lower triangular, the product of two diagonal matrices is also both, hence diagonal. More explicitly,

$$\operatorname{diag}(d_1,\ldots,d_n) \operatorname{diag}(e_1,\ldots,e_n) = \operatorname{diag}(d_1e_1,\ldots,d_ne_n).$$

This formula also shows that two diagonal matrices of the same order commute. It does not imply that arbitrary upper or lower triangular matrices commute.

There is a second consequence hidden in the same index structure. If $A$ and $B$ are both lower triangular, then for a diagonal position $(i,i)$,

$$(AB)_{ii}=\sum_{k=1}^{n}a_{ik}b_{ki}=a_{ii}b_{ii}.$$

Indeed, when $k\lt i$, the factor $b_{ki}$ is above the diagonal of $B$ and is zero; when $k\gt i$, the factor $a_{ik}$ is above the diagonal of $A$ and is zero. Only $k=i$ can contribute. The same diagonal-entry formula holds for products of upper triangular matrices. This observation will later connect triangular shape with determinants and invertibility.

Finally, the elementary-matrix theorem follows from how matrix multiplication forms rows. The rows of $I_n$ are the standard row vectors. Performing $\mathcal R$ on those rows produces the rows of $E$. When $E$ multiplies $A$, the same coefficients form the rows of $EA$ from the rows of $A$. Checking the three elementary operation types gives respectively a row interchange, a nonzero row scaling, and a row replacement.

Read structure before calculating

Worked example 3.4.5

(Identify every applicable matrix family)

Consider

$$D = \begin{bmatrix} 3 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{bmatrix}, \qquad U = \begin{bmatrix} 1 & 2 & -3 \\ 0 & 4 & 1 \\ 0 & 0 & 2 \end{bmatrix}.$$

The matrix $D$ is diagonal because all six off-diagonal entries are zero. It is therefore both upper triangular and lower triangular. It is also symmetric because $D^T=D$.

The matrix $U$ is upper triangular because its entries with $i\gt j$ are zero. It is not diagonal, since entries such as $u_{12}=2$ above the diagonal are nonzero, and it is not lower triangular for the same reason.

The identity matrix

$$I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

is an especially important diagonal matrix. Hence it is upper triangular, lower triangular, and symmetric. None of these classifications requires a separate calculation once the defining zero pattern has been read correctly.

Worked example 3.4.6

(Transpose a triangular matrix and track its indices)

Let

$$U = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}.$$

Then

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

The zero $u_{31}=0$, for example, becomes $(U^T)_{13}=0$. In general, every below-diagonal position $i\gt j$ of $U$ becomes the above-diagonal position $(j,i)$ of $U^T$. Thus the displayed calculation illustrates the index proof rather than replacing it.

As a quick numerical closure check, take

$$L_1=\begin{bmatrix}1&0\\2&3\end{bmatrix}, \qquad L_2=\begin{bmatrix}-1&0\\4&5\end{bmatrix}.$$

Then

$$L_1L_2 = \begin{bmatrix} -1&0\\ 10&15 \end{bmatrix},$$

which is lower triangular. This one example is evidence, while the arbitrary index proof establishes the theorem for every order $n$.

Worked example 3.4.7

(Diagonal matrices and an elementary matrix: rows versus columns)

Let

$$A= \begin{bmatrix} 1&-2&0\\ 3&4&5 \end{bmatrix}, \quad D_2=\operatorname{diag}(2,-1), \quad D_3=\operatorname{diag}(4,1,-2).$$

The dimensions determine which diagonal matrix can appear on each side: $D_2A$ is $2\times2$ times $2\times3$, whereas $AD_3$ is $2\times3$ times $3\times3$. Direct multiplication gives

$$D_2A= \begin{bmatrix} 2&-4&0\\ -3&-4&-5 \end{bmatrix}, \qquad AD_3= \begin{bmatrix} 4&-2&0\\ 12&4&-10 \end{bmatrix}.$$

Left multiplication by a diagonal matrix scales rows: the two diagonal entries of $D_2$ multiply rows 1 and 2. Right multiplication scales columns: the three diagonal entries of $D_3$ multiply columns 1, 2, and 3.

Now form the elementary matrix

$$E= \begin{bmatrix} 1&0\\ -3&1 \end{bmatrix},$$

obtained from $I_2$ by $R_2\leftarrow R_2-3R_1$. Then

$$EA= \begin{bmatrix} 1&-2&0\\ 0&10&5 \end{bmatrix}.$$

The first row is unchanged and the second is $-3[1,-2,0]+[3,4,5]=[0,10,5]$, exactly as the elementary left multiplication theorem predicts. Right multiplication by $E$ would not even be defined here, because the inner dimensions $3$ and $2$ do not match.

Consequences and boundary cases

The zero pattern also survives addition and scalar multiplication. If $A$ and $B$ are lower triangular of order $n$, then for every $i\lt j$,

$$(A+B)_{ij}=a_{ij}+b_{ij}=0+0=0.$$

For any scalar $c$, $(cA)_{ij}=ca_{ij}=0$ in the same positions. Thus the sum, difference, and scalar multiple of lower triangular matrices remain lower triangular; the upper triangular and diagonal families have the analogous property. These statements require the matrices being added to have the same dimensions. They are simpler than the multiplication theorem because addition compares matching entries, whereas multiplication mixes an entire row with an entire column.

Closure must not be confused with commutativity. Consider the two upper triangular matrices

$$P=\begin{bmatrix}1&1\\0&1\end{bmatrix}, \qquad Q=\begin{bmatrix}2&0\\0&3\end{bmatrix}.$$

Both products remain upper triangular, as the theorem promises, but

$$PQ=\begin{bmatrix}2&3\\0&3\end{bmatrix}, \qquad QP=\begin{bmatrix}2&2\\0&3\end{bmatrix}.$$

Hence $PQ\ne QP$. Diagonal matrices commute with one another because their products are computed entry by entry on the diagonal; a diagonal matrix does not necessarily commute with a nondiagonal triangular matrix.

Nor does the theorem cover a product of different triangular types. For

$$U=\begin{bmatrix}1&1\\0&1\end{bmatrix}, \qquad L=\begin{bmatrix}1&0\\1&1\end{bmatrix},$$

we obtain

$$UL=\begin{bmatrix}2&1\\1&1\end{bmatrix},$$

which is neither upper nor lower triangular. The closure theorem assumes that both factors belong to the same triangular family; a mixed product must be analysed from its actual entries.

Repeated multiplication gives another immediate consequence. If $T$ is upper triangular, then every positive power $T^r$ is upper triangular: start with $T^1=T$ and apply multiplication closure at each step. The same is true for lower triangular and diagonal matrices. This is an induction argument; it does not say that unrelated operations preserve the same orientation. Transposition, for instance, turns upper triangular shape into lower triangular shape unless the matrix has further restrictions.

Identity and elementary matrices in context

The identity matrix behaves in matrix multiplication as the scalar $1$ does in ordinary multiplication, but its order must match the side on which it appears. If $A$ is $m\times n$, then

$$I_mA=A,\qquad AI_n=A.$$

The two identity matrices need not have the same size when $m\ne n$. Entrywise, let $\delta_{ij}=1$ when $i=j$ and $\delta_{ij}=0$ otherwise. Then

$$(I_mA)_{ij}=\sum_{k=1}^{m}\delta_{ik}a_{kj}=a_{ij}, \qquad (AI_n)_{ij}=\sum_{k=1}^{n}a_{ik}\delta_{kj}=a_{ij}.$$

This explains why identity matrices occur in inverse problems: a square matrix $A$ is invertible precisely when another matrix $A^{-1}$ satisfies $A^{-1}A=AA^{-1}=I_n$. Triangular shape alone does not guarantee invertibility; a triangular matrix with a zero diagonal entry is singular.

Every elementary matrix is invertible because its defining row operation can be undone by another elementary row operation. The detailed construction of all three types, products of several elementary matrices, and their inverses continues in 3.5 Elementary row-operation matrices.

Common mistakes

Common mistake

(Using “triangular” as if it meant “diagonal”)

Every diagonal matrix is both upper triangular and lower triangular. A matrix that is only upper triangular or only lower triangular may have nonzero entries off the diagonal, so the converse is false.

Common mistake

(Testing the wrong side of the diagonal)

For an upper triangular matrix, entries with $i\gt j$—below the diagonal—must vanish. For a lower triangular matrix, entries with $i\lt j$—above the diagonal—must vanish.

Common mistake

(Assuming triangular means invertible)

A triangular matrix may have zero diagonal entries. Triangular shape by itself does not imply invertibility; the later determinant criterion will require every diagonal entry to be nonzero.

Common mistake

(Ignoring dimensions or the side of multiplication)

For $A\in\mathbb F^{m\times n}$, the valid identity equations are $I_mA=A$ and $AI_n=A$. Likewise, left multiplication by an elementary matrix performs a row operation; right multiplication, when defined, acts on columns instead.

Common mistake

(Treating an elementary matrix as an arbitrary matrix)

An elementary matrix is obtained from an identity matrix by exactly one elementary row operation. A product of several elementary matrices is invertible, but need not itself be elementary.

From matrix shape to algebra

Special matrices turn visible zero patterns into reusable algebra. A diagonal matrix has all off-diagonal entries equal to zero and is therefore both upper and lower triangular. Transposition exchanges the upper and lower triangular families. Multiplication preserves each triangular family: for a lower triangular product, the split $k\le i$ or $k\gt i$ forces one factor in every term above the diagonal to be zero.

Identity matrices preserve compatible matrices on either side, with $I_m$ acting on the rows of an $m\times n$ matrix and $I_n$ acting on its columns. Diagonal left and right multipliers scale rows and columns respectively. An elementary matrix goes further by recording one reversible row operation, and left multiplication applies that operation to every column of the target matrix at once.

Quick checks

Quick check

(Why is every diagonal matrix automatically upper triangular and lower triangular?)

Answer by referring to the off-diagonal entries.

Solution · Answer

In a diagonal matrix, every off-diagonal entry is zero. Thus all entries below the diagonal satisfy the upper triangular condition, and all entries above the diagonal satisfy the lower triangular condition.

Quick check

(What does the identity matrix do when you multiply a compatible matrix by it?)

State the effect in words and remember that the order depends on the side.

Solution · Answer

It leaves the matrix unchanged: if $A$ is $m\times n$, then $I_mA=A$ and $AI_n=A$.

Exercises

Quick check

(Why does the product of two lower triangular $3 \times 3$ matrices remain lower triangular?)

Do not multiply every entry. Explain why the entries above the diagonal stay zero.

Solution · Guided solution

For an entry $(i,j)$ above the diagonal, $i\lt j$. In $(AB)_{ij}=\sum\limits_{k=1}^{3}a_{ik}b_{kj}$, if $k\le i$, then $k\lt j$ and $b_{kj}=0$; if $k\gt i$, then $a_{ik}=0$. Every summand is therefore zero, so every entry above the diagonal of $AB$ is zero. The same argument works with $3$ replaced by any $n$.

Quick check

(Give one reason the identity matrix belongs to more than one matrix family.)

Name at least two families and justify the classification from the entries.

Solution · Guided solution

The identity matrix is diagonal because every off-diagonal entry is zero. Therefore it is also both upper triangular and lower triangular. It is symmetric as well, since $I_n^T=I_n$.

Related notes

Review 3.3 Transpose, symmetric, and skew-symmetric matrices for the transpose rules used above.

Continue to 3.5 Elementary row-operation matrices for the full algebra of elementary matrices.

Supplementary exercises with complete solutions

Question 12

All matrices and unspecified scalar parameters in this solution are over the real numbers. Matrix dimensions are positive integers; O denotes the zero matrix of the displayed compatible size.

Problem

$$A=\begin{pmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{pmatrix},\quad B=\begin{pmatrix}-1&2&4\\1&-2&-4\\-1&2&4\end{pmatrix}.$$

Compute (a) AB,BA; (b) verify both matrices are idempotent; (c) for real p,q verify the square formula and prove the power formula below by induction for every positive integer n.

Solution

$$\text{(a)}\quad AB=BA=O_{3\times3},\qquad \text{(b)}\quad A^2=\begin{pmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{pmatrix}=A,\quad B^2=\begin{pmatrix}-1&2&4\\1&-2&-4\\-1&2&4\end{pmatrix}=B.$$ $$\text{(c.i)}\quad (pA+qB)^2=p^2A^2+pqAB+pqBA+q^2B^2=p^2A+q^2B.$$ $$\text{(c.ii)}\quad (pA+qB)^n=p^nA+q^nB\quad(n\ge1).$$ $$\begin{aligned}(pA+qB)^{k+1}&=(p^kA+q^kB)(pA+qB)\\&=p^{k+1}A^2+p^kqAB+q^kpBA+q^{k+1}B^2\\&=p^{k+1}A+q^{k+1}B.\end{aligned}$$

For an efficient verification of (a),(b), put u=(−1,1,−1)ᵀ and v=(1,−2,−4)ᵀ. Then B=uvᵀ, vᵀu=1, and A=I−B. Consequently B²=u(vᵀu)vᵀ=B, A²=I−2B+B²=A, AB=BA=B−B²=0. Expanding the square gives (c.i). For (c.ii), n=1 is immediate. Assuming the formula at k, multiplication on the right gives the displayed induction step. This covers every real p,q, including zero.

Question 13

All matrices and unspecified scalar parameters in this solution are over the real numbers. Matrix dimensions are positive integers; O denotes the zero matrix of the displayed compatible size.

Problem

$$r\in\mathbb R,\quad A=\begin{pmatrix}3-2r&2-2r&2-2r\\-1+r&r&-1+r\\-3+r&-3+r&-2+r\end{pmatrix}.$$

Verify A is involutory, meaning A²=I₃.

Solution

$$u=\begin{pmatrix}2-2r\\r-1\\r-3\end{pmatrix},\quad v=\begin{pmatrix}1\\1\\1\end{pmatrix},\quad A=I_3+uv^T,\quad v^Tu=(2-2r)+(r-1)+(r-3)=-2.$$ $$A^2=I_3+2uv^T+u(v^Tu)v^T=I_3+2uv^T-2uv^T=I_3.$$

The equality A=I₃+uvᵀ follows by writing out its entries. Associativity gives (uvᵀ)²=u(vᵀu)vᵀ. The scalar vᵀu equals −2 for every real r, so the computation proves involutoricity without dividing by any parameter.

Question 18

All matrices and unspecified scalar parameters in this solution are over the real numbers. Matrix dimensions are positive integers; O denotes the zero matrix of the displayed compatible size.

Problem

$$a\in\mathbb R,\qquad A=\begin{pmatrix}0&a&a^2&a^3\\0&0&a&a^2\\0&0&0&a\\0&0&0&0\end{pmatrix}.$$

Compute A²,A³,A⁴.

Solution

$$A^2=\begin{pmatrix}0&0&a^2&2a^3\\0&0&0&a^2\\0&0&0&0\\0&0&0&0\end{pmatrix},\qquad A^3=\begin{pmatrix}0&0&0&a^3\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix},\qquad A^4=O_{4\times4}.$$

In A², the only possible nonzero entries are (1,3), (1,4), (2,4): their sums are a·a, a·a²+a²·a, a·a. Multiplying this result by A leaves only (1,4)=a²·a in A³. Its sole nonzero column position is 4, but row 4 of A is zero; hence A⁴=0. The formulas also hold for a=0.

Question 21

All matrices and unspecified scalar parameters in this solution are over the real numbers. Matrix dimensions are positive integers; O denotes the zero matrix of the displayed compatible size.

Problem

$$A\in M_{n\times n}(\mathbb R),\qquad \alpha,\beta,\gamma\in\mathbb R.$$

For n≥1: (a) prove that A=αIₙ or A=βIₙ or A=γIₙ implies (A−αIₙ)(A−βIₙ)(A−γIₙ)=O. (b) Disprove the converse by a counterexample.

Solution

$$\text{(b)}\quad n=2,\quad \alpha=0,\ \beta=1,\ \gamma=2,\quad A=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$ $$A(A-I_2)(A-2I_2)=\begin{pmatrix}0(-1)(-2)&0\\0&1(0)(-1)\end{pmatrix}=O,\qquad A\notin\{0I_2,1I_2,2I_2\}.$$

In (a), the relevant factor is the zero matrix in each of the three cases; a product containing a compatible zero factor is zero. In (b), diagonal multiplication proves the required zero product. A has two different diagonal entries, whereas each scalar identity has equal diagonal entries. Thus it equals none of the three matrices, disproving the universal converse.

Question 22

All matrices and unspecified scalar parameters in this solution are over the real numbers. Matrix dimensions are positive integers; O denotes the zero matrix of the displayed compatible size.

Problem

$$A,B\in M_{n\times n}(\mathbb R),\quad AB=A,\quad BA=B.$$

Prove A and B are both idempotent.

Solution

$$A^2=(AB)A=A(BA)=AB=A,\qquad B^2=(BA)B=B(AB)=BA=B.$$

Use the two hypotheses and associativity in exactly the displayed order. Each matrix has square equal to itself, which is the definition of idempotence. No cancellation is needed.

Question 23

All matrices and unspecified scalar parameters in this solution are over the real numbers. Matrix dimensions are positive integers; O denotes the zero matrix of the displayed compatible size.

Problem

$$A\in M_{n\times n}(\mathbb R).$$

For n≥1: (a) prove A²=Iₙ implies (Iₙ+A)(Iₙ−A)=O; (b) determine whether the converse is true, with justification.

Solution

$$(I_n+A)(I_n-A)=I_n-A+A-A^2=I_n-A^2.$$ $$(I_n+A)(I_n-A)=O\quad\Longleftrightarrow\quad I_n-A^2=O\quad\Longleftrightarrow\quad A^2=I_n.$$

The first identity uses distributivity and multiplication by Iₙ. If A is involutory it gives zero, proving (a). The same identity also gives the reverse implication, so (b) is true.

Question 24

All matrices and unspecified scalar parameters in this solution are over the real numbers. Matrix dimensions are positive integers; O denotes the zero matrix of the displayed compatible size.

Problem

$$A\in M_{n\times n}(\mathbb R),\qquad A^2=O.$$

Use induction to prove A(Iₙ+A)^m=A for every positive integer m.

Solution

$$\text{Base:}\quad A(I_n+A)=A+A^2=A.$$ $$\text{Step:}\quad A(I_n+A)^{k+1}=\bigl[A(I_n+A)^k\bigr](I_n+A)=A(I_n+A)=A.$$

The base case is m=1. For the step assume the desired equality holds at k≥1; associativity and the induction hypothesis give the displayed computation. Mathematical induction now proves the assertion for all positive m.

Question 25

All matrices and unspecified scalar parameters in this solution are over the real numbers. Matrix dimensions are positive integers; O denotes the zero matrix of the displayed compatible size.

Problem

$$a,b,c,d\in\mathbb R,\quad a\ne1,\qquad Q=\begin{pmatrix}a&b&c&d\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}.$$

Compute (a) Q²,Q³,Q⁴ explicitly; (b) guess an explicit formula for Q^n for every positive integer n and prove it by induction.

Solution

$$Q^2=\begin{pmatrix}a^2&(1+a)b&(1+a)c&(1+a)d\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix},$$ $$Q^3=\begin{pmatrix}a^3&(1+a+a^2)b&(1+a+a^2)c&(1+a+a^2)d\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix},$$ $$Q^4=\begin{pmatrix}a^4&(1+a+a^2+a^3)b&(1+a+a^2+a^3)c&(1+a+a^2+a^3)d\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}.$$ $$S_n=\sum_{j=0}^{n-1}a^j=\frac{a^n-1}{a-1},\qquad Q^n=\begin{pmatrix}a^n&S_nb&S_nc&S_nd\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}.$$

Repeated multiplication gives (a). For (b), n=1 has S₁=1 and reproduces Q. Suppose the formula holds at n=k. In Q^kQ, the first entry is $a^{k+1}$, and each other first-row entry is (a^k+S_k) times b,c,d respectively; the bottom three rows stay those of I₄. Since $S_{k+1}$=S_k+a^k, the formula holds at k+1. Finally (a−1)S_n=a^n−1 by cancellation in the finite sum; division is valid because a≠1.

Question 26

All matrices and unspecified scalar parameters in this solution are over the real numbers. Matrix dimensions are positive integers; O denotes the zero matrix of the displayed compatible size.

Problem

$$A\in M_{3\times3}(\mathbb R),\qquad A^3-3A^2+4A-2I_3=O.$$

Use induction to prove: for every integer n>2 there are real numbers bₙ,cₙ,dₙ such that A^n=bₙA²+cₙA+dₙI₃.

Solution

$$b_3=3,\quad c_3=-4,\quad d_3=2.$$ $$\begin{aligned}A^{k+1}&=(b_kA^2+c_kA+d_kI_3)A=b_kA^3+c_kA^2+d_kA\\&=(3b_k+c_k)A^2+(-4b_k+d_k)A+2b_kI_3.\end{aligned}$$ $$b_{k+1}=3b_k+c_k,\qquad c_{k+1}=-4b_k+d_k,\qquad d_{k+1}=2b_k.$$

The given polynomial identity supplies the base case n=3. Assuming the representation at k≥3, multiply by A and substitute A³=3A²−4A+2I₃. The displayed recurrence defines real coefficients for the next exponent, proving existence for every integer n>2. Uniqueness of these coefficients is neither claimed nor needed.