6.7 Matrix subspaces, basis, and dimension

A matrix can itself be a vector in a vector space. The relevant operations are matrix addition and scalar multiplication, so all matrices under discussion must have the same size. Throughout this note, scalars and entries are real, and $m,n$ are positive integers. We write $M_{m,n}(\mathbb R)$ for the space of real $m\times n$ matrices.

The main questions remain familiar: which matrices can a list generate, which generators are redundant, and how many matrices are needed in a basis? Entry constraints sometimes reveal the answer directly. For a less transparent generating list, we can stack the entries and use row reduction.

Matrices as vectors

Definition 6.7.1

(Matrix span and matrix basis)

For $A_1,\ldots,A_q\in M_{m,n}(\mathbb R)$, their span is

$$\operatorname{Span}\{A_1,\ldots,A_q\} =\{c_1A_1+\cdots+c_qA_q:c_1,\ldots,c_q\in\mathbb R\}.$$

A list in a subspace $W$ is a basis of $W$ if it spans $W$ and is linearly independent: the equality $c_1A_1+\cdots+c_qA_q=0$ forces every $c_j=0$. The dimension of $W$ is the number of matrices in a basis. The zero subspace has the empty basis and dimension zero.

Matrix span is a subspace: the zero matrix is obtained with zero coefficients, and adding or scaling combinations gives another combination of the same matrices. A basis provides a unique coefficient list for every matrix in the subspace. Spanning gives existence; subtracting two representations and using independence gives uniqueness. Conversely, unique representations imply spanning and force the zero matrix's coefficients to be zero.

Concept lensAlgebraic

The whole matrix is one vector

In $M_{2,3}(\mathbb R)$, one vector has six entries arranged as a rectangle. Its shape does not make the dimension two or three. Dimension counts how many independent matrices are needed to generate the space, while matrix rank concerns the column space of one matrix. These are different questions. For example, the identity matrix of size two has matrix rank two, but the subspace spanned by that one nonzero matrix has dimension one: every element is a scalar multiple of the identity.

A matrix subspace only needs closure under addition and scalar multiplication. Closure under matrix multiplication is not part of its definition; rectangular matrices may not even be multipliable with each other. The zero vector here is the zero matrix of the fixed size, and coefficients remain real numbers.

Matrix units give coordinates

For each position $(i,j)$, let $E_{ij}$ have entry one at that position and zero elsewhere. The ambient size is fixed when using this notation. For example, in $M_{2,2}(\mathbb R)$,

$$E_{11}=\begin{bmatrix}1&0\\0&0\end{bmatrix},\quad E_{12}=\begin{bmatrix}0&1\\0&0\end{bmatrix},\quad E_{21}=\begin{bmatrix}0&0\\1&0\end{bmatrix},\quad E_{22}=\begin{bmatrix}0&0\\0&1\end{bmatrix}.$$

Theorem 6.7.2

(Basis and dimension of the full matrix space)

The matrices $E_{ij}$, where $1\leq i\leq m$ and $1\leq j\leq n$, form a basis of $M_{m,n}(\mathbb R)$. Consequently,

$$\dim M_{m,n}(\mathbb R)=mn.$$

Proof

(Spanning and independence by comparing entries)

Every matrix $A=[a_{ij}]$ satisfies

$$A=\sum_{i=1}^m\sum_{j=1}^n a_{ij}E_{ij}.$$

This proves spanning: the entry at position $(i,j)$ on the right is precisely $a_{ij}$. If a combination $\sum\limits_{i,j}c_{ij}E_{ij}$ equals zero, its $(i,j)$ entry is $c_{ij}$, so every coefficient vanishes. This proves independence. There are $mn$ basis matrices, giving the dimension.

Thus counting entry positions works here because we have proved that these entries can be prescribed independently. In a constrained subspace, counting all displayed entries may overcount the actual degrees of freedom.

Column stacking preserves linear relations

Define $\operatorname{vec}(A)$ by stacking the columns of $A$, from left to right, into one column. For instance,

$$\operatorname{vec}\!\begin{bmatrix}a&b\\c&d\end{bmatrix} =\begin{bmatrix}a\\c\\b\\d\end{bmatrix}.$$

The order must be used consistently for every matrix. The result has $mn$ entries and can be reshaped uniquely into the original $m\times n$ matrix.

Theorem 6.7.3

(The coordinate dictionary for matrix bases)

For fixed-size matrices $A_1,\ldots,A_q,B$ and real coefficients $c_j$,

$$B=\sum_{j=1}^q c_jA_j \quad\Longleftrightarrow\quad \operatorname{vec}(B)=\sum_{j=1}^q c_j\operatorname{vec}(A_j).$$

In particular, a list of matrices is linearly independent exactly when its stacked columns are independent. For a subspace $W$, write $\operatorname{vec}(W)=\{\operatorname{vec}(A):A\in W\}$. A list in $W$ is a basis of $W$ exactly when its stacked columns form a basis of $\operatorname{vec}(W)$.

Proof

(No equation is lost when entries are rearranged)

At every position, the matrix equation asserts $b_{rs}=\sum\limits_j c_j(A_j)_{rs}$. Stacking simply places these same scalar equations in a fixed order, establishing both directions. Applying this equivalence to $B=0$ preserves all homogeneous coefficient relations and hence independence. Applying it to every $B\in W$ preserves spanning. The same identity shows that $\operatorname{vec}(W)$ is closed under linear combinations and contains zero, so it is a subspace. Together these facts prove the basis assertion.

To find a basis of a supplied matrix span, form

$$U=[\,\operatorname{vec}(A_1)\ \cdots\ \operatorname{vec}(A_q)\,].$$

This matrix has $mn$ rows and $q$ columns. Reduce $U$, record its pivot-column indices, and select the original matrices with those indices. The original-column theorem from 6.6 Column space, row space, and rank then transfers through the dictionary. If every supplied matrix is zero, there are no pivot columns and the resulting basis is empty.

Worked example 6.7.4

(Extracting a basis from a dependent matrix family)

Let

$$A_1=\begin{bmatrix}0&-1\\0&2\end{bmatrix},\quad A_2=\begin{bmatrix}1&-2\\0&7\end{bmatrix},\quad A_3=\begin{bmatrix}-2&3\\0&-12\end{bmatrix}.$$

Column stacking gives

$$U=\begin{bmatrix}0&1&-2\\0&0&0\\-1&-2&3\\2&7&-12\end{bmatrix}.$$

Swap the first and third rows, negate the first, and subtract twice the first from the fourth. Then swap the second and third rows:

$$U\longrightarrow \begin{bmatrix}1&2&-3\\0&1&-2\\0&0&0\\0&3&-6\end{bmatrix}.$$

Subtract twice the second row from the first and three times the second from the fourth:

$$\operatorname{rref}(U)= \begin{bmatrix}1&0&1\\0&1&-2\\0&0&0\\0&0&0\end{bmatrix}.$$

The pivot indices are one and two. Hence $A_1,A_2$ form a basis of the original matrix span, whose dimension is two. The third column records

$$A_3=A_1-2A_2 =\begin{bmatrix}-2&3\\0&-12\end{bmatrix}.$$

Every combination can therefore be rewritten as

$$c_1A_1+c_2A_2+c_3A_3=(c_1+c_3)A_1+(c_2-2c_3)A_2.$$

Independence of $A_1,A_2$ can also be checked directly: the top-left entry forces the coefficient of $A_2$ to vanish, and the top-right entry then forces the coefficient of $A_1$ to vanish.

Common mistake

(Reshaping the reduced columns changes the matrices)

The second pivot column of the RREF is $(0,1,0,0)^T$. Reshaping it gives $E_{21}$, which is outside the original span: every original generator has lower-left entry zero. Row reduction preserves relations among columns, but need not preserve their span. Use the pivot indices to return to $A_1,A_2$, rather than reshaping the reduced columns.

Upper triangular and diagonal subspaces

Let $UT_n(\mathbb R)$ consist of square matrices whose entries below the diagonal are zero. These zero constraints survive addition and scalar multiplication, and the zero matrix satisfies them, so this is a subspace. Its general expansion is

$$A=\sum_{1\leq i\leq j\leq n}a_{ij}E_{ij}.$$

The allowed matrix units span, and entry comparison proves their independence exactly as before. Counting the allowed positions gives

$$\dim UT_n(\mathbb R)=n+(n-1)+\cdots+1=\frac{n(n+1)}2.$$

For diagonal matrices, only $E_{11},\ldots,E_{nn}$ remain, giving dimension $n$. Lower triangular matrices have the same dimension as upper triangular matrices: transpose takes the basis units $E_{ij}$ to $E_{ji}$.

Worked example 6.7.5

(Reading six coordinates in an upper triangular matrix)

A general element of $UT_3(\mathbb R)$ is

$$\begin{bmatrix}a&b&c\\0&d&e\\0&0&f\end{bmatrix} =aE_{11}+bE_{12}+cE_{13}+dE_{22}+eE_{23}+fE_{33}.$$

Each of the six parameters is arbitrary and occupies a distinct allowed entry. A zero combination forces all six to vanish, so these six matrices are a basis. For the matrix with $(a,b,c,d,e,f)=(1,-2,3,4,0,5)$, the coordinate list in this displayed order is $(1,-2,3,4,0,5)$. The zero coefficient of $E_{23}$ in this particular matrix does not remove $E_{23}$ from a basis of the whole subspace.

Symmetric and skew-symmetric subspaces

Let $Sym_n(\mathbb R)$ be the matrices satisfying $A^T=A$, and let $Skew_n(\mathbb R)$ be those satisfying $A^T=-A$. Both are subspaces: transpose distributes over addition and scalar multiplication, so each defining equation is preserved under linear combinations.

For symmetric matrices, entries below the diagonal repeat those above it. The diagonal entries remain arbitrary, giving

$$A=\sum_{i=1}^n a_{ii}E_{ii} +\sum_{i\lt j}a_{ij}(E_{ij}+E_{ji}).$$

These matrices span $Sym_n(\mathbb R)$. In a zero combination, the diagonal entries recover the first coefficients, and each upper-diagonal entry recovers exactly one remaining coefficient. All must be zero. Therefore this is a basis and

$$\dim Sym_n(\mathbb R)=n+\frac{n(n-1)}2=\frac{n(n+1)}2.$$

Worked example 6.7.6

(A basis of the symmetric matrices of size two)

Every symmetric matrix of size two has the form

$$\begin{bmatrix}a&b\\b&c\end{bmatrix} =a\begin{bmatrix}1&0\\0&0\end{bmatrix} +b\begin{bmatrix}0&1\\1&0\end{bmatrix} +c\begin{bmatrix}0&0\\0&1\end{bmatrix}.$$

The three displayed matrices span. If their combination with coefficients $\alpha,\beta,\gamma$ is zero, comparison of positions $(1,1),(1,2),(2,2)$ gives $\alpha=\beta=\gamma=0$. They form a basis, so the dimension is three. Although $b$ appears twice in the original array, the symmetry constraint makes those two entries a single choice.

The following sequence separates the three parameters into their fixed basis matrices.

Skew-symmetry fixes the diagonal and pairs the entries

For a real skew-symmetric matrix, $a_{ii}=-a_{ii}$ gives $2a_{ii}=0$ and hence $a_{ii}=0$. Every entry below the diagonal is the negative of its partner above it. Thus

$$A=\sum_{i\lt j}a_{ij}(E_{ij}-E_{ji}).$$

The same upper-entry comparison proves independence, so

$$\dim Skew_n(\mathbb R)=\frac{n(n-1)}2.$$

In particular, every matrix in $Skew_3(\mathbb R)$ is

$$\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix},$$

with basis $E_{12}-E_{21},E_{13}-E_{31},E_{23}-E_{32}$ and dimension three. When $n=1$, the space is zero and this basis list is empty, consistently giving dimension zero.

Every square matrix has a unique symmetric and skew part

Theorem 6.7.7

(Unique symmetric and skew-symmetric decomposition)

For every $A\in M_{n,n}(\mathbb R)$ there is exactly one pair $S\in Sym_n(\mathbb R)$ and $K\in Skew_n(\mathbb R)$ with $A=S+K$. It is

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

Proof

(Existence by formulas and uniqueness by zero intersection)

Transpose linearity and $(A^T)^T=A$ give $S^T=S$ and $K^T=-K$. Adding the formulas gives $S+K=A$, proving existence. For uniqueness, if $A=S_1+K_1=S_2+K_2$, then $X=S_1-S_2=K_2-K_1$ is both symmetric and skew-symmetric. Hence $X=X^T=-X$, so $2X=0$ and $X=0$. Both parts therefore agree. Division by two is valid over the real field used throughout this note.

This proves more than the numerical identity $n(n+1)/2+n(n-1)/2=n^2$: every matrix can actually be assembled from the two subspaces, and their overlap contains only zero. Equal dimension counts alone would not establish either fact.

Worked example 6.7.8

(Computing both parts of a matrix)

For

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

the formulas give

$$S=\begin{bmatrix}0&2&4\\2&2&5\\4&5&5\end{bmatrix},\qquad K=\begin{bmatrix}0&1&2\\-1&0&-1\\-2&1&0\end{bmatrix}.$$

For example, the entries three and one in positions $(1,2),(2,1)$ contribute the average two to $S$ and the opposite pair one, minus one to $K$. Checking all entries gives $S+K=A$. The skew part has basis coefficients $(1,2,-1)$ relative to $E_{12}-E_{21},E_{13}-E_{31},E_{23}-E_{32}$, so its signs can also be checked through the earlier basis expansion.

Dimension bounds transfer with the coordinates

Since stacking preserves bases, it also preserves dimension. In particular, a matrix subspace of $M_{m,n}(\mathbb R)$ has dimension at most $mn$. More than $mn$ matrices of this fixed size must be dependent: their stacked coefficient matrix has more columns than rows, so its homogeneous system has a free variable and hence a nonzero solution. The same coefficients give a nontrivial zero combination of the original matrices.

More precisely, if $W$ has known dimension $d$ and a list consists of exactly $d$ matrices belonging to $W$, then either independence or spanning suffices to establish that it is a basis. This follows by applying the corresponding basis theorem to the stacked subspace. The membership hypothesis matters: a list of $d$ independent matrices outside $W$ is not a basis of $W$ merely because its length matches the dimension.

These facts also clarify the meaning of “dimension counts parameters.” A complete parametrization proves spanning, but it can still assign several parameter lists to the same matrix. To identify the parameter count with dimension, uniqueness must be checked, for example by comparing entries or solving the homogeneous coefficient equations. The earlier symmetric and triangular expansions pass this check; the three-generator dependent family does not.

Choosing a method

For a space defined by entry constraints, write its general matrix, separate its genuinely free parameters, and prove that the resulting fixed matrices span and are independent. For a supplied generating family, stack columns, reduce the coordinate matrix, and return to the original pivot-indexed matrices. These are two ways of answering the same basis question.

Common mistake

(Counting parameters before proving independence)

A parametrization can be redundant. The three matrices in the dependent-family example give three coefficients but a space of dimension two. Counting parameters determines dimension only after the associated matrices are shown to be independent. Conversely, a particular matrix with many zero coefficients does not lower the dimension of its ambient subspace.

Quick checks

Quick check

(What is the dimension of the real matrix space with two rows and three columns?)

Use the matrix units.

Solution · Answer

There are $2\cdot3=6$ basis matrices $E_{ij}$, so $\dim M_{2,3}(\mathbb R)=6$.

Quick check

(Why must a real skew-symmetric matrix have zero diagonal?)

Apply its defining equation at a diagonal entry.

Solution · Answer

Transpose leaves $a_{ii}$ in place, so $a_{ii}=-a_{ii}$. Thus $2a_{ii}=0$ and $a_{ii}=0$ over the reals.

Exercises

Quick check

(Find a basis and dimension for the diagonal matrices of size three.)

Write the general diagonal matrix.

Solution · Guided solution

Every diagonal matrix is $aE_{11}+bE_{22}+cE_{33}$. The three units span and are independent by comparison of diagonal entries. They form a basis and the dimension is three.

Quick check

(In the dependent-family example, express twice the first generator plus the third in the basis of the first two generators.)

Use the relation obtained from the third reduced column.

Solution · Guided solution

Since $A_3=A_1-2A_2$, we obtain $2A_1+A_3=3A_1-2A_2$. Its basis coordinates are $(3,-2)$, and the resulting matrix is $\begin{bmatrix}-2&1\\0&-8\end{bmatrix}$.

Quick check

(If a real matrix is both symmetric and skew-symmetric, what is it? What is the dimension of the intersection of these subspaces?)

Compare the two transpose equations.

Solution · Guided solution

The equations give $A=A^T=-A$, hence $A=0$. Their intersection is the zero subspace, with empty basis and dimension zero.

Related notes

This note uses 6.5 Basis and dimension and 3.3 Transpose, symmetric, and skew-symmetric matrices. The coordinate computation uses the original-column rule of 6.6; the extra step here is translating back to matrices of the original size.