Matrix Algebra

Column and Row Spaces

Column space. Let be an $m\times n$ matrix. Then we can write $A = [\mathbf{a}_1 \cdots \mathbf{a}_n]$ with column vectors $\mathbf{a}_1,\ldots,\mathbf{a}_n$ in $\mathbb{R}^m$ . The subspace of $\mathbb{R}^m$ spanned by the columns $\mathbf{a}_1,\ldots,\mathbf{a}_n$ is called the column space of , and denoted by $\mathrm{col}\,A$ .

Dimension of column space. Recall that the pivot columns are obtained when the matrix is reduced to an REF, and that they correspond to basic variables for the homogeneous equation $A\mathbf{x} = \mathbf{0}$ . Now in the context of column space, we can introduce the matrix consisting of the column vectors $\mathbf{a}_i$ 's which correspond to the pivot columns of $A = [\mathbf{a}_1 \cdots \mathbf{a}_n]$ .

Any column vector $\mathbf{a}_i$ of can be expressed as a linear combination of the columns in . This implies that $\mathrm{col}\,A = \mathrm{col}\,A'$ .
The homogeneous equation $A'\mathbf{x'} = \mathbf{0}$ has the unique solution $\mathbf{x'} = \mathbf{0}$ . Thus, the columns of are linearly independent.

Therefore, we have shown that the column vectors of form a basis for $\mathrm{col}\,A$ .

Rank. We can equivalently define the rank of a matrix by

$\displaystyle \mathrm{rank}\,A = \dim(\mathrm{col}\,A)$

which is also identified as the number of pivot columns in

EXAMPLE 4. Find a basis for the column space of the matrix

$\displaystyle A = \begin{bmatrix} 1 & 4 & 0 & 2 &-1 \\ 3 &12 & 1 & 5 & 5 \\ 2 & 8 & 1 & 3 & 2 \\ 5 &20 & 2 & 8 & 8 \end{bmatrix}$

Matlab/Octave. The function rank(A) returns the rank of a matrix A.

Row space. In a similar manner we can express $A = [\mathbf{b}_1 \cdots \mathbf{b}_m]^T$ with column vectors $\mathbf{b}_1,\ldots,\mathbf{b}_m$ in $\mathbb{R}^n$ , where $\mathbf{b}_1^T,\ldots,\mathbf{b}_m^T$ become the rows of . The subspace of $\mathbb{R}^n$ spanned by the rows $\mathbf{b}_1,\ldots,\mathbf{b}_m$ is called the row space of , and denoted by $\mathrm{row}\,A$ .

Dimension of row space. If is obtained from by applying a row operation, any element of $\mathrm{row}\,A'$ is expressed as a linear combination of the rows of . Since can be obtained from by the reversed row operation, we can observe that

$\displaystyle \mathrm{row}\,A = \mathrm{row}\,A'$

In fact, $\mathrm{row}\,A = \mathrm{row}\,EA$ holds for any invertible $m\times m$ -matrix

. When a REF is obtained from

by a series of row operations, the subspace spanned by the rows of the REF is equal to $\mathrm{row}\,A$ . Furthermore, it is easily observed that the nonzero rows of the REF are linearly independent, and therefore, that they form a basis for $\mathrm{row}\,A$ . Consequently, we can find

$\displaystyle \dim(\mathrm{row}\,A) = \mathrm{rank}\,A$

EXAMPLE 5. Find a basis for the row space of the matrix

$\displaystyle A = \begin{bmatrix} -2 &-5 & 8 & 0 &-17 \\ 1 & 3 & -5 & 1 & 5 \\ 3 &11 &-19 & 7 & 1 \\ 1 & 7 &-13 & 5 & -3 \end{bmatrix}$

Also, find a basis for the null space of

Rank theorem. Let be an $m\times n$ -matrix. Then we have obtained the following fundamental relations among $\mathrm{col}\,A$ , $\mathrm{row}\,A$ , and $\mathrm{null}\,A$ .

$\dim(\mathrm{col}\,A) = \dim(\mathrm{row}\,A) = \mathrm{rank}\,A$
$\mathrm{rank}\,A + \dim(\mathrm{null}\,A) = n$