Column Space

Definition

The column space of a matrix, $A$, often denoted by $C(A)$, is a vector space spanned by the column vectors of $A$.

The column-space of an $m \times n$ matrix $A$ is a subspace of $\R^m$. The dimensions of the column space are the number of linearly independent columns, that is, the rank of the matrix $A$.

Examples

1. The column space of $\begin{bmatrix}1 &0\\0&1\end{bmatrix}$ is $\R^2$.

2. The column space of $\begin{bmatrix}1 &2\\2&4\end{bmatrix}$ is a one-dimensional subspace of $\R^2$. This is because there’s a single linearly independent vector as $2\bold{c_1} =\bold{c_2}$. Any of these columns can be a basis for the subspace.

3. The column space of $\begin{bmatrix}0 &0\\0&0\end{bmatrix}$ is a zero-dimensional subspace of $\R^2$.

4. The column space of an $n \times n$ invertible matrix $A$ is $\R^n$. This is because all the columns of such a matrix are linearly independent. That is, $rank(A)=n$.

The columns of every $n \times n$ invertible matrix form a basis of $R^n$.

Basis of column space

A basis of the column space of a matrix, $A$, is a set of linearly independent columns of $A$. One algorithm for computing the basis is as follows:

1. Take the transpose of the matrix to convert the columns into rows.

2. Compute the $rref$ (reduced row echelon form) of the transposed matrix.

3. The set of non-zero rows in $rref$ is a linearly-independent set. These form the basis of the column space.

Note that the basis found by the algorithm above may not contain the original column vectors, but their linear combinations as elementary row operations have transformed the matrix into $rref$.

Let’s try it out by using the following code: