Mathwizurd.com is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

# What is it?

The column space is the span of the columns of a vector.

## How do you find it?

MathJax TeX Test Page To demonstrate how to find the column space of a matrix, it is good to demonstrate through an example $$A = \begin{bmatrix}2 & 4 & -2 & 1\\-2 & -5 & 7 & -5\\3 & 7 & -8 & 5.5\end{bmatrix}$$ We want to see when we get a solution for $Ax = b$. This is possible if we create an augmented matrix. $$\begin{bmatrix}2 & 4 & -2 & 1 & b_1\\-2 & -5 & 7 & -5 & b_2\\3 & 7 & -8 & 5.5 & b_3\end{bmatrix} \to \begin{bmatrix}2 & 4 & -2 & 1 & b_1\\0 & -1 & 5 & -4 & b_1 + b_2\\3 & 7 & -8 & 5.5 & b_3\end{bmatrix}$$ $$\to \begin{bmatrix}2 & 4 & -2 & 1 & b_1\\0 & -1 & 5 & -4 & b_1 + b_2\\0 & 1 & -5 & 4 & b_3 - \frac{3}{2}b_1\end{bmatrix} \to \begin{bmatrix}2 & 4 & -2 & 1 & b_1\\0 & -1 & 5 & -4 & b_1 + b_2 \\0 & 0 & 0 & 0 & b_3 + b_2 - \frac{1}{2}b_1\end{bmatrix}$$ This only works when the augmented matrix is consistent. That means each equation is value (No $0 = 3$ or something). So, you know $$b_1 = 2b_2 + 2b_3 \to \begin{bmatrix}2b_2 + 2b_3\\b_2\\b_3\end{bmatrix} = b_2\begin{bmatrix}2\\1\\0\end{bmatrix} + b_3\begin{bmatrix}2\\0\\1\end{bmatrix}$$ $b_2$ and $b_3$ can be anything. The only restriction is it has to be a real number. So, the column space of A is $$\text{Span}\left(\begin{bmatrix}2\\1\\0\end{bmatrix}, \begin{bmatrix}2\\0\\1\end{bmatrix}\right)$$

## Better way

There’s a better way to find it.

MathJax TeX Test Page The column space of a matrix is equal to the span of the pivot columns in the matrix's RREF. Consider the matrix from above again. $$\begin{bmatrix}2 & 4 & -2 & 1\\-2 & -5 & 7 & -5\\3 & 7 & -8 & 5.5\end{bmatrix} \to \begin{bmatrix}1 & 0 & 9 & -7.5\\0 & 1 & -5 & 4\\0 & 0 & 0 & 0\end{bmatrix}$$ So, the pivot columns are the first two columns. Therefore Column space of A equals the span of $\begin{bmatrix}2 \\ -2 \\ 3\end{bmatrix}$ and $\begin{bmatrix}4 \\ -5 \\ 7 \end{bmatrix}$. To show that these are the same, I will show that you can express the vectors in the first basis as a linear combination of the second basis. $$\begin{bmatrix}2 \\1\\0\end{bmatrix} = 7*\begin{bmatrix}2\\-2\\3\end{bmatrix}-3*\begin{bmatrix}4\\-5\\7\end{bmatrix}$$ Additionally, $$\begin{bmatrix}2 \\0\\1\end{bmatrix} = 5*\begin{bmatrix}2\\-2\\3\end{bmatrix}-2*\begin{bmatrix}4\\-5\\7\end{bmatrix}$$

# How rank relates to this

The rank is the dimension of the column space, meaning it’s the number of vectors in the basis of the column space.