# How to know if a vector is in the column space?

## Question:

How to know if a vector is in the column space?

## Column Space

The column space is the vector space formed by the columns of a matrix.

To find a basis of a vector space, we need to find a set of linearly independent vectors that span the vector space.

A set of vectors is linearly independent if the vectors, placed in a matrix, gives a row echelon form with all the columns being pivot columns.

To determine if a vector is in the column space, we need to verify that an augmented matrix is consistent. An augmented matrix is consistent if there is no pivot column on the last column.

A vector {eq}\displaystyle \vec{b} {/eq} is in the column space of a given matrix, {eq}\displaystyle A, \boxed{\text{if it is a linear combination of the columns of } A}. {/eq}

To verify that {eq}\displaystyle \vec{v} {/eq} is in the column space of {eq}\displaystyle A {/eq} we check if the augmented matrix system {eq}\displaystyle \left[A|\vec{b}\right] {/eq} is consistent, or there is no pivot column in the last column.

For example, the vector {eq}\displaystyle \vec{v}=\left[\begin{array}{c} 2\\\ 3\\ 4 \end{array}\right] {/eq} is in the column space of matrix {eq}\displaystyle A=\left[\begin{array}{c} 1&0&-3&2\\\ -2&9&-1&0\\ -4&-1&0&2\\ \end{array}\right] {/eq}

because the following augmented matrix is consistent

{eq}\displaystyle \left[A|\vec{v}\right]=\left[\begin{array}{cccc|c} 1&0&-3&2&2\\\ -2&9&-1&0&3\\ -4&-1&0&2&4\\ \end{array}\right]\overset{2R_1+R_2}{\implies}\left[\begin{array}{cccc|c} 1&0&-3&2&2\\\ 0&9&-7&4&7\\ -4&-1&0&2&4\\ \end{array}\right]\overset{4R_1+R_3}{\implies}\left[\begin{array}{cccc|c} 1&0&-3&2&2\\\ 0&9&-7&4&7\\ 0&-1&-12&10&12\\ \end{array}\right]\\\\ \overset{\frac{1}{9}R_2+R_3}{\implies}\left[\begin{array}{cccc|c} \boxed{1}&0&-3&2&2\\\ 0&\boxed{9}&-7&4&7\\ 0&0&\boxed{-115/9}&94/9&115/9\\ \end{array}\right]\\ \implies \text{ the system is consistent, because the last column is not a pivot column, so the vector is in the column space}. {/eq}