# Find a basis for the space spanned by the given vectors: v1, v2, v3, v4, and v5.

$v_1 = \begin{bmatrix} 2 \\ -1 \\ 0 \\ 5 \end{bmatrix}, v_2 = \begin{bmatrix} -8 \\ -3 \\ 3 \\ 6 \end{bmatrix}, v_3 = \begin{bmatrix} 0 \\ 4 \\ 2 \\ 3 \end{bmatrix}, v_4 = \begin{bmatrix} 7 \\ 1 \\ 11 \\ 1 \end{bmatrix}, v_5 = \begin{bmatrix} 2 \\ 1 \\ -3 \\ 0 \end{bmatrix}$

This question aims to find the column space of the given vectors forming a matrix.

The concepts needed to solve this question are column space, homogeneous equation of vectors, and linear transformations. A vector’s column space is written as Col A, which is the set of all possible linear combinations or range of the given matrix.

The collective matrix given by the vectors is calculated to be:

$\begin {bmatrix} 2 & -8 & 0 & 7 & 2 \\ -1 & -3 & 4 & 1 & 1 \\ 0 & 3 & 2 & 11 & -3 \\ 5 & 6 & 3 & 1 & 0 \end {bmatrix}$

We can calculate the row echelon form of the matrix using the row operations. The row echelon form of the matrix is calculated to be:

$\begin {bmatrix} 2 & -8 & 0 & 7 & 2 \\ 0 & -7 & 4 & 4.5 & 2 \\ 0 & 0 & 3.7 & 13 & -2.14 \\ 0 & 0 & 0 & -62 & 12.7 \end {bmatrix}$

Observing the above row echelon form of the matrix, we can see that it contains 4 pivot columns. Thus, those pivot columns correspond to the column space of the matrix. The basis for the space spanned by the given 5 vectors is given as:

$\begin{bmatrix} 2 \\ -1 \\ 0 \\ 5 \end{bmatrix}, \begin{bmatrix} -8 \\ -3 \\ 3 \\ 6 \end{bmatrix}, \begin{bmatrix} 0 \\ 4 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} 7 \\ 1 \\ 11 \\ 1 \end{bmatrix}$

## Numerical Result

The basis for the space spanned by the vectors that formed a matrix of 4×5 is calculated to be:

$\begin{bmatrix} 2 \\ -1 \\ 0 \\ 5 \end{bmatrix}, \begin{bmatrix} -8 \\ -3 \\ 3 \\ 6 \end{bmatrix}, \begin{bmatrix} 0 \\ 4 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} 7 \\ 1 \\ 11 \\ 1 \end{bmatrix}$

## Example

Find the column space spanned by the 3×3 matrix given below. Each column in the matrix represents a vector.

$\begin {bmatrix} 2 & -1 & 0 \\ -1 & -3 & 5 \\ 0 & 2 & 2 \end {bmatrix}$

The row echelon form of the matrix is calculated using row operations as:

$\begin {bmatrix} 2 & -1 & 0 \\ 0 & -3.5 & 5 \\ 0 & 0 & 4.8 \end {bmatrix}$

This row echelon form of the matrix represents three pivot columns corresponding to the matrix’s column space. The column space of the given 3×3 matrix is given as:

$\begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ -3 \\ 2 \end{bmatrix}, \begin{bmatrix} 0 \\ 5 \\ 2 \end{bmatrix}$