Question

Find a basis of the image of the matrices in Exercise 27 through 33.

$\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{1}\\ \mathbf{1}& \mathbf{2}& \mathbf{5}\\ \mathbf{1}& \mathbf{3}& \mathbf{7}\end{array}\mathbf{\right]}$

Short answer

the basis of the image of matrix A is

$\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right],\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right],\left[\begin{array}{l}1\\ 5\\ 7\end{array}\right]$

Step-by-step solution

Given that

Given a matrix

$\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 5\\ 1& 3& 7\end{array}\right]$

Find redundant column vectors to find the basis of the image

The column vectors of the matrix A are:

${\overrightarrow{v}}_1=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$, ${\overrightarrow{v}}_2=\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]$and${\overrightarrow{v}}_3=\left[\begin{array}{l}1\\ 5\\ 7\end{array}\right]$

Here, ${\overrightarrow{v}}_1$and${\overrightarrow{v}}_2$ both are non-zero vectors.

Also,${\overrightarrow{v}}_2$ is not a scaler multiple of ${\overrightarrow{v}}_1$.

So, the vectors ${\overrightarrow{v}}_1$and${\overrightarrow{v}}_2$ are not redundant vectors.

Find whether the vector v→3is redundant or not

Assume that${\overrightarrow{v}}_3$ is redundant.

So, it can be written as a linear combination of ${\overrightarrow{v}}_1$and${\overrightarrow{v}}_2$.

Let${\overrightarrow{v}}_3=c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2$

Consider the augmented matrix

$\left[\begin{array}{cc}1& 1\\ 1& 2\\ 1& 3\end{array}|\begin{array}{l}1\\ 5\\ 7\end{array}\right]$

Apply Row operations as follows:

$\begin{array}{l}\left[\begin{array}{cc}1& 1\\ 1& 2\\ 1& 3\end{array}|\begin{array}{l}1\\ 5\\ 7\end{array}\right]\underrightarrow{R_2:R_2-R_1}\left[\begin{array}{cc}1& 1\\ 0& 1\\ 1& 3\end{array}|\begin{array}{l}1\\ 4\\ 7\end{array}\right]\\ \left[\begin{array}{cc}1& 1\\ 0& 1\\ 1& 3\end{array}|\begin{array}{l}1\\ 4\\ 7\end{array}\right]\underrightarrow{R_3:R_3-R_1}\left[\begin{array}{cc}1& 1\\ 0& 1\\ 0& 2\end{array}|\begin{array}{l}1\\ 4\\ 6\end{array}\right]\\ \left[\begin{array}{cc}1& 1\\ 0& 1\\ 0& 2\end{array}|\begin{array}{l}1\\ 4\\ 6\end{array}\right]\underrightarrow{R_3:R_3-2R_2}\left[\begin{array}{cc}1& 1\\ 0& 1\\ 0& 0\end{array}|\begin{array}{l}1\\ 4\\ -2\end{array}\right]\end{array}$

Thus the matrix implies:

$\begin{array}{c}{c}_1+c_2=1\\ 0\cdot c_1+c_2=4\\ 0\cdot c_1+0\cdot c_2=-2\end{array}$

The last equation implies$0=-2$ which is clearly false.

Hence, our assumption was wrong.

Thus, ${\overrightarrow{v}}_3$is not redundant.

Find basis of the image matrix A

Since, the three vectors are linearly independent, therefore the basis of the image of matrix A is

$\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right],\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right],\left[\begin{array}{l}1\\ 5\\ 7\end{array}\right]$