Question

In Exercises 1 through 20, find the redundant column vectors of the given matrix A “by inspection.” Then find a basis of the image of A and a basis of the kernel of A.

20. $\left[\begin{array}{ccccc}1& 0& 5& 3& -3\\ 0& 0& 0& 1& 3\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right]$

Short answer

The redundant column vectors of matrix A are ${\overleftrightarrow{v}}_2,{\overrightarrow{v}}_3\mathrm{and}{\overrightarrow{v}}_4$.

The basis of the image of A = $\left\{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ 3\\ 0\\ 0\end{array}\right]\right\}$.

The basis of the kernel of A = role="math" localid="1659419993266" $\left\{\left[\begin{array}{c}0\\ -1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}5\\ 0\\ -1\\ 0\\ 0\end{array}\right],\begin{array}{c}\left[\begin{array}{c}4\\ 0\\ 0\\ -1\\ \frac{1}{3}\end{array}\right]\end{array}\right\}$.

Step-by-step solution

Finding the redundant vectors

Let${\overrightarrow{v}}_1=\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right],{\overrightarrow{v}}_3=\left[\begin{array}{c}5\\ 0\\ 0\\ 0\end{array}\right]{\overrightarrow{v}}_4=\left[\begin{array}{c}3\\ 1\\ 0\\ 0\end{array}\right]{\overrightarrow{v}}_5=\left[\begin{array}{c}-3\\ 3\\ 0\\ 0\end{array}\right]$

Here we have ${\overrightarrow{\mathrm{v}}}_2=0.,{\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_3=5{\overrightarrow{\mathrm{v}}}_1\mathrm{and}{\overrightarrow{\mathrm{v}}}_4=4{\overrightarrow{\mathrm{v}}}_1+\frac{1}{3}{\overrightarrow{\mathrm{v}}}_5$.

$\Rightarrow {\overrightarrow{\mathrm{v}}}_2,{\overrightarrow{\mathrm{v}}}_3\mathrm{and}{\overrightarrow{\mathrm{v}}}_4$are redundant vectors and ${\overrightarrow{\mathrm{v}}}_1\mathrm{and}{\overrightarrow{\mathrm{v}}}_5$are non-redundant vectors.

  Finding the basis of the image of A

The non-redundant column vectors of A form the basis of the image of A.

Since column vectors ${\overrightarrow{\mathrm{v}}}_1\mathrm{and}{\overrightarrow{\mathrm{v}}}_5$are non-redundant vectors of matrix A, thus the basis of the image A = $\left\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_5\right\}i.e\left\{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ 3\\ 0\\ 0\end{array}\right]\right\}$.

  Finding the basis of the kernel of A

Since the vectors ${\overrightarrow{\mathrm{v}}}_2,{\overrightarrow{\mathrm{v}}}_3\mathrm{and}{\overrightarrow{\mathrm{v}}}_4$are redundant vectors such that

$$\begin{gathered} {\overrightarrow{v}}_2=0.{\overrightarrow{v}}_1,{\overrightarrow{v}}_3=5{\overrightarrow{v}}_{1}and{\overrightarrow{v}}_4=4{\overrightarrow{v}}_1+\frac{1}{3}{\overrightarrow{v}}_5 \\ \Rightarrow 0.{\overrightarrow{v}}_1-{\overrightarrow{v}}_2=0,5{\overrightarrow{v}}_1-{\overrightarrow{v}}_3=0and4{\overrightarrow{v}}_1-{\overrightarrow{v}}_4+\frac{1}{3}{\overrightarrow{V}}_5=0 \end{gathered}$$

Thus the vectors in the kernel of A are ${\overrightarrow{w}}_1=\left[\begin{array}{c}0\\ -1\\ 0\\ 0\\ 0\end{array}\right],{\overrightarrow{w}}_2=\left[\begin{array}{c}5\\ 0\\ -1\\ 0\\ 0\end{array}\right],{\overrightarrow{w}}_3=\begin{array}{c}\left[\begin{array}{c}4\\ 0\\ 0\\ -1\\ \frac{1}{3}\end{array}\right]\end{array}$.

Final Answer

The redundant column vectors of matrix A are ${\overrightarrow{\mathrm{v}}}_2,{\overrightarrow{\mathrm{v}}}_3\mathrm{and}{\overrightarrow{\mathrm{v}}}_4$.

The basis of the image of A = $A=\left\{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ 3\\ 0\\ 0\end{array}\right]\right\}$.

The basis of the kernel of A = $\left\{\left[\begin{array}{c}0\\ -1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}5\\ 0\\ -1\\ 0\\ 0\end{array}\right],\begin{array}{c}\left[\begin{array}{c}4\\ 0\\ 0\\ -1\\ \frac{1}{3}\end{array}\right]\end{array}\right\}$.