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.

17. $\left[\begin{array}{ccccc}0& 1& 2& 0& 3\\ 0& 0& 0& 1& 4\end{array}\right]$

Short answer

The redundant column vectors of matrix A are ${\stackrel{\rightharpoonup }{\mathrm{v}}}_1,{\stackrel{\rightharpoonup }{\mathrm{v}}}_3\mathrm{and}{\stackrel{\rightharpoonup }{\mathrm{v}}}_5$.

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

The basis of the kernel of A =localid="1659425873346" $\left\{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 2\\ -1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 3\\ 0\\ 4\\ -1\end{array}\right]\right\}$

Step-by-step solution

Finding the redundant vectors

Let ${\stackrel{\rightharpoonup }{v}}_1=\left[\begin{array}{c}0\\ 0\end{array}\right]{\stackrel{\rightharpoonup }{v}}_2=\left[\begin{array}{c}1\\ 0\end{array}\right],{\stackrel{\rightharpoonup }{v}}_3=\left[\begin{array}{c}2\\ 0\end{array}\right],{\stackrel{\rightharpoonup }{v}}_4=\left[\begin{array}{c}0\\ 1\end{array}\right],{\stackrel{\rightharpoonup }{v}}_5=\left[\begin{array}{c}3\\ 4\end{array}\right]$

Here we have ${\stackrel{\rightharpoonup }{v}}_1=0.{\stackrel{\rightharpoonup }{v}}_2,{\stackrel{\rightharpoonup }{v}}_3=2.{\stackrel{\rightharpoonup }{v}}_2\mathrm{and}{\stackrel{\rightharpoonup }{\mathrm{v}}}_5,=3{\stackrel{\rightharpoonup }{\mathrm{v}}}_2+4{\stackrel{\rightharpoonup }{\mathrm{v}}}_4$

$\Rightarrow {\stackrel{\rightharpoonup }{v}}_2$are${\stackrel{\rightharpoonup }{v}}_4$ non-redundant vectors androle="math" localid="1659424640461" ${\stackrel{\rightharpoonup }{v}}_1,{\stackrel{\rightharpoonup }{v}}_3\mathrm{and}{\stackrel{\rightharpoonup }{\mathrm{v}}}_5$ are 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 ${\stackrel{\rightharpoonup }{v}}_2\mathrm{and}{\stackrel{\rightharpoonup }{\mathrm{v}}}_4$are non-redundant vectors of matrix A, thus the basis of the image A =$\left\{{\stackrel{\rightharpoonup }{v}}_2,{\stackrel{\rightharpoonup }{\mathrm{v}}}_4\right\}$i.e. $\left\{\left[\begin{array}{c}1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\end{array}\right]\right\}$

  Finding the basis of the kernel of A

Since the vectors ${\stackrel{\rightharpoonup }{v}}_1,{\stackrel{\rightharpoonup }{v}}_3\mathrm{and}{\stackrel{\rightharpoonup }{\mathrm{v}}}_5$are redundant vectors such that

role="math" localid="1659425809735" $$\begin{gathered} {\stackrel{\rightharpoonup }{\mathrm{v}}}_1=0.{\stackrel{\rightharpoonup }{\mathrm{v}}}_2,{\stackrel{\rightharpoonup }{\mathrm{v}}}_3=2.{\stackrel{\rightharpoonup }{\mathrm{v}}}_2\mathrm{and}{\stackrel{\rightharpoonup }{\mathrm{v}}}_5=3{\stackrel{\rightharpoonup }{\mathrm{v}}}_2+4{\stackrel{\rightharpoonup }{\mathrm{v}}}_4 \\ \Rightarrow {\stackrel{\rightharpoonup }{\mathrm{v}}}_1-0.{\stackrel{\rightharpoonup }{\mathrm{v}}}_2=0,2{\stackrel{\rightharpoonup }{\mathrm{v}}}_2-{\stackrel{\rightharpoonup }{\mathrm{v}}}_3=0\mathrm{and}3{\stackrel{\rightharpoonup }{\mathrm{v}}}_2+3{\stackrel{\rightharpoonup }{\mathrm{v}}}_2+4{\stackrel{\rightharpoonup }{\mathrm{v}}}_4-{\stackrel{\rightharpoonup }{\mathrm{v}}}_5=0 \\ \mathrm{Thus}\mathrm{the}\mathrm{vectors}\mathrm{in}\mathrm{kernel}\mathrm{of}\mathrm{A}\mathrm{is}{\stackrel{\rightharpoonup }{\mathrm{w}}}_1=\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right],{\stackrel{\rightharpoonup }{\mathrm{w}}}_2=\left[\begin{array}{c}0\\ 2\\ -1\\ 0\\ 0\end{array}\right],{\stackrel{\rightharpoonup }{\mathrm{w}}}_3=\left[\begin{array}{c}0\\ 3\\ 0\\ 4\\ -1\end{array}\right]. \end{gathered}$$

Final Answer

The redundant column vectors of matrix A are ${\stackrel{\rightharpoonup }{v}}_1,{\stackrel{\rightharpoonup }{v}}_3\mathrm{and}{\stackrel{\rightharpoonup }{v}}_5$

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

The basis of the kernel of A =$\left\{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 2\\ -1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 3\\ 0\\ 4\\ -1\end{array}\right]\right\}$