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Found in: Page 143

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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

The redundant column vectors of matrix A are ${\stackrel{↔}{v}}_{2},{\stackrel{\to }{v}}_{3}\mathrm{and}{\stackrel{\to }{v}}_{4}$.

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" $\left\{\left[\begin{array}{c}0\\ -1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}5\\ 0\\ -1\\ 0\\ 0\end{array}\right],\begin{array}{c}\left[\begin{array}{c}4\\ 0\\ 0\\ -1\\ \frac{1}{3}\end{array}\right]\end{array}\right\}$.

See the step by step solution

## Step 1: Finding the redundant vectors

Let ${\stackrel{\to }{v}}_{1}=\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],{\stackrel{\to }{v}}_{2}=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right],{\stackrel{\to }{v}}_{3}=\left[\begin{array}{c}5\\ 0\\ 0\\ 0\end{array}\right]{\stackrel{\to }{v}}_{4}=\left[\begin{array}{c}3\\ 1\\ 0\\ 0\end{array}\right]{\stackrel{\to }{v}}_{5}=\left[\begin{array}{c}-3\\ 3\\ 0\\ 0\end{array}\right]$

Here we have ${\stackrel{\to }{\mathrm{v}}}_{2}=0.,{\stackrel{\to }{\mathrm{v}}}_{1},{\stackrel{\to }{\mathrm{v}}}_{3}=5{\stackrel{\to }{\mathrm{v}}}_{1}\mathrm{and}{\stackrel{\to }{\mathrm{v}}}_{4}=4{\stackrel{\to }{\mathrm{v}}}_{1}+\frac{1}{3}{\stackrel{\to }{\mathrm{v}}}_{5}$.

$⇒{\stackrel{\to }{\mathrm{v}}}_{2},{\stackrel{\to }{\mathrm{v}}}_{3}\mathrm{and}{\stackrel{\to }{\mathrm{v}}}_{4}$are redundant vectors and ${\stackrel{\to }{\mathrm{v}}}_{1}\mathrm{and}{\stackrel{\to }{\mathrm{v}}}_{5}$are non-redundant vectors.

## Step 2:   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 ${\stackrel{\to }{\mathrm{v}}}_{1}\mathrm{and}{\stackrel{\to }{\mathrm{v}}}_{5}$are non-redundant vectors of matrix A, thus the basis of the image A = $\left\{{\stackrel{\to }{v}}_{1},{\stackrel{\to }{v}}_{5}\right\}i.e\left\{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ 3\\ 0\\ 0\end{array}\right]\right\}$.

## Step 3:   Finding the basis of the kernel of A

Since the vectors ${\stackrel{\to }{\mathrm{v}}}_{2},{\stackrel{\to }{\mathrm{v}}}_{3}\mathrm{and}{\stackrel{\to }{\mathrm{v}}}_{4}$are redundant vectors such that

${\stackrel{\to }{v}}_{2}=0.{\stackrel{\to }{v}}_{1},{\stackrel{\to }{v}}_{3}=5{\stackrel{\to }{v}}_{1}and{\stackrel{\to }{v}}_{4}=4{\stackrel{\to }{v}}_{1}+\frac{1}{3}{\stackrel{\to }{v}}_{5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒0.{\stackrel{\to }{v}}_{1}-{\stackrel{\to }{v}}_{2}=0,5{\stackrel{\to }{v}}_{1}-{\stackrel{\to }{v}}_{3}=0and4{\stackrel{\to }{v}}_{1}-{\stackrel{\to }{v}}_{4}+\frac{1}{3}{\stackrel{\to }{V}}_{5}=0$

Thus the vectors in the kernel of A are ${\stackrel{\to }{w}}_{1}=\left[\begin{array}{c}0\\ -1\\ 0\\ 0\\ 0\end{array}\right],{\stackrel{\to }{w}}_{2}=\left[\begin{array}{c}5\\ 0\\ -1\\ 0\\ 0\end{array}\right],{\stackrel{\to }{w}}_{3}=\begin{array}{c}\left[\begin{array}{c}4\\ 0\\ 0\\ -1\\ \frac{1}{3}\end{array}\right]\end{array}$.

The redundant column vectors of matrix A are ${\stackrel{\to }{\mathrm{v}}}_{2},{\stackrel{\to }{\mathrm{v}}}_{3}\mathrm{and}{\stackrel{\to }{\mathrm{v}}}_{4}$.

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 = $\left\{\left[\begin{array}{c}0\\ -1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}5\\ 0\\ -1\\ 0\\ 0\end{array}\right],\begin{array}{c}\left[\begin{array}{c}4\\ 0\\ 0\\ -1\\ \frac{1}{3}\end{array}\right]\end{array}\right\}$.