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Q16E

Expert-verified
Found in: Page 143

### Linear Algebra With Applications

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

# 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.16.$\left[\begin{array}{ccccc}1& -2& 0& -1& 0\\ 0& 0& 1& 5& 0\\ 0& 0& 0& 0& 1\end{array}\right]$

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

The basis of the image of A =$\left\{\left[1\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}0\right],\left[0\phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}0\right],\left[0\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}1\right]\right\}$

The basis of the kernel of A = $\left\{\left[2\phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}0\right],\left[1\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}-5\phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}0\right]\right\}$

See the step by step solution

## Step 1: Finding the redundant vectors

Let $\stackrel{\to }{{v}_{1}}=\left[1\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}0\right],\stackrel{\to }{{v}_{2}}=\left[-2\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}0\right],\stackrel{\to }{{v}_{3}}=\left[0\phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}0\right],\stackrel{\to }{{v}_{4}}=\left[-1\phantom{\rule{0ex}{0ex}}5\phantom{\rule{0ex}{0ex}}0\right],\stackrel{\to }{{v}_{5}}=\left[0\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}1\right]\phantom{\rule{0ex}{0ex}}$

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

$⇒\stackrel{\to }{{v}_{2}}$ are $\stackrel{\to }{{v}_{4}}$ redundant vectors and $\stackrel{\to }{{v}_{1}},\stackrel{\to }{{v}_{3}}\mathrm{and}\stackrel{\to }{{\mathrm{v}}_{5}}$ 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 $\stackrel{\to }{{v}_{1}},\stackrel{\to }{{v}_{3}}\mathrm{and}\stackrel{\to }{{\mathrm{v}}_{5}}$ non-redundant vectors of matrix A, thus the basis of the image A =$\left\{\left[1\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}0\right],\left[0\phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}0\right],\left[0\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}1\right]\right\}$ i.e. .

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

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

$\stackrel{\to }{{v}_{2}}=2\stackrel{\to }{{v}_{1}}\mathrm{and}\stackrel{\to }{{\mathrm{v}}_{4}}=-\stackrel{\to }{{\mathrm{v}}_{1}}+5\stackrel{\to }{{\mathrm{v}}_{3}}\phantom{\rule{0ex}{0ex}}⇒2\stackrel{\to }{{\mathrm{v}}_{1}}+\stackrel{\to }{{\mathrm{v}}_{2}}+0\mathrm{and}\stackrel{\to }{{\mathrm{v}}_{1}}-5\stackrel{\to }{{\mathrm{v}}_{3}}+\stackrel{\to }{{\mathrm{v}}_{4}}=0$

Thus the vectors in kernel of A is $\stackrel{\to }{{w}_{1}}=\left[2\phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}0\right],\stackrel{\to }{{w}_{2}}=\left[1\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}-5\phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}0\right]$.

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

The basis of the image of A = $\left\{\left[1\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}0\right],\left[0\phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}0\right],\left[0\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}1\right]\right\}$

The basis of the kernel of A = $\left\{\left[2\phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}0\right],\left[1\phantom{\rule{0ex}{0ex}}0\phantom{\rule{0ex}{0ex}}-5\phantom{\rule{0ex}{0ex}}1\phantom{\rule{0ex}{0ex}}0\right]\right\}$