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Found in: Page 144

### Linear Algebra With Applications

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

# Let V be the subspace of ${{\mathbf{ℝ}}}^{{\mathbf{4}}}$ defined by the equation ${{\mathbf{x}}}_{{\mathbf{1}}}{\mathbf{-}}{{\mathbf{x}}}_{{\mathbf{2}}}{\mathbf{+}}{\mathbf{2}}{{\mathbf{x}}}_{{\mathbf{3}}}{\mathbf{+}}{\mathbf{4}}{{\mathbf{x}}}_{{\mathbf{4}}}{\mathbf{=}}{\mathbf{0}}$Find a linear transformation T from ${{\mathbf{ℝ}}}^{{\mathbf{4}}}$to ${{\mathbf{ℝ}}}^{{\mathbf{4}}}$ such that ${\mathbf{ker}}{\mathbf{\left(}}{\mathbf{T}}{\mathbf{\right)}}{\mathbf{=}}\left\{\stackrel{\to }{0}\right\}$ and im(T) = V. Describe T by its matrix A.

The basis of the subspace V of ${\mathrm{ℝ}}^{4}$ defined by the equation${x}_{1}-{x}_{2}+2{x}_{3}+4{x}_{4}=0$ is $\left\{\left[\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-2\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right]\right\}$ and the matrix of the linear transformation T from to is

$A{=}^{\left[\begin{array}{ccc}1& -2& -4\\ 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$.

See the step by step solution

## Step 1: Finding the basis

${x}_{1}-{x}_{2}+2{x}_{3}+4{x}_{4}=0$

Let V be a subspace ${\mathrm{ℝ}}^{4}$such that

localid="1659931383607" $V=\left\{\left({x}_{1},{x}_{2},{x}_{3},{x}_{4}\right)\in {\mathrm{ℝ}}^{4};{x}_{1}-{x}_{2}+2{x}_{3}+4{x}_{4}=0\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒V=\left\{\left({x}_{1}-2{x}_{3}-4{x}_{4},{x}_{2},{x}_{3},{x}_{4}\right)\in {\mathrm{ℝ}}^{4};{x}_{1}={x}_{2}-2{x}_{3}-4{x}_{4}\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒V=\left\{{x}_{2}\left[\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\right]+{x}_{3}\left[\begin{array}{c}-2\\ 0\\ 1\\ 0\end{array}\right]+{x}_{4}\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right]\right\}$ $⇒BasisofV=\left\{\left[\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-2\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right]\right\}$.

## Step 2: Finding the matrix of the linear transformation

Since the basis of the subspace V= $\left\{\left[\begin{array}{c}\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\end{array}\right],\left[\begin{array}{c}-2\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right]\right\}$, thus the dimension of the subspace V is three.

Also, the vectors in the basis of V are linearly independent so, we have

$lm\left(T\right)=V\mathrm{and}Ker\left(T\right)=\left\{\right\}$

Thus if T is a linear transformation from${\mathrm{ℝ}}^{4}$to ${\mathrm{ℝ}}^{4}$such that $T\left(\stackrel{\to }{x}\right)=A\stackrel{\to }{x}$ then

$A{=}^{\left[\begin{array}{ccc}\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}& \begin{array}{c}-2\\ 0\\ 1\\ 0\end{array}& \begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\end{array}\right]}$

The basis of the subspace V of ${\mathrm{ℝ}}^{4}$ defined by the equation${x}_{1}-{x}_{2}+2{x}_{3}+4{x}_{4}=0$is localid="1659870465922" $\left\{\left[\begin{array}{c}\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\end{array}\right],\left[\begin{array}{c}-2\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right]\right\}$ and the matrix of the linear transformation T from ${\mathrm{ℝ}}^{3}$to ${\mathrm{ℝ}}^{4}$is
localid="1659870529535" $A{=}^{\left[\begin{array}{ccc}\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}& \begin{array}{c}-2\\ 0\\ 1\\ 0\end{array}& \begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\end{array}\right]}$.