Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

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

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

$A =^{[\begin{matrix} 1 & - 2 & - 4 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Finding the basis

$x_{1} - x_{2} + 2 x_{3} + 4 x_{4} = 0$

Let V be a subspace $ℝ^{4}$ such that

localid="1659931383607" $V = \{(x_{1}, x_{2}, x_{3}, x_{4}) \in ℝ^{4}; x_{1} - x_{2} + 2 x_{3} + 4 x_{4} = 0\} \Rightarrow V = \{(x_{1} - 2 x_{3} - 4 x_{4}, x_{2}, x_{3}, x_{4}) \in ℝ^{4}; x_{1} = x_{2} - 2 x_{3} - 4 x_{4}\} \Rightarrow V = \{x_{2} [\begin{matrix} 1 \\ 1 \\ 0 \\ 0 \end{matrix}] + x_{3} [\begin{matrix} - 2 \\ 0 \\ 1 \\ 0 \end{matrix}] + x_{4} [\begin{matrix} - 4 \\ 0 \\ 0 \\ 1 \end{matrix}]\}$ $\Rightarrow B a s i s o f V = \{[\begin{matrix} 1 \\ 1 \\ 0 \\ 0 \end{matrix}], [\begin{matrix} - 2 \\ 0 \\ 1 \\ 0 \end{matrix}], [\begin{matrix} - 4 \\ 0 \\ 0 \\ 1 \end{matrix}]\}$ .

Step 2: Finding the matrix of the linear transformation

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

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

$l m (T) = V and K e r (T) = \{\}$

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

$A =^{[\begin{matrix} \begin{matrix} 1 \\ 1 \\ 0 \\ 0 \end{matrix} & \begin{matrix} - 2 \\ 0 \\ 1 \\ 0 \end{matrix} & \begin{matrix} - 4 \\ 0 \\ 0 \\ 1 \end{matrix} \end{matrix}]}$

Step 3: Final Answer

The basis of the subspace V of $ℝ^{4}$ defined by the equation $x_{1} - x_{2} + 2 x_{3} + 4 x_{4} = 0$ is localid="1659870465922" $\{[\begin{matrix} \begin{matrix} 1 \\ 1 \\ 0 \\ 0 \end{matrix} \end{matrix}], [\begin{matrix} - 2 \\ 0 \\ 1 \\ 0 \end{matrix}], [\begin{matrix} - 4 \\ 0 \\ 0 \\ 1 \end{matrix}]\}$ and the matrix of the linear transformation T from $ℝ^{3}$ to $ℝ^{4}$ is

localid="1659870529535" $A =^{[\begin{matrix} \begin{matrix} 1 \\ 1 \\ 0 \\ 0 \end{matrix} & \begin{matrix} - 2 \\ 0 \\ 1 \\ 0 \end{matrix} & \begin{matrix} - 4 \\ 0 \\ 0 \\ 1 \end{matrix} \end{matrix}]}$ .

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Linear Algebra With Applications

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Short Answer

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

Step by Step Solution

Step 1: Finding the basis

Step 2: Finding the matrix of the linear transformation

Step 3: Final Answer

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Linear Algebra With Applications

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Short Answer

Let V be the subspace of ℝ4 defined by the equation x1-x2+2x3+4x4=0Find a linear transformation T from ℝ4to ℝ4 such that ker(T)={0→} and im(T) = V. Describe T by its matrix A.

Step by Step Solution

Step 1: Finding the basis

Step 2: Finding the matrix of the linear transformation

Step 3: Final Answer

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