Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

For which constant k is a linear transformation $T (M) = [\begin{matrix} 2 & 3 \\ 0 & 4 \end{matrix}] M - M [\begin{matrix} 3 & 0 \\ 0 & k \end{matrix}]$ is an isomorphism form $R^{2 \times 2}$ to $R^{2 \times 2}$ .

The solution is an isomorphism when ${k \in R - {2, 4}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step1: Definition of Linear Transformation

Consider two linear spaces V and W. A function T is said to be linear transformation if the following holds.

$T (f + g) = T (f) + T (g) T (kf) = kT (f)$

For all elements $f, g$ of V and k is scalar.

A linear transformation $T : V \to W$ is said to be an isomorphism if and only if $\ker (T) = {0}$ and $im (T) = W$ or $\dim (V) = \dim (W)$ .

Step2: Explanation of the solution

The given transformation as follows.

$T (M) = [\begin{matrix} 2 & 3 \\ 0 & 4 \end{matrix}] M - M [\begin{matrix} 3 & 0 \\ 0 & k \end{matrix}]$ , from $R^{2 \times 2}$ to $R^{2 \times 2}$ .

By using the definition of linear transformation as follows.

$T (A + B) = T (A) + T (B) T (k A) = k T (A)$

Now, to check the first condition as follows.

Let A and B be arbitrary matrices from $R^{2 \times 2}$ and as follows.

$T (A + B) = [\begin{matrix} 2 & 3 \\ 0 & 4 \end{matrix}] (A + B) (A - B) [\begin{matrix} 3 & 0 \\ 0 & k \end{matrix}] = [\begin{matrix} 2 & 3 \\ 0 & 4 \end{matrix}] A + [\begin{matrix} 2 & 3 \\ 0 & 4 \end{matrix}] B - A [\begin{matrix} 3 & 0 \\ 0 & k \end{matrix}] - B [\begin{matrix} 3 & 0 \\ 0 & k \end{matrix}] = [\begin{matrix} 2 & 3 \\ 0 & k \end{matrix}] A - A [\begin{matrix} 3 & 0 \\ 0 & k \end{matrix}] + [\begin{matrix} 2 & 3 \\ 0 & 4 \end{matrix}] B - B [\begin{matrix} 3 & 0 \\ 0 & k \end{matrix}] T (A + B) = T (A) + T (B)$

Similarly, to check the second condition as follows.

Let $α$ be an arbitrary scalar, and $A \in R^{2 \times 2}$ as follows.

$T (α A) = [\begin{matrix} 2 & 3 \\ 0 & 4 \end{matrix}] α A - α A [\begin{matrix} 3 & 0 \\ 0 & k \end{matrix}] = α ([\begin{matrix} 2 & 3 \\ 0 & 4 \end{matrix}] A - A [\begin{matrix} 3 & 0 \\ 0 & k \end{matrix}]) T (αA) = αT (A)$

Thus,T is a linear transformation.

Step3: Properties of isomorphism

A linear transformation $T : V \to W$ is isomorphism if and only if $\ker (t) = {0}$ and localid="1659426664071" $Im (t) = W$

Now, check if $\ker (t) = \{0\}$ as follows.

$k e r (T) = \{A \in R^{2 \times 2} | T (A) = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]\}$

Consider a matrix A as follows.

$A = [\begin{matrix} a & b \\ c & d \end{matrix}]$

The next equation as follows.

$T (A) = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 2 & 3 \\ 0 & 4 \end{matrix}] [\begin{matrix} a & b \\ c & d \end{matrix}] - [\begin{matrix} a & b \\ c & d \end{matrix}] [\begin{matrix} 3 & 0 \\ 0 & k \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 2 a + 3 c & 2 b + 3 d \\ 0 a + 4 c & 0 d + 4 d \end{matrix}] - [\begin{matrix} 3 a + 0 b & 0 a + k b \\ 3 c + 0 d & 0 c + k d \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 2 a + 3 c & - 3 a - 0 d \\ 0 a + 4 c & - 3 c - 3 d \end{matrix} \begin{matrix} 2 b + 3 d & - 0 a - k d \\ 0 b + 4 d & - 0 c - k d \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

Simplify further as follows.

$[\begin{matrix} 2 a + 3 c - 3 a - 0 d & 2 b + 3 d - 0 a - k b \\ 0 a + 4 c - 3 c - 0 d & 0 d + 4 d - 0 c - k d \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} a + 3 c & 2 b - k b + 3 d \\ c & 4 d - k d \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} - a + 3 c & (2 - k) b + 3 d \\ c & (4 - k) d \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

Equating the corresponding entries as follows.

$c = 0$ and $a + 3 c = 0$

Solve and find the values as follows.

$a = 0, c = 0$

Substitute the value 0 for and 0 for in $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ as follows.

$\begin{array}{l} A = [\begin{matrix} a & b \\ c & d \end{matrix}] \\ = [\begin{matrix} 0 & b \\ 0 & d \end{matrix}] \end{array}$

For $k = 4$ the solution as follows.

$T ([\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}]) = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] k e r (T) \neq \{[\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]\} k e r (T) \neq \{0\}$

Therefore,T to be an isomorphism for $k \neq 4$ .

Similarly, for T to be an isomorphism for $k \neq 2$ .

Thus, is a linear transformation and is an isomorphism when $k \in R - \{2, 4\}$

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Short Answer

For which constant k is a linear transformation $T (M) = [\begin{matrix} 2 & 3 \\ 0 & 4 \end{matrix}] M - M [\begin{matrix} 3 & 0 \\ 0 & k \end{matrix}]$ is an isomorphism form $R^{2 \times 2}$ to $R^{2 \times 2}$ .

Step by Step Solution

Step1: Definition of Linear Transformation

Step2: Explanation of the solution

Step3: Properties of isomorphism

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Step by Step Solution

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For which constant k is a linear transformation $T (M) = [\begin{matrix} 2 & 3 \\ 0 & 4 \end{matrix}] M - M [\begin{matrix} 3 & 0 \\ 0 & k \end{matrix}]$ is an isomorphism form $R^{2 \times 2}$ to $R^{2 \times 2}$ .