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Q33E

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Linear Algebra and its Applications
Found in: Page 191
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Let \({M_{2 \times 2}}\) be the vector space of all \(2 \times 2\) matrices, and define \(T:{M_{2 \times 2}} \to {M_{2 \times 2}}\) by \(T\left( A \right) = A + {A^T}\), where \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\).

  1. Show that \(T\)is a linear transformation.
  2. Let \(B\) be any element of \({M_{2 \times 2}}\) such that \({B^T} = B\). Find an \(A\) in \({M_{2 \times 2}}\) such that \(T\left( A \right) = B\).
  3. Show that the range of \(T\) is the set of \(B\) in \({M_{2 \times 2}}\) with the property that \({B^T} = B\).
  4. Describe the kernel of \(T\).

  1. It is proved that \(T\) is a linear transformation.
  2. \(T\left( A \right) = B\)
  3. It is proved that the range of \(T\) is the set of \(B\) in \({M_{2 \times 2}}\) with the property \({B^T} = B\).
  4. The kernel of \(T\) is \(\left\{ {\left( {\begin{array}{*{20}{c}}0&b\\{ - b}&0\end{array}} \right):b\,\,{\mathop{\rm real}\nolimits} } \right\}\).
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Show that \(T\) is a linear transformation

a)

Let \(A\) and \(B\) be any matrix in \({M_{2 \times 2}}\). Then,

\(\begin{array}{c}T\left( {A + B} \right) = \left( {A + B} \right) + {\left( {A + B} \right)^T}\\ = A + B + {A^T} + {B^T}\\ = \left( {A + {A^T}} \right) + \left( {B + {B^T}} \right)\\ = T\left( A \right) + T\left( B \right).\end{array}\)

Let \(c\) be any scalar. Then,

\(\begin{array}{c}T\left( {cA} \right) = {\left( {cA} \right)^T}\\ = c{\left( A \right)^T}\\ = cT\left( A \right).\end{array}\)

Therefore, \(T\) is a linear transformation.

Thus, it is proved that \(T\) is a linear transformation.

Step 2: Determine that \(A\) in \({M_{2 \times 2}}\) such that \(T\left( A \right) = B\)

b)

Consider \(B\) is an element of \({M_{2 \times 2}}\) with \({B^T} = B\), and \(A = \frac{1}{2}B\). Then,

\(\begin{array}{c}T\left( A \right) = A + {A^T}\\ = \frac{1}{2}B + {\left( {\frac{1}{2}B} \right)^T}\\ = \frac{1}{2}B + \frac{1}{2}{B^T}\\ = \frac{1}{2}B + \frac{1}{2}B\\ = B.\end{array}\)

Thus, \(T\left( A \right) = B\).

Step 3: Show that the range of \(T\) is the set of \(B\) in \({M_{2 \times 2}}\) with the property \({B^T} = B\)

c)

Part b demonstrates that the range of \(T\) contains the set of all \(B\) in \({M_{2 \times 2}}\) with \({B^T} = B\).

It must be demonstrated that \(B\) in the range of \(T\) has this property.

Suppose \(B\) is in the range of \(T\), then \(B = T\left( A \right)\) for some \(A\) in \({M_{2 \times 2}}\). Therefore,\(B = A + {A^T}\) and

\(\begin{array}{c}{B^T} = {\left( {A + {A^T}} \right)^T}\\ = {A^T} + {\left( {{A^T}} \right)^T}\\ = {A^T} + A\\ = A + {A^T}\\ = B.\end{array}\)

Thus, \(B\) has the property \({B^T} = B\).

Hence, it is proved that the range of \(T\) is the set of \(B\) in \({M_{2 \times 2}}\), with the property \({B^T} = B\).

Step 4: Describe the kernel of T

d)

Consider \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\) is in the kernel of T. Then \(T\left( A \right) = A + {A^T} = 0\), and

\(\begin{array}{c}A + {A^T} = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right) + \left( {\begin{array}{*{20}{c}}a&c\\b&d\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{2a}&{c + b}\\{b + c}&{2d}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right).\end{array}\)

By solving, you get \({\mathop{\rm a}\nolimits} = d = 0\) and \({\mathop{\rm c}\nolimits} = - {\mathop{\rm b}\nolimits} \).

Therefore, the kernel of \(T\) is \(\left\{ {\left( {\begin{array}{*{20}{c}}0&b\\{ - b}&0\end{array}} \right):b\,\,{\mathop{\rm real}\nolimits} } \right\}\).

Most popular questions for Math Textbooks

Let be a linear transformation from a vector space \(V\) \(T:V \to W\)in to vector space \(W\). Prove that the range of T is a subspace of . (Hint: Typical elements of the range have the form \(T\left( {\mathop{\rm x}\nolimits} \right)\) and \(T\left( {\mathop{\rm w}\nolimits} \right)\) for some \({\mathop{\rm x}\nolimits} ,\,{\mathop{\rm w}\nolimits} \)in \(V\).)\(W\)

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