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Found in: Page 191

### 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

# 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 $$A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)$$.Show that $$T$$is a linear transformation.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 $$T\left( A \right) = B$$.Show that the range of $$T$$ is the set of $$B$$ in $${M_{2 \times 2}}$$ with the property that $${B^T} = B$$.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 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 $$\left\{ {\left( {\begin{array}{*{20}{c}}0&b\\{ - b}&0\end{array}} \right):b\,\,{\mathop{\rm real}\nolimits} } \right\}$$.