91Ó°ÊÓ

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.

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

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 \({B^T} = B\).

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