Question

Find the matrix of the linear transformation $\mathit{L}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{=}{\mathit{A}}^{\mathbf{T}}$ from ${\mathbf{ℝ}}^{\mathbf{2}\mathbf{\times }\mathbf{2}}\mathbf{}\mathbf{to}\mathbf{}{\mathbf{ℝ}}^{\mathbf{2}\mathbf{\times }\mathbf{2}}$towith respect to the basis$\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\mathbf{,}\mathbf{}\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\mathbf{,}\mathbf{}\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\mathbf{,}\mathbf{}\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$

Short answer

The matrix B is $\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& -1\end{array}\right]$.

Step-by-step solution

Determine the matrix B with respect to the basis.

Consider the linear transformation $L\left(A\right)=A^T$with respect to the basis .

role="math" localid="1660126784675" $\left\{\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right],\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\right\}$

Substitute the value $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$for A in the equation $L\left(A\right)=A^T$as follows.

role="math" localid="1660126694162" $$\begin{gathered} L\left(A\right)=A^T \\ L\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)={\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]}^T \\ =\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right] \\ L\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=1.A_1+0.A_2+0.A_3+0.A_4 \end{gathered}$$

Substitute the value $\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$for A in the equation $L\left(A\right)=A^T$as follows.

role="math" localid="1660126618413" $$\begin{gathered} L\left(A\right)=A^T \\ L\left(\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right)={\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]}^T \\ =\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right] \\ L\left(\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right)=1.A_1+0.A_2+0.A_3+0.A_4 \end{gathered}$$

Substitute the value $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$for in the equation $L\left(A\right)=A^T$as follows.

$$\begin{gathered} L\left(A\right)=A^T \\ L\left(\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\right)={\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]}^T \\ =\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right] \\ L\left(\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\right)=1.A_1+0.A_2+1.A_3+0.A_4 \end{gathered}$$

Substitute the value for in the equation as follows.

$$\begin{gathered} L\left(A\right)=A^T \\ L\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)={\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]}^T \\ =\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right] \\ L\left(\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\right)=1.A_1+0.A_2+0.A_3-0.A_4 \end{gathered}$$

Therefore, the matrix corresponding to the basis $\left\{\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right],\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\right\}$is $\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& -1\end{array}\right]$.

Hence, the matrix corresponding to the basis is $\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& -1\end{array}\right]$.