Question

Find the matrix of the linear transformation$\mathit{L}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{=}\mathit{A}\mathbf{-}{\mathit{A}}^{\mathbf{T}}$ 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,role="math" localid="1660127697622" $\mathbf{\left[}\begin{array}{cc}\mathbf{1}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}\end{array}\mathbf{\right]}\mathbf{,}\mathbf{}\mathbf{\left[}\begin{array}{cc}\mathbf{0}& \mathbf{0}\\ \mathbf{0}& \mathbf{1}\end{array}\mathbf{\right]}\mathbf{,}\mathbf{\left[}\begin{array}{cc}\mathbf{0}& \mathbf{1}\\ \mathbf{1}& \mathbf{0}\end{array}\mathbf{\right]}\mathbf{,}\mathbf{}\mathbf{\left[}\begin{array}{cc}\mathbf{0}& \mathbf{1}\\ \mathbf{-}\mathbf{1}& \mathbf{0}\end{array}\mathbf{\right]}$

Short answer

The matrix is $\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 2\end{array}\right]$.

Step-by-step solution

Determine the matrix  with respect to the basis.

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

role="math" localid="1660128818135" $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right],\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$

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

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

Further, simplify the equation as follows.

role="math" localid="1660128433867" $$\begin{gathered} L\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ L\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=0.A_1+0.A_2+0.A_3+0.A_4 \end{gathered}$$

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

role="math" localid="1660128539548" $$\begin{gathered} L\left(A\right)=A-A^T \\ L\left(\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]-{\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]}^T \\ =\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right] \\ L\left(\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

role="math" localid="1660128740922" $$\begin{gathered} L\left(\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ L\left(\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right)=0.A_1+0.A_2+0.A_3+0.A_4 \end{gathered}$$

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

role="math" localid="1660128684525" $$\begin{gathered} L\left(A\right)=A-A^T \\ L\left(\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\right)=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]-{\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]}^T \\ =\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]-\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right] \\ L\left(\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

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

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

$$\begin{gathered} L\left(A\right)=A-A^T \\ L\left(\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\right)=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]-{\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]}^T \\ =\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]-\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right] \\ L\left(\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\right)=\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} L\left(\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\right)=\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right] \\ L\left(\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\right)=0.A_1+0.A_2+0.A_3+2.A_4 \end{gathered}$$

Therefore, the matrix B 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 $\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 2\end{array}\right]$.

Hence, the matrix corresponding to the basis is $\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 2\end{array}\right]$.