Considere a transformação linear L: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} dada por
L\left(\left[\begin{array}{l} x \\\ y \end{array}\right]\right)=\left[\begin{array}{cc} 0 & 1 \\\\ 1 & -1 \\\\ 1 & 0 \end{array}\right]\left[\begin{array}{l} x \\\ y \end{array}\right]
Calcule a matriz [L]_{\mathscr{U}}^{\mathscr{B}}, onde \mathscr{B}=\left\{(1,-1)^{\top},(0,2)^{\top}\right\} e \mathscr{U}=\left\{(1,0,-1)^{\top},(0,1,2)^{\top},(1,2,0)^{\top}\right\}.
Passo 1
O enunciado dá a transformação linear L:\mathbb{R}^2\to\mathbb{R}^3 com matriz padrão
as bases do domínio e do contradomínio são \mathscr{B}=\{b_1,b_2\} com b_1=(1,-1)^\top, b_2=(0,2)^\top e \mathscr{U}=\{u_1,u_2,u_3\} com u_1=(1,0,-1)^\top, u_2=(0,1,2)^\top, u_3=(1,2,0)^\top.
Para achar [L]_{\mathscr{U}}^{\mathscr{B}}, aplicamos L em cada b_j e escrevemos o resultado em coordenadas na base \mathscr{U}.
Expressão de L(b_1)=(-1,2,1)^\top\) na base \(\mathscr{U}\). Queremos \(c_1,c_2,c_3\) tais que
\[c_1u_1+c_2u_2+c_3u_3 =c_1\begin{pmatrix}1\\\ 0\\\ -1\end{pmatrix} +c_2\begin{pmatrix}0\\\ 1\\\ 2\end{pmatrix} +c_3\begin{pmatrix}1\\\ 2\\\ 0\end{pmatrix} =\begin{pmatrix}-1\\\ 2\\\ 1\end{pmatrix}.
Logo a primeira coluna de [L]_{\mathscr{U}}^{\mathscr{B}} é \bigl(-\tfrac{7}{3},-\tfrac{2}{3},\tfrac{4}{3}\bigr)^\top.
Passo 4
Expressão de L(b_2)=(2,-2,0)^\top\) na base \(\mathscr{U}\). Queremos \(d_1,d_2,d_3\) tais que
\[d_1u_1+d_2u_2+d_3u_3 =\begin{pmatrix}2\\\ -2\\\ 0\end{pmatrix},