Sejam M_{2 \times 2} o espaço das matrizes quadradas de ordem 2 e
\begin{aligned} & \beta=\left\{\left(\begin{array}{cc} 1 & 0 \\\\ 0 & 0 \end{array}\right),\left(\begin{array}{cc} 0 & 1 \\\\ 0 & 0 \end{array}\right),\left(\begin{array}{cc} 0 & 0 \\\\ 1 & 0 \end{array}\right),\left(\begin{array}{cc} 0 & 0 \\\\ 0 & 1 \end{array}\right)\right\}, \\\ & \beta^{\prime}=\left\{\left(\begin{array}{cc} 1 & 0 \\\\ 0 & 1 \end{array}\right),\left(\begin{array}{cc} -1 & 0 \\\\ 1 & 0 \end{array}\right),\left(\begin{array}{cc} 1 & 1 \\\\ 0 & 0 \end{array}\right),\left(\begin{array}{cc} 0 & 0 \\\\ -1 & 1 \end{array}\right)\right\} \text { bases de } M_{2 \times 2} \text {. Encon- } \\\ & \text { tre }[I]_{\beta^{\prime}}^{\beta} \mathrm{e}[I]_{\beta}^{\beta^{\prime}} \end{aligned}

Passo 1

O enunciado nos dá o espaço M_{2\times2} com duas bases
\beta=\{E_{11},E_{12},E_{21},E_{22}\}, \quad \beta'=\{B_{1},B_{2},B_{3},B_{4}\},
onde
E_{11}=\begin{pmatrix}1&0\\\ 0&0\end{pmatrix},\; E_{12}=\begin{pmatrix}0&1\\\ 0&0\end{pmatrix},\; E_{21}=\begin{pmatrix}0&0\\\ 1&0\end{pmatrix},\; E_{22}=\begin{pmatrix}0&0\\\ 0&1\end{pmatrix},
B_{1}=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix},\; B_{2}=\begin{pmatrix}-1&0\\\ 1&0\end{pmatrix},\; B_{3}=\begin{pmatrix}1&1\\\ 0&0\end{pmatrix},\; B_{4}=\begin{pmatrix}0&0\\\ -1&1\end{pmatrix}.
Queremos a matriz de mudança de coordenadas de \beta' para \beta, que chamamos [I]_{\beta'}^{\beta}, e sua inversa [I]_{\beta}^{\beta'}.

Passo 2

Cada coluna de [I]_{\beta'}^{\beta} é o vetor de coordenadas de B_j na base \beta. Como \beta é a base canônica de matrizes, basta ler os coeficientes de
B_j = a_{1j}E_{11}+a_{2j}E_{12}+a_{3j}E_{21}+a_{4j}E_{22}.
Para cada j: B_{1}=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}\Rightarrow(1,0,0,1), B_{2}=\begin{pmatrix}-1&0\\\ 1&0\end{pmatrix}\Rightarrow(-1,0,1,0), B_{3}=\begin{pmatrix}1&1\\\ 0&0\end{pmatrix}\Rightarrow(1,1,0,0), B_{4}=\begin{pmatrix}0&0\\\ -1&1\end{pmatrix}\Rightarrow(0,0,-1,1).

Passo 3

Assim
[I]_{\beta'}^{\beta} = \begin{pmatrix} 1 & -1 & 1 & 0\\\\ 0 & 0 & 1 & 0\\\\ 0 & 1 & 0 & -1\\\\ 1 & 0 & 0 & 1 \end{pmatrix}.

Passo 4

A matriz [I]_{\beta}^{\beta'} é a inversa de [I]_{\beta'}^{\beta}. Calculando \bigl([I]_{\beta'}^{\beta}\bigr)^{-1} por eliminação de Gauss, obtemos
[I]_{\beta}^{\beta'} = \begin{pmatrix} \tfrac12 & -\tfrac12 & \tfrac12 & \tfrac12\\[6pt] -\tfrac12 & \tfrac12 & \tfrac12 & \tfrac12\\[6pt] 0 & 1 & 0 & 0\\[6pt] -\tfrac12 & \tfrac12 & -\tfrac12 & \tfrac12 \end{pmatrix}.]

Resposta

[I]_{\beta'}^{\beta}=\begin{pmatrix} 1 & -1 & 1 & 0\\\\ 0 & 0 & 1 & 0\\\\ 0 & 1 & 0 & -1\\\\ 1 & 0 & 0 & 1 \end{pmatrix}, [I]_{\beta}^{\beta'}=\begin{pmatrix} \tfrac12 & -\tfrac12 & \tfrac12 & \tfrac12\\\\ -\tfrac12 & \tfrac12 & \tfrac12 & \tfrac12\\\\ 0 & 1 & 0 & 0\\\\ -\tfrac12 & \tfrac12 & -\tfrac12 & \tfrac12 \end{pmatrix}.

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