Seja V o espaço vetorial das matrizes simétricas 2 × 2. Mostre que dim V = 3 e exiba uma base de V

\(x = {-2 \pm \sqrt{2^2-4.54} \over 25}\)

Para encontrarmos a dimensão dessa matriz, primeiro devemos encontrar sua base:

\(\begin{align} & A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} \\ {{a}_{21}} & {{a}_{22}} \\ \end{matrix} \right] \\ & {{e}_{1}}=\left[ \begin{matrix} 1 & 0 \\ 0 & 0 \\ \end{matrix} \right] \\ & {{e}_{2}}=\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \\ \end{matrix} \right] \\ & {{e}_{3}}=\left[ \begin{matrix} 0 & 0 \\ 1 & 0 \\ \end{matrix} \right] \\ & {{e}_{4}}=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & A={{a}_{11}}\cdot {{e}_{1}}+{{a}_{12}}\cdot {{e}_{2}}+{{a}_{21}}\cdot {{e}_{3}}+{{a}_{22}}\cdot {{e}_{4}} \\ & A=\left[ \begin{matrix} {{a}_{11}} & 0 \\ 0 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 0 & {{a}_{12}} \\ 0 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 0 & 0 \\ {{a}_{21}} & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 0 & 0 \\ 0 & {{a}_{22}} \\ \end{matrix} \right] \\ & Base={{a}_{11}}(1,0,0,0)+{{a}_{12}}(0,1,0,0)+{{a}_{21}}(0,0,1,0)+{{a}_{22}}(0,0,0,1) \\ \\ \end{align}\ \)

Portanto, concluimos que \(\boxed{\dim {\text{ V = 3}}}\).