6.4 Matriz de mudança de coordenadas

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Se conhecemos as coordenadas do vetor $\overrightarrow{x}$ em uma base $\mathcal{ℬ}=\left\{\right.{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,\dots ,{\overrightarrow{v}}_n\left\}\right.$:

$$\left[\right.\overrightarrow{x}{\left]\right.}_{\mathcal{ℬ}}={\left[\begin{array}{c}\hfill b_1\hfill \\ \hfill b_2\hfill \\ \hfill ⋮\hfill \\ \hfill b_n\hfill \\ \hfill \hfill \end{array}\right]}_{\mathcal{ℬ}}=b_1{\overrightarrow{v}}_1+b_2{\overrightarrow{v}}_2+\cdots +b_n{\overrightarrow{v}}_n,$$

(6.38)

podemos obter as coordenadas usuais (na base canônica) de $\overrightarrow{x}$ de forma sistemática. Em outras palavras, queremos encontrar $x_1,x_2,\dots ,x_n$ tais que

$$\overrightarrow{x}=\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \\ \hfill ⋮\hfill \\ \hfill x_n\hfill \\ \hfill \hfill \end{array}\right]=x_1{\overrightarrow{e}}_1+x_2{\overrightarrow{e}}_2+\cdots +x_n{\overrightarrow{e}}_n.$$

(6.39)

Uma maneira é considerar a matriz cujas colunas são os vetores ${\overrightarrow{v}}_i$:

$$A=\left[\begin{array}{cccc}\hfill |\hfill & \hfill |\hfill & \hfill \hfill & \hfill |\hfill \\ \hfill {\overrightarrow{v}}_1\hfill & \hfill {\overrightarrow{v}}_2\hfill & \hfill \cdots \hfill & \hfill {\overrightarrow{v}}_n\hfill \\ \hfill |\hfill & \hfill |\hfill & \hfill \hfill & \hfill |\hfill \\ \hfill \hfill \end{array}\right]=\left[\begin{array}{cccc}\hfill v_{11}\hfill & \hfill v_{12}\hfill & \hfill \cdots \hfill & \hfill v_{1n}\hfill \\ \hfill v_{21}\hfill & \hfill v_{22}\hfill & \hfill \cdots \hfill & \hfill v_{2n}\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill \hfill & \hfill ⋮\hfill \\ \hfill v_{n1}\hfill & \hfill v_{n2}\hfill & \hfill \cdots \hfill & \hfill v_{nn}\hfill \\ \hfill \hfill \end{array}\right].$$

(6.40)

Funciona porque $$\begin{array}{llll}\hfill b_1{\overrightarrow{v}}_1+b_2{\overrightarrow{v}}_2+\cdots +b_n{\overrightarrow{v}}_n& =\sum _{j=1}^n{b}_j{\overrightarrow{v}}_j=\sum _{j=1}^n{b}_j\sum _{i=1}^n{v}_{ij}{\overrightarrow{e}}_i=\sum _{i=1}^n\left(\right.\sum _{j=1}^n{b}_j{v}_{ij}\left)\right.{\overrightarrow{e}}_i& \hfill & \\ \hfill & =\left(\right.\sum _{j=1}^n{b}_j{v}_{1j}\left)\right.{\overrightarrow{e}}_1+\left(\right.\sum _{j=1}^n{b}_j{v}_{2j}\left)\right.{\overrightarrow{e}}_2+\cdots +\left(\right.\sum _{j=1}^n{b}_j{v}_{nj}\left)\right.{\overrightarrow{e}}_n.& \hfill & \end{array}$$

Assim, devemos ter

$$\left\{\begin{array}{ccc}{x}_1\hfill & \hfill =\hfill & v_{11}{b}_1+v_{12}{b}_2+\cdots +v_{1n}{b}_n\hfill \\ x_2\hfill & \hfill =\hfill & v_{21}{b}_1+v_{22}{b}_2+\cdots +v_{2n}{b}_n\hfill \\ \hfill & \hfill ⋮\hfill & \hfill \\ x_n\hfill & \hfill =\hfill & v_{n1}{b}_1+v_{n2}{b}_2+\cdots +v_{nn}{b}_n\hfill \\ \hfill \end{array}\right.\text{i.e}\overrightarrow{x}=A{\left[\overrightarrow{x}\right]}_{\mathcal{ℬ}}.$$

(6.41)

A matriz $A$ construida como acima é chamada de matriz de mudança de coordenadas da base $\mathcal{ℬ}$ para a base usual (canônica) $\left\{{\overrightarrow{e}}_1,{\overrightarrow{e}}_2,\dots ,{\overrightarrow{e}}_n\right\}$.

A matriz de mudança de coordenadas da base canônica para a base $\mathcal{ℬ}$ pode ser obtida a partir da inversa de $A$:

$${\left[\overrightarrow{x}\right]}_{\mathcal{ℬ}}=A^{-1}\cdot \overrightarrow{x}.$$

(6.42)

(Notamos que a matriz $A$ é necessariamente invertível, já que suas colunas são elementos de uma base; em particular, são linearmente independentes).

Exercícios resolvidos

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Exercícios

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