Ensino Superior ⇒ Álgebra Linear 1 - Mudança de base Tópico resolvido

[kagenizio]

Seja [tex3]E=[/tex3]{[tex3]\vec{i},\vec{j},\vec{k}[/tex3]} uma base ortonormal de [tex3]V^{3}[/tex3]. Sendo [tex3]\vec{u}=\frac{1}{\sqrt{3}}(\vec{i}+\vec{j}-\vec{k}),\vec{v}=\frac{1}{\sqrt{2}}(\vec{j}+\vec{k})[/tex3] e [tex3]\vec{w}=\frac{1}{\sqrt{6}}(2\vec{i}-\vec{j}-\vec{k})[/tex3], prove que [tex3]F=[/tex3]{[tex3]\vec{u},\vec{v},\vec{w}[/tex3]} é uma base ortonormal de [tex3]V^{3}[/tex3] e calcule as coordenadas do vetos [tex3]\vec{t}=3\vec{i}-2\vec{j}-\vec{k}[/tex3] em relação à base [tex3]F[/tex3].

Resposta

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[tex3]\vec{a}=\left(\frac{2}{\sqrt{3}},-\frac{3}{\sqrt{2}},\frac{7}{\sqrt{6}}\right)_{F}[/tex3]

[joaopcarvMOD]

Esta questão é da segunda lista de Álgebra Linear 1 da Politécnica USP, mas no caso tem-se [tex3]\mathsf{\vec{w} \ = \ \dfrac{1}{\sqrt{6}} \cdot \bigg(2\cdot \hat{i} \ - \ \hat{j} \ \color{\red}+} \mathsf{\ \hat{k}\bigg).}[/tex3]

Supondo [tex3]\mathsf{E}[/tex3] a base canônica [tex3]\mathsf{\hat{i} \ = \ (1,0,0), \hat{j} \ = \ (0,1,0), \hat{k} \ = \ (0,0,1)}[/tex3], as coordenadas dos vetores já estão dadas em relação a essa base [tex3]\mathsf{\hat{i}, \hat{j}, \ \hat{k}.}[/tex3]

Para provar que [tex3]\mathsf{F \ \in \ \mathbb{V}^3 \ = \ \{\vec{u}, \ \vec{v}, \vec{w} \}}[/tex3] é uma base ortonormal, é necessário que:

[tex3]\mathsf{\hookrightarrow \ ||\vec{u}|| \ = \ ||\vec{v}|| \ = \ ||\vec{w}|| \ = 1;}[/tex3]

[tex3]\mathsf{||\vec{u}|| = \ \sqrt{\bigg(\dfrac{1}{\sqrt{3}}\bigg)^2 \ + \ \bigg(\dfrac{1}{\sqrt{3}}\bigg)^2 \ + \ \bigg(- \dfrac{1}{\sqrt{3}}\bigg)^2} \ = \ \boxed{\mathsf{1}}}[/tex3]

[tex3]\mathsf{||\vec{v}|| = \ \sqrt{0 \ + \ \bigg(\dfrac{1}{\sqrt{2}}\bigg)^2 \ + \ \bigg(\dfrac{1}{\sqrt{2}}\bigg)^2} \ = \ \boxed{\mathsf{1}}}[/tex3]

[tex3]\mathsf{||\vec{w}|| = \ \sqrt{\bigg(\dfrac{2}{\sqrt{6}}\bigg)^2 \ + \ \bigg(-\dfrac{1}{\sqrt{6}}\bigg)^2 \ + \ \bigg(\dfrac{1}{\sqrt{6}}\bigg)^2} \ = \ \boxed{\mathsf{1}}}[/tex3]

[tex3]\mathsf{\hookrightarrow \ \vec{u} \ \perp \ \vec{v} \ \perp \ \vec{v} \ \Leftrightarrow \ \vec{u} \bullet \vec{v} \ = \ \vec{v} \bullet \vec{w} \ = \ \vec{u} \bullet \vec{w} \ = \ 0.}[/tex3]

[tex3]\mathsf{ \vec{u} \bullet \vec{v} \ = \ \dfrac{1}{\sqrt{3}}\cdot 0 \ + \ \dfrac{1}{\sqrt{3}} \cdot \dfrac{1}{\sqrt{2}} \ - \dfrac{1}{\sqrt{3}} \cdot \dfrac{1}{\sqrt{2}} \ = \ \boxed{\mathsf{0}}}[/tex3]

[tex3]\mathsf{ \vec{v} \bullet \vec{w} \ = \ 0 \cdot \dfrac{2}{\sqrt{6}} \ - \ \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{\sqrt{6}} \ + \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{\sqrt{6}} \ = \ \boxed{\mathsf{0}}}[/tex3]

[tex3]\mathsf{ \vec{u} \bullet \vec{w} \ = \ \dfrac{1}{\sqrt{3}} \cdot \dfrac{2}{\sqrt{6}} \ - \ \dfrac{1}{\sqrt{3}} \cdot \dfrac{1}{\sqrt{6}} \ - \dfrac{1}{\sqrt{3}} \cdot \dfrac{1}{\sqrt{6}} \ = \ \boxed{\mathsf{0}}}[/tex3]

Então, provamos que [tex3]\mathsf{F}[/tex3] é uma base ortonormal de [tex3]\mathbb{V}^{\mathsf{3}}[/tex3], sendo os três vetores geradores LI.

Agora, temos achar a matriz de mudança de base de [tex3]\mathsf{E}[/tex3] para [tex3]\mathsf{F}[/tex3]. Para isso, exprimindo a matriz dos vetores geradores de [tex3]\mathsf{F}[/tex3] em relação à base [tex3]\mathsf{E}[/tex3], temos:

[tex3]\mathsf{M_{FE} \ = \ \begin{pmatrix}
\mathsf{\frac{1}{\sqrt{3}}} & 0 & \mathsf{\frac{2}{\sqrt{6}}} \\
\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{6}}} \\
-\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{6}}} \\
\end{pmatrix}}[/tex3]

A inversa dessa matriz é [tex3]\mathsf{M_{EF}}[/tex3], ou seja, converte vetores construídos na base [tex3]\mathsf{E}[/tex3] para a base [tex3]\mathsf{F}[/tex3]. Para descobrirmos essa inversa, podemos usar escalonamento da matriz aumentada, mas usarei aqui a relação:

[tex3]\mathsf{A^{-1} \ = \ \dfrac{1}{det \ A} \ \cdot \ A_C^t}[/tex3], em que [tex3]\mathsf{A_C^t}[/tex3] é a transposta da matriz dos cofatores de [tex3]\mathsf{A}[/tex3].

No caso, teremos: [tex3]\mathsf{M_{EF} \ = \ \dfrac{1}{det \ M_{FE}} \ \cdot \ M_{FE_{C}}^t}[/tex3]

Calculando primeiramente [tex3]\mathsf{det \ M_{FE} :}[/tex3]

[tex3]\mathsf{det \ M_{FE} \ = \ (-1)^{(1+1)} \cdot \underbrace{\begin{vmatrix} \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{- \frac{1}{\sqrt{6}}} \\ \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{6}}} \end{vmatrix}}_{menor \ associado \ de \ a_{1,1}} \cdot \dfrac{1}{\sqrt{3}} \ + \ (-1)^{(1+2)} \cdot \underbrace{\begin{vmatrix} \mathsf{\frac{1}{\sqrt{3}}} & \mathsf{- \frac{1}{\sqrt{6}}} \\ \mathsf{- \frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{6}}} \end{vmatrix}}_{menor \ associado \ de \ a_{1,2}} \cdot \ 0 \ + \ (-1)^{(1+3)} \cdot \underbrace{\begin{vmatrix} \mathsf{\frac{1}{\sqrt{3}}} & \mathsf{ \frac{1}{\sqrt{2}}} \\ \mathsf{-\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{2}}} \end{vmatrix}}_{menor \ associado \ de \ a_{1,3}} \cdot \dfrac{2}{\sqrt{6}}}[/tex3]

[tex3]\boxed{\mathsf{det \ M_{FE} \ = \ 1}}[/tex3]

Para calcular um cofator de um elemento [tex3]\mathsf{a_{i,j}}[/tex3], temos: [tex3]\mathsf{C_{a_{i,j}} \ = \ -1^{(i \ + \ j)} \ \cdot \ det \ M_{i,j}}[/tex3], onde a matriz [tex3]\mathsf{M_{i,j}}[/tex3] é o "menor associado", ou seja, é a matriz formada quando eliminamos a linha e a coluna de [tex3]\mathsf{a_{i,j}.}[/tex3]

Ou seja, teremos a matriz de cofatores [tex3]\mathsf{M_{FE_{C}}}[/tex3]:

[tex3]\mathsf{M_{FE_{C}} \ = \ \begin{pmatrix}
\mathsf{\frac{1}{\sqrt{3}}} & 0 & \mathsf{\frac{2}{\sqrt{6}}} \\
\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{-\frac{1}{\sqrt{6}}} \\
-\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{6}}} \\
\end{pmatrix}}[/tex3]

Transpondo essa matriz: [tex3]\mathsf{M_{FE_{C}^t} \ = \ \begin{pmatrix}
\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{3}}} & \mathsf{-\frac{1}{\sqrt{3}}} \\
0 & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{2}}} \\
\mathsf{\frac{2}{\sqrt{6}}} & \mathsf{-\frac{1}{\sqrt{6}}} & \mathsf{\frac{1}{\sqrt{6}}} \\
\end{pmatrix}}[/tex3]

Por fim: [tex3]\mathsf{M_{EF} \ = \ \dfrac{1}{det \ M_{FE}} \ \cdot \ M_{FE_{C}}^t}[/tex3]

[tex3]\mathsf{M_{EF} \ = \ \dfrac{1}{1} \ \cdot \ \begin{pmatrix}
\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{3}}} & \mathsf{-\frac{1}{\sqrt{3}}} \\
0 & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{2}}} \\
\mathsf{\frac{2}{\sqrt{6}}} & \mathsf{-\frac{1}{\sqrt{6}}} & \mathsf{\frac{1}{\sqrt{6}}} \\
\end{pmatrix}}[/tex3]

[tex3]\boxed{\mathsf{M_{EF}} \ = \ \begin{pmatrix}
\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{3}}} & \mathsf{-\frac{1}{\sqrt{3}}} \\
0 & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{2}}} \\
\mathsf{\frac{2}{\sqrt{6}}} & \mathsf{-\frac{1}{\sqrt{6}}} & \mathsf{\frac{1}{\sqrt{6}}} \\
\end{pmatrix}}[/tex3]

Então, para o vetor [tex3]\mathsf{\vec{t} \ = \ \bigg(3\cdot \hat{i} \ - \ 2\cdot \hat{j} \ - \ \hat{k} \bigg)_E}[/tex3], suas coordenadas [tex3]\mathsf{(x, \ y, z)_F}[/tex3] são dadas pela operação com seus vetores coluna nas bases [tex3]\mathsf{E, \ F}[/tex3]:

[tex3]\begin{pmatrix}
\mathsf{x} \\
\mathsf{y} \\
\mathsf{z} \\
\end{pmatrix}_\mathsf{F} \ = \ \begin{pmatrix}
\mathsf{\frac{1}{\sqrt{3}}} & \mathsf{\frac{1}{\sqrt{3}}} & \mathsf{-\frac{1}{\sqrt{3}}} \\
0 & \mathsf{\frac{1}{\sqrt{2}}} & \mathsf{\frac{1}{\sqrt{2}}} \\
\mathsf{\frac{2}{\sqrt{6}}} & \mathsf{-\frac{1}{\sqrt{6}}} & \mathsf{\frac{1}{\sqrt{6}}} \\
\end{pmatrix} \ \mathsf{\cdot} \ \begin{pmatrix}
\mathsf{3} \\
\mathsf{-2} \\
\mathsf{-1} \\
\end{pmatrix}_{\mathsf{E}}[/tex3]

[tex3]\mathsf{x \ = \ 3 \cdot \dfrac{1}{\sqrt{3}} \ - \ 2 \cdot \dfrac{1}{\sqrt{3}} \ + \ 1 \cdot \dfrac{1}{\sqrt{3}} \ = \ \boxed{\mathsf{\dfrac{2}{\sqrt{3}}}}}[/tex3]
[tex3]\mathsf{y \ = \ 3 \cdot 0 \ - \ 2 \cdot \dfrac{1}{\sqrt{2}} \ - \ 1 \cdot \dfrac{1}{\sqrt{2}} \ = \ \boxed{\mathsf{-\dfrac{3}{\sqrt{2}}}}}[/tex3]
[tex3]\mathsf{z \ = \ 3 \cdot \dfrac{2}{\sqrt{6}} \ + \ 2 \cdot \dfrac{1}{\sqrt{6}} \ - \ 1 \cdot \dfrac{1}{\sqrt{6}} \ = \ \boxed{\mathsf{\dfrac{7}{\sqrt{6}}}}}[/tex3]

Logo, [tex3]\boxed{\boxed{\mathsf{\vec{t} \ = \ \bigg(\dfrac{2}{\sqrt{3}}, -\dfrac{3}{\sqrt{2}}, \dfrac{7}{\sqrt{6}}\bigg)_F}}}[/tex3]