(2 pontos) Seja W o subespaço gerado pelos vetores \left\{\boldsymbol{u}_{\mathbf{1}}, \boldsymbol{u}_{\mathbf{2}}\right\} abaixo. Escreva \boldsymbol{y} como uma soma de um vetor em W(\hat{\boldsymbol{y}}) e outro ortogonal à W(\boldsymbol{z}).
\boldsymbol{y}=\left[\begin{array}{r} -5 \\\\ 5 \\\\ 1 \\\\ -5 \end{array}\right] ; \quad \boldsymbol{u}_{\mathbf{1}}=\left[\begin{array}{l} -3 \\\\ -2 \\\\ -3 \\\\ -2 \end{array}\right] ; \quad \boldsymbol{u}_{\mathbf{2}}=\left[\begin{array}{r} -2 \\\\ 3 \\\\ -2 \\\\ 3 \end{array}\right]

Resposta

Temos:
  • os vetores \boldsymbol{u}_{\mathbf{1}} e \boldsymbol{u}_{\mathbf{2}} são ortogonais;
  • \hat{y}=\frac{y \cdot u 1}{u 1 \cdot u 1} u 1+\frac{y \cdot u 2}{u 2 \cdot u 2} u 2
  • \hat{\boldsymbol{y}}=\frac{12}{26}\left[\begin{array}{l}-3 \\\\ -2 \\\\ -3 \\\\ -2\end{array}\right]+\frac{8}{26}\left[\begin{array}{l}-3 \\\\ -2 \\\\ -3 \\\\ -2\end{array}\right]=\left[\begin{array}{r}-2 \\\\ 0 \\\\ -2 \\\\ 0\end{array}\right]
  • \boldsymbol{z}=\boldsymbol{y}-\hat{\boldsymbol{y}}=\left[\begin{array}{r}-5 \\\\ 5 \\\\ 1 \\\\ -5\end{array}\right]-\left[\begin{array}{r}-2 \\\\ 0 \\\\ -2 \\\\ 0\end{array}\right]=\left[\begin{array}{r}-3 \\\\ 5 \\\\ 3 \\\\ -5\end{array}\right]

Passo 1

O enunciado pede para decompor o vetor
\displaystyle \boldsymbol{y}=\begin{pmatrix}-5\\\ 5\\\ 1\\\ -5\end{pmatrix}
na soma de uma componente em
\displaystyle W=\operatorname{span}\{\boldsymbol{u}_1,\boldsymbol{u}_2\}
e outra componente ortogonal a W.
O conteúdo é projeção ortogonal em \mathbb{R}^4 usando produto interno.

Passo 2

Primeiro verificamos que
\boldsymbol{u}_{1}=\begin{pmatrix}-3\\\ -2\\\ -3\\\ -2\end{pmatrix} \quad\text{e}\quad \boldsymbol{u}_{2}=\begin{pmatrix}-2\\\ 3\\\ -2\\\ 3\end{pmatrix}
são ortogonais, pois
\boldsymbol{u}_1\cdot\boldsymbol{u}_2 =(-3)(-2)+(-2)(3)+(-3)(-2)+(-2)(3)=6-6+6-6=0.
Como são ortogonais, a projeção de \boldsymbol{y} em W é
\hat{\boldsymbol{y}} =\frac{\boldsymbol{y}\cdot\boldsymbol{u}_1}{\boldsymbol{u}_1\cdot\boldsymbol{u}_1}\,\boldsymbol{u}_1 +\frac{\boldsymbol{y}\cdot\boldsymbol{u}_2}{\boldsymbol{u}_2\cdot\boldsymbol{u}_2}\,\boldsymbol{u}_2.
Calculemos cada produto:
\boldsymbol{y}\cdot\boldsymbol{u}_1 =(-5)(-3)+5(-2)+1(-3)+(-5)(-2)=15-10-3+10=12,
\boldsymbol{u}_1\cdot\boldsymbol{u}_1 =9+4+9+4=26,
\boldsymbol{y}\cdot\boldsymbol{u}_2 =(-5)(-2)+5(3)+1(-2)+(-5)(3)=10+15-2-15=8,
\boldsymbol{u}_2\cdot\boldsymbol{u}_2 =4+9+4+9=26.
Logo os coeficientes são \tfrac{12}{26}=\tfrac{6}{13} e \tfrac{8}{26}=\tfrac{4}{13}.

Passo 3

Agora calculamos explicitamente
\hat{\boldsymbol{y}} =\frac{6}{13}\begin{pmatrix}-3\\\ -2\\\ -3\\\ -2\end{pmatrix} +\frac{4}{13}\begin{pmatrix}-2\\\ 3\\\ -2\\\ 3\end{pmatrix}.
Separando termos:
\frac{6}{13}\begin{pmatrix}-3\\\ -2\\\ -3\\\ -2\end{pmatrix} =\begin{pmatrix}-18/13\\\ -12/13\\\ -18/13\\\ -12/13\end{pmatrix}, \quad \frac{4}{13}\begin{pmatrix}-2\\\ 3\\\ -2\\\ 3\end{pmatrix} =\begin{pmatrix}-8/13\\\ 12/13\\\ -8/13\\\ 12/13\end{pmatrix}.
Somando:
\hat{\boldsymbol{y}} =\begin{pmatrix}-18/13-8/13\\\ -12/13+12/13\\\ -18/13-8/13\\\ -12/13+12/13\end{pmatrix} =\begin{pmatrix}-2\\\ 0\\\ -2\\\ 0\end{pmatrix}.

Passo 4

A parte ortogonal a W é
\boldsymbol{z}=\boldsymbol{y}-\hat{\boldsymbol{y}} =\begin{pmatrix}-5\\\ 5\\\ 1\\\ -5\end{pmatrix} -\begin{pmatrix}-2\\\ 0\\\ -2\\\ 0\end{pmatrix} =\begin{pmatrix}-3\\\ 5\\\ 3\\\ -5\end{pmatrix}.
Para conferir, note que
\boldsymbol{z}\cdot\boldsymbol{u}_1=0 e
\boldsymbol{z}\cdot\boldsymbol{u}_2=0,
logo \boldsymbol{z} é ortogonal a W.

Resposta

\hat{\boldsymbol{y}}=\begin{pmatrix}-2\\\ 0\\\ -2\\\ 0\end{pmatrix}, \quad \boldsymbol{z}=\begin{pmatrix}-3\\\ 5\\\ 3\\\ -5\end{pmatrix}.

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