Sejam \alpha=\{(1,0),(0,1)\}, \beta=\{(-1,1),(1,1)\} e \gamma=\{(3,-1),(1,3)\} bases ordenadas de \mathbb{R}^{2}.
(a) Ache as seguintes matrizes de mudança de base: [I]_{\alpha}^{\beta},[I]_{\beta}^{\alpha}, \mathrm{e}[I]_{\gamma}^{\beta}.
(b) Seja u=(2,3). Determine: [u]_{\alpha},[u]_{\beta} \mathrm{e}[u]_{\gamma}.
(c) Seja v \in \mathbb{R}^{2} tal que [v]_{\beta}=\left[\begin{array}{c}3 \\\\ -1\end{array}\right]. Determine:
[v]_{\alpha} e [v]_{\gamma}.

Passo 1

O enunciado nos dá três bases ordenadas de \mathbb{R}^2: \alpha=\{(1,0),(0,1)\}, \beta=\{(-1,1),(1,1)\} e \gamma=\{(3,-1),(1,3)\}.
Pede-se em (a) as matrizes de mudança de base [I]_{\alpha}^{\beta}, [I]_{\beta}^{\alpha} e [I]_{\gamma}^{\beta};
em (b) as coordenadas de u=(2,3) nessas três bases;
e em (c) para v dado por [v]_{\beta}=\bigl(3,\,-1\bigr)^T, encontrar [v]_{\alpha} e [v]_{\gamma}.
O conteúdo envolve expressar vetores de uma base em outra e montar a matriz cuja coluna j são as coordenadas do j-ésimo vetor da base de partida na base de chegada.

Passo 2

Para achar [I]_{\alpha}^{\beta}\), expressamos cada vetor de \(\alpha\) em coordenadas da base \(\beta\). 1) \((1,0)=x\,(-1,1)+y\,(1,1)\) \[\begin{cases} -x + y = 1,\\\ \;x + y = 0, \end{cases}
daí x=-\tfrac12\], \[y=\tfrac12. Logo [\, (1,0)\,]_{\beta}=(-\tfrac12,\tfrac12)^T.
  1. (0,1)=x\,(-1,1)+y\,(1,1)
\begin{cases} -x + y = 0,\\\ \;x + y = 1, \end{cases}
daí x=\tfrac12\], \(y=\tfrac12. Logo [\, (0,1)\,]_{\beta}=(\tfrac12,\tfrac12)^T.
Montando as colunas,
[I]_{\alpha}^{\beta} =\begin{pmatrix} -\tfrac12 & \tfrac12\\\ \;\tfrac12 & \tfrac12 \end{pmatrix}.

Passo 3

Para [I]_{\beta}^{\alpha}, basta expressar cada vetor de \beta em coordenadas da base \alpha. Como \alpha é a base padrão, as coordenadas são os próprios componentes:
(-1,1)\mapsto \begin{pmatrix}-1\\\ 1\end{pmatrix},\quad (1,1)\mapsto \begin{pmatrix}1\\\ 1\end{pmatrix}.
Então
[I]_{\beta}^{\alpha} =\begin{pmatrix} -1 & 1\\\ \;1 & 1 \end{pmatrix},
que de fato é a inversa de [I]_{\alpha}^{\beta}.

Passo 4

Para [I]_{\gamma}^{\beta}\), expressamos cada vetor de \(\gamma\) em coordenadas da base \(\beta\). 1) \((3,-1)=x\,(-1,1)+y\,(1,1)\) \[\begin{cases} -x + y = 3,\\\ \;x + y = -1, \end{cases} \quad\implies\quad (x,y)=(-2,1).
  1. ((1,3)=x,(-1,1)+y,(1,1)]
    \begin{cases} -x + y = 1,\\\ \;x + y = 3, \end{cases} \quad\implies\quad (x,y)=(1,2).
    Logo
    [I]_{\gamma}^{\beta} =\begin{pmatrix} -2 & 1\\\ \;1 & 2 \end{pmatrix}.

Passo 5

Agora u=(2,3).
  • Em \alpha (base padrão) as coordenadas são triviais:
[u]_{\alpha}=\begin{pmatrix}2\\\ 3\end{pmatrix}.
  • Em \beta, usamos [u]_{\beta}=[I]_{\alpha}^{\beta}\,[u]_{\alpha}:
\begin{pmatrix} -\tfrac12 & \tfrac12\\\ \;\tfrac12 & \tfrac12 \end{pmatrix} \begin{pmatrix}2\\\ 3\end{pmatrix} =\begin{pmatrix} -1 + 1{,}5\\\\ 1 + 1{,}5 \end{pmatrix} =\begin{pmatrix}\tfrac12\\tfrac52\end{pmatrix}.
  • Em \gamma, expressamos (2,3)=a\,(3,-1)+b\,(1,3):
\begin{cases} 3a + b = 2,\\\\ -a + 3b = 3, \end{cases} \quad\implies\quad \bigl(a,b\bigr)=\Bigl(\tfrac{3}{10},\,\tfrac{11}{10}\Bigr).
Logo
[u]_{\gamma} =\begin{pmatrix}\tfrac{3}{10}\\[0.3em]\tfrac{11}{10}\end{pmatrix}.]

Passo 6

Finalmente, seja v tal que [v]_{\beta}=(3,-1)^T.
  • Em \alpha, basta combinar: v=3\,(-1,1)-1\,(1,1)=(-3,3)+(-1,-1)=(-4,2).
    Logo
[v]_{\alpha} =\begin{pmatrix}-4\\\ 2\end{pmatrix}.
  • Em \gamma, resolvemos (-4,2)=a\,(3,-1)+b\,(1,3):
\begin{cases} 3a + b = -4,\\\\ -a + 3b = 2, \end{cases} \quad\implies\quad \bigl(a,b\bigr)=\Bigl(-\tfrac{7}{5},\,\tfrac{1}{5}\Bigr).
Então
[v]_{\gamma} =\begin{pmatrix}-\tfrac{7}{5}\\[0.3em]\tfrac{1}{5}\end{pmatrix}.]

Resposta

(a)
[I]_{\alpha}^{\beta} =\begin{pmatrix} -\tfrac12 & \tfrac12\\\ \;\tfrac12 & \tfrac12 \end{pmatrix}, \quad [I]_{\beta}^{\alpha} =\begin{pmatrix} -1 & 1\\\ \;1 & 1 \end{pmatrix}, \quad [I]_{\gamma}^{\beta} =\begin{pmatrix} -2 & 1\\\ \;1 & 2 \end{pmatrix}.
(b)
[u]_{\alpha}=\begin{pmatrix}2\\\ 3\end{pmatrix},\quad [u]_{\beta}=\begin{pmatrix}\tfrac12\\[0.2em]\tfrac52\end{pmatrix},\quad [u]_{\gamma}=\begin{pmatrix}\tfrac{3}{10}\\[0.2em]\tfrac{11}{10}\end{pmatrix}.
(c)
[v]_{\alpha}=\begin{pmatrix}-4\\\ 2\end{pmatrix},\quad [v]_{\gamma}=\begin{pmatrix}-\tfrac{7}{5}\\[0.2em]\tfrac{1}{5}\end{pmatrix}.]

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