a) (1.0) Verifique \alpha=\{(1,1,-1,0,0),(2,-1,0,0,0),(3,2,0,0,0),(0,0,0,1,0),(0,0,0,0,1)\}
é uma base de ( \mathbb{R}^{5}, \mathbb{R},+, \cdot ) (Padrão)
b) (\mathbf{0 . 7 5}) \operatorname{Se}[\mathbf{I}]_{\beta}^{\alpha}=\left[\begin{array}{ccccc}1 & 1 & 0 & 0 & 0 \\\\ 0 & -1 & 1 & 0 & 0 \\\\ 1 & 0 & -1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 0 & 1\end{array}\right], determine a base \beta.
c) (\mathbf{0 . 7 5}) \mathrm{E} se [\mathbf{v}]_{\beta}=\left[\begin{array}{c}-1 \\\\ 2 \\\\ 3 \\\\ 0 \\\\ 0\end{array}\right], determine [\mathbf{v}]_{\alpha}

Passo 1

O enunciado trata de mudanças de base em \mathbb{R}^5.
a) Queremos verificar que \alpha=\{v_1,v_2,v_3,v_4,v_5\} com
v_1=(1,1,-1,0,0),\quad v_2=(2,-1,0,0,0),\quad v_3=(3,2,0,0,0),\quad v_4=(0,0,0,1,0),\quad v_5=(0,0,0,0,1)
é uma base, ou seja, é um conjunto de 5 vetores linearmente independentes em \mathbb{R}^5.
b) Dada a matriz de mudança de coordenadas de \beta para \alpha,
[I]_{\beta}^{\alpha} =\begin{pmatrix} 1 & 1 & 0 & 0 & 0\\\\ 0 & -1 & 1 & 0 & 0\\\\ 1 & 0 & -1 & 0 & 0\\\\ 0 & 0 & 0 & 1 & 0\\\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix},
cada coluna dá as coordenadas de b_j (o j-ésimo vetor de \beta) na base \alpha.
c) Para um vetor \mathbf{v} com [\mathbf{v}]_{\beta} conhecido, basta multiplicar [\mathbf{v}]_{\alpha}=[I]_{\beta}^{\alpha}\,[\mathbf{v}]_{\beta}.

Passo 2

Verificamos a independência linear de \alpha. Seja
a_1v_1+a_2v_2+a_3v_3+a_4v_4+a_5v_5=0.
Escrevendo componente a componente:
1ª coordenada: a_1+2a_2+3a_3=0
2ª coordenada: a_1 - a_2+2a_3=0
3ª coordenada: -a_1=0 \implies a_1=0
4ª coordenada: a_4=0
5ª coordenada: a_5=0
Com a_1=0, as duas primeiras ficam
2a_2+3a_3=0,\quad -a_2+2a_3=0.
Da segunda, a_2=2a_3. Substituindo na primeira, 4a_3+3a_3=7a_3=0\implies a_3=0\implies a_2=0.
Logo todos os a_i=0, provando independência. Como temos 5 vetores num espaço de dimensão 5, \alpha é base.

Passo 3

Cada coluna j de [I]_{\beta}^{\alpha} é [\;b_j\;]_{\alpha}. Assim:
Coluna 1= \bigl(1,0,1,0,0\bigr)^T \implies b_1=1\cdot v_1+0\cdot v_2+1\cdot v_3=(1,1,-1,0,0)+(3,2,0,0,0)=(4,3,-1,0,0).
Coluna 2= \bigl(1,-1,0,0,0\bigr)^T \implies b_2=v_1-v_2=(1,1,-1,0,0)-(2,-1,0,0,0)=(-1,2,-1,0,0).
Coluna 3= \bigl(0,1,-1,0,0\bigr)^T \implies b_3=v_2-v_3=(2,-1,0,0,0)-(3,2,0,0,0)=(-1,-3,0,0,0).
Coluna 4= \bigl(0,0,0,1,0\bigr)^T \implies b_4=v_4=(0,0,0,1,0).
Coluna 5= \bigl(0,0,0,0,1\bigr)^T \implies b_5=v_5=(0,0,0,0,1).
Portanto
\beta=\{\,b_1,b_2,b_3,b_4,b_5\} =\{(4,3,-1,0,0),\,(-1,2,-1,0,0),\,(-1,-3,0,0,0),\,(0,0,0,1,0),\,(0,0,0,0,1)\}.

Passo 4

Dado [\mathbf{v}]_{\beta}=(-1,2,3,0,0)^T, temos
[\mathbf{v}]_{\alpha} =[I]_{\beta}^{\alpha}\,[\mathbf{v}]_{\beta} =\begin{pmatrix} 1 & 1 & 0 & 0 & 0\\\\ 0 & -1 & 1 & 0 & 0\\\\ 1 & 0 & -1 & 0 & 0\\\\ 0 & 0 & 0 & 1 & 0\\\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}-1\\\ 2\\\ 3\\\ 0\\\ 0\end{pmatrix} =\begin{pmatrix} -1+2\\\\ -2+3\\\\ -1-3\\\\ 0\\\\ 0 \end{pmatrix} =\begin{pmatrix}1\\\ 1\\\ -4\\\ 0\\\ 0\end{pmatrix}.

Resposta

a) \alpha é base de \mathbb{R}^5 pois é conjunto de 5 vetores linearmente independentes.
b) \displaystyle \beta=\{(4,3,-1,0,0),\,(-1,2,-1,0,0),\,(-1,-3,0,0,0),\,(0,0,0,1,0),\,(0,0,0,0,1)\}.
c) \displaystyle [\mathbf{v}]_{\alpha}=\begin{pmatrix}1\\\ 1\\\ -4\\\ 0\\\ 0\end{pmatrix}.

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