121 views
3 votes
3 votes
Let $\mathcal{B}=\left\{\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}\right\}$ and $\mathcal{C}=\left\{\mathbf{c}_{1}, \mathbf{c}_{2}, \mathbf{c}_{3}\right\}$ be two bases of $\mathbb{R}^{3}$, and suppose that
$$
\mathbf{c}_{1}=\mathbf{b}_{1}-\mathbf{b}_{3}, \quad \mathbf{c}_{2}=3 \mathbf{b}_{1}-\mathbf{b}_{2}+\mathbf{b}_{3}, \quad \mathbf{c}_{3}=\mathbf{b}_{1}+2 \mathbf{b}_{2}+\mathbf{b}_{3} .
$$
Consider the vector $\mathbf{x}=2 \mathbf{c}_{1}+\mathbf{c}_{2}-3 \mathbf{c}_{3}$. Find $[\mathbf{x}]_{\mathcal{B}}$, that is, the components of $\mathbf{x}$ in the basis $\mathcal{B}$.
  1. $\left[\begin{array}{r}2 \\ 1 \\ -3\end{array}\right]$
  2. $\left[\begin{array}{r}2 \\ -7 \\ -4\end{array}\right]$
  3. $\left[\begin{array}{l}1 \\ 2 \\ 1\end{array}\right]$
  4. $\left[\begin{array}{r}3 \\ -1 \\ 1\end{array}\right]$

1 Answer

1 votes
1 votes
(a) Find the change of basis matrix $P_{\mathcal{B} \leftarrow \mathcal{C}}$.

$$
\begin{gathered}
{\left[\mathbf{c}_{1}\right]_{\mathcal{B}}=\left[\begin{array}{r}
1 \\
0 \\
-1
\end{array}\right], \quad\left[\mathbf{c}_{2}\right]_{\mathcal{B}}=\left[\begin{array}{r}
3 \\
-1 \\
1
\end{array}\right], \quad\left[\mathbf{c}_{3}\right]_{\mathcal{B}}=\left[\begin{array}{l}
1 \\
2 \\
1
\end{array}\right],} \\
P_{\mathcal{B} \leftarrow \mathcal{C}}=\left[\left[\mathbf{c}_{1}\right]_{\mathcal{B}},\left[\mathbf{c}_{2}\right]_{\mathcal{B}},\left[\mathbf{c}_{3}\right]_{\mathcal{B}}\right]=\left[\begin{array}{rrr}
1 & 3 & 1 \\
0 & -1 & 2 \\
-1 & 1 & 1
\end{array}\right] .
\end{gathered}
$$

(b) Consider the vector $\mathbf{x}=2 \mathbf{c}_{1}+\mathbf{c}_{2}-3 \mathbf{c}_{3}$. Find $[\mathbf{x}]_{\mathcal{B}}$, that is, the components of $\mathbf{x}$ in the basis $\mathcal{B}$.

$$
\begin{gathered}
{[\mathbf{x}]_{\mathcal{C}}=\left[\begin{array}{r}
2 \\
1 \\
-3
\end{array}\right], \quad[\mathbf{x}]_{\mathcal{B}}=P_{B \leftarrow \mathcal{C}}[\mathbf{x}]_{\mathcal{C}}} \\
{[\mathbf{x}]_{\mathcal{B}}=\left[\begin{array}{rrr}
1 & 3 & 1 \\
0 & -1 & 2 \\
-1 & 1 & 1
\end{array}\right]\left[\begin{array}{r}
2 \\
1 \\
-3
\end{array}\right]=\left[\begin{array}{r}
2 \\
-7 \\
-4
\end{array}\right]}
\end{gathered}
$$
Answer:

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