I'm supposed to use a change of basis to express a linear combination of some vectors.
Consider the invertible $3 \times 3$ matrix $A$ below and its inverse $$ A=\begin{bmatrix}
1 & -1 & 2\\
-1 & 2 & -3 \\
1 & 1 & 1\\
\end{bmatrix},\hspace{5mm}A^{-1}=\begin{bmatrix}
5 & 3 & -1\\
-2 & -1 & 1 \\
-3 & -2 & 1\\
\end{bmatrix}$$
Suppose that $B=\{\vec{u}_1,\vec{u}_2,\vec{u}_3\}$ and $C=\{\vec{v}_1,\vec{v}_2,\vec{v}_3\}$ are bases for a 3-dimensional vector space $V$, and suppose that $P_{C\leftarrow B}=A$.
a) Express $2\vec{u}_1+3\vec{u}_2+5\vec{u}_3$ as a linear combination of $\vec{v}_1,\vec{v}_2,\vec{v}_3$.
b) Express $\vec{v}_1+2\vec{v}_2+3\vec{v}_3$ as a linear combination of $\vec{u}_1,\vec{u}_2,\vec{u}_3$.
My reasoning:
a) Since $P_{C\leftarrow B}=A$, this implies $P_{B\leftarrow C}=A^{-1}$
I assume that since I need to calculate $2\vec{u}_1+3\vec{u}_2+5\vec{u}_3$, I need to write the $u$'s in terms of $v$'s.
I also know that $P_{B\leftarrow C}=A^{-1}$ so this means:
$$\begin{bmatrix}
5 & 3 & -1\\
-2 & -1 & 1 \\
-3 & -2 & 1\\
\end{bmatrix} \begin{bmatrix}
v_1\\
v_2\\
v_3\\
\end{bmatrix}=\begin{bmatrix}
u_1\\
u_2\\
u_3\\
\end{bmatrix}$$
So this means, I get:
$5v_1+3v_2-v_3=u_1$
$-2v_1-v_2+v_3=u_2$
$-3v_1-2v_2+v_3=u_3$
This implies:
$10v_1+6v_2-2v_3=2u_1$
$-6v_1-3v_2+3v_3=3u_2$
$-15v_1-10v_2+5v_3=5u_3$
So:
$(10v_1+6v_2-2v_3)+(-6v_1-3v_2+3v_3)+(-15v_1-10v_2+5v_3)=-11v_1-7v_2+6v_2$
I assume that I would do the exact same thing for part b), except I'd use $P_{C\leftarrow B}=A$ I think.
Is my thought process correct? It would be great if someone could verify. Thanks a lot!
Best Answer
You seem to confuse scalars $u_1,u_2,u_3$ with vectors.
Let's keep $u_i$ and $v_i$ denoting vectors and use $x_i$, $y_i$ for coordinates if they occur.
$P_{C\leftarrow B}=A$ means that $A[u_i]_B=[v_i]_B$, that is, since $[u_i]_B=e_i$ the $i$th standard basis vector, the $i$th column of $A$ is $[v_i]_B$, so we have $$v_1=u_1-u_2+u_3\\ v_2=-u_1+2u_2+u_3\\ v_3=2u_1-3u_2+u_3$$ Can you finish b) from here?
Indeed, for a) you need to do the same with $A^{-1}$.