Question from linear algebra textbook:
Bases B = {(1, 1), (1, -1)} = {u1, u2} and B` = {(2, 0), (3, 1)} = {u`1, u`2} are given.
a) Find the change of basis from the standard basis to each of these.
This part seems straightforward. With {e1, e2} forming the R2 identity matrix, u1 = e1 + e2 and u2 = e1 - e2. Similarly, u`1 = 2e1 and u`2 = 3e1 + e2.
b) Use a) to find the change of basis from B to B` with T = R2 identity matrix I
For this, I first found the inverse of B` to be {(1/2, -3/2), (0, 1)}. So, the change-of-basis M = B`^-1 * I * B:
{(1/2, -3/2), (0, 1)} * {(1, 0), (0, 1)} * {(1, 1), (1, -1)} = {(1/2, -1/2), (1/2, -5/2)}
I am fairly confident in the above. But, I'm struggling with part c, which is:
c) Given that w = 3u1 + 4u2, use b) to express w as a linear combination of u`1 and u`2.
Fairly unsure where to start here. I think the result w is {(3, 3), (4, -4)}. But, I'm not seeing a linear combination of the u`1 and u`2 vectors that results in w, nor do I see the connection with the result from b).
Once you have the matrix M which represent the change of basis from B to B' you can compute
$$(a,b)=M(3,4)$$
such that
$$au_1'+bu_2'=3u_1+4u_2$$
It seems there is something wrong in your calculation, indeed
For point 1 we have
$$M_{SB}=\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \implies M_{BS}=M^{-1}_{SB}=\begin{pmatrix} \frac12 & \frac12 \\ \frac12 & -\frac12 \end{pmatrix}$$
$$M_{SB'}=\begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix} \implies M_{B'S}=M^{-1}_{SB'}=\begin{pmatrix} \frac12 & -\frac32\\ 0 & 1\end{pmatrix}$$
For point 2 we have
$$v_S=M_{SB}v_B \quad v_B'=M_{B'S}v_S \implies v_B'=M_{B'S}M_{SB}v_B $$
thus
$$M_{B'B}=M_{B'S}M_{SB}=\begin{pmatrix}-1 & 2 \\ 1 & -1 \end{pmatrix}$$
For point 3 we have
$$M_{B'B} \begin{pmatrix} 3\\4\end{pmatrix}=\begin{pmatrix}-1 & 2 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 3\\4\end{pmatrix}=\begin{pmatrix} 5\\-1\end{pmatrix}$$
and indeed
$$3u_1+4u_2=5u_1'-u_2'=\begin{pmatrix} 7\\-1\end{pmatrix}$$