Problem:
Let $B = (w_1,…,w_n)$ and $C = (u_1,…,u_n)$ be bases in linear space V.
$P$ is a transition matrix from basis $A$ to $B$.
$Q$ is a transition matrix from basis $B$ to $C$.
Prove that $PQ$ is a transition matrix from $A$ to $C$.
My attempt:
$P[\alpha_1,…,\alpha_n]_A = [\beta_1,…,\beta_n]_B$
$Q[\beta_1,…,\beta_n]_B = [\gamma_1,…,\gamma_n]_C$
Where $\alpha_i,\beta_i,\gamma_i$ are scalars denoting position of a vector $[v]$ in given basis. Then:
$[v]_C=[\gamma_1,…,\gamma_n]_C = Q[\beta_1,…,\beta_n]_B = Q(P[\alpha_1,…,\alpha_n]_A) = QP[v]_A$ WRONG
Attempt 2:
Let $P = ([w_1]_A,..,[w_n]_A)^{-1}$ be a transition matrix from basis A to B, where each column is filled with corresponding vector $w_i\in B$ in base A. The same with Q: $Q = ([u_1]_B,..,[u_n]_B)^{-1}$
The transition matrix we're looking for is $M = ([u_1]_A,..,[u_n]_A)^{-1} =_{(1)} (P^{-1} * Q^{-1})^{-1} = QP$
wrong again??
In transition (1) I used the fact that $P^{-1}$ is a transition from basis $B$ to $A$, and that $Q^{-1}$ is the set of base vectors of $C$.
I will be grateful for any help as I'm very frustrated with this problem.
Best Answer
Given a vector $[x_1, \ldots, x_n]^T$ expressed in basis $A$ (which is to say, our vector is equal to $x_1a_1+\cdots+x_na_n$), then $P^{-1}[x_1, \ldots, x_n]^T$ is going to be the same vector expressed in basis $B$.
Similarly, given a vector $[y_1,\ldots, y_n]^T$ expressed in basis $B$, $Q^{-1}[y_1, \ldots, y_n]^T$ is the same vector expressed in basis $C$.
Now note that this means $Q^{-1}P^{-1}[x_1, \ldots, x_n]^T$ is the first vector expressed in basis $C$. So to translate from $A$ to $C$ we multiply by $Q^{-1}P^{-1}=(PQ)^{-1}$, which makes $PQ$ the transition matrix from $A$ to $C$.
Which is to say, your attempt 1 was very close to the real solution. It's just that we multiply by the inverse of the transition matrix when transitioning between bases. At least according to some conventions for transition matrices.