Why does this transition from $B$ to $B'$?

Question

This question is related to the question Transition matrix which rotates counter clockwise (don't panic, I'll get around to approving the answer soon).

The text now says that if the vector sets $$ B=\{v_1, v_2, ..., v_n \} \\ B'=\{u_1, u_2, ..., u_n \} $$ are bases and if $$ v_1 = c_{11}u_1 + c_{21}u_2 + ... + c_{n1}u_n \\ v_2 = c_{12}u_1 + c_{22}u_2 + ... + c_{n2}u_n \\ \vdots v_n = c_{1n}u_1 + c_{2n}u_2 + ... + c_{nn}u_n $$ then the transition matrix $$ Q = \begin{bmatrix} c_{11} & c_{12} & ... & c_{1n} \\ c_{21} & c_{22} & ... & c_{2n} \\ \vdots & \vdots & & \vdots \\ c_{n1} & c_{n2} & ... & c_{nn} \\ \end{bmatrix} $$ is called the transition matrix from $B$ to $B'$. The proof is rather algebraic and not very enlightening, but isn't that name backwards?

(Also, is there a tag for embarrassing questions?)

Answer 1

I think you misunderstand what is meant by "transition matrix." What it means it that if you have a vector $X$ expressed in terms of basis $B,$ then to get the representation in terms of $B'$ you multiply by $Q.$

To convince yourself of this, suppose $$X=a_1v_1+a_2v_2+\dots+a_nv_n.$$

To express $X$ in terms of $B'$ you just have to substitute the expressions for the $v_k$ into the expression for $X$ and collect like terms. Let's just concentrate on $u_1$. What will be the coefficient of $u_1$ in the expression for $X$? Well, from $v_1$ we'll get $a_1c_{11},$ from $v_2$ we'll get $a_2c_{12},$ in general, from $v_k$ we'll get $a_kc_{1k}.$ Altogether, we get the first element of $$ \begin{bmatrix} c_{11} & c_{12} & ... & c_{1n} \\ c_{21} & c_{22} & ... & c_{2n} \\ \vdots & \vdots &\ddots & \vdots \\ c_{n1} & c_{n2} & ... & c_{nn} \\ \end{bmatrix} \begin{bmatrix} a_1\\ a_2\\ \vdots\\ a_n \end{bmatrix} $$

and a similar argument works for the other elements of course. This is what the proof in your book is doing, but it's probably a good deal more compact. Try to read it again now.

Answer 2

Assume you have a vector $x$ in coordinates $(x_i)$ regarding $B$ $$ x = \sum_i x_i v_i $$ what coordinates $(x_i')$ would it have regarding $B'$? $$ x = \sum_i x_i' u_i $$ We already know what coordinates the basis vectors $v_i$ of $B$ have regarding $B'$, they form the matrix $Q^\top$: $$ v_j = \sum_i c_{ij} u_i $$ So $$ x = \sum_j x_j v_j = \sum_j x_j \sum_i c_{ij} u_i = \sum_i \sum_j c_{ij} x_j u_i $$ which means $$ x_i' = \sum_j c_{ij} x_j \iff \\ (x') = Q (x) $$ So $Q$ transforms a vector of coordinates regarding $B$ into a vector of coordinates regarding $B'$, designating the same vector.