let $V$ and $W$ be finite-dimensional vector spaces having ordered bases, and let $%T:V W$ be linear. Then for each

There's a theorem proved in my book that I keep having to refer to, but I do not understand what it's saying at all. I tried talking to my professor, and got some tutoring, but both were a waste of time. Can anyone maybe shed light on it?

Theorem 2.14 (Linear Algebra, 4th Ed, Friedman)
Let $V$ and $W$ be finite-dimensional vector spaces having ordered bases $\beta$ and $\gamma$, respectively, and let $T:V\rightarrow W$ be linear. Then for each $u \in V$ we have $$[T(u)]_\gamma = > [T]_\gamma^\beta[u]_\beta$$

Proof. Fix $u \in V$, and define the linear transformations $f: F\rightarrow V$ by $f(a)=au$ and $g:F\rightarrow W$ by $g(a)=aT(u)$ for all $a \in F$. Let $\alpha = \{1\}$ be the standard ordered basis for $F$. Notice that $g = Tf$. Identifying column vectors as matrices and using Theorem 2.11, we obtain $$[T(u)]_\gamma = [g(1)]_\gamma = > [g]_\alpha^\gamma=[Tf]_\alpha^\gamma = > [T]_\beta^\gamma[f]_\alpha^\beta = [T]_\beta^\gamma > [f(1)]_\beta=[T]_\beta^\gamma[u]_\beta$$

What is that even saying? I thought I was getting the idea of the little "ordered bases" notations after $[T]$ but apparently not? I understand the notation $[T]_\beta^\gamma$ is the transformation of a function in the basis $\beta$ to the basis $\gamma$... that's about all though I guess.

• Are you asking what the theorem is saying? It's saying that if you start with $u$ in $V$, apply $T$ to it to get an element of $W$, and then find the coordinates of this element of $W$ with respect to the basis $\gamma$, you get the same answer you would get if you first found the coordinates of $u$ with respect to the basis $\beta$ of $V$, and then multiplied that by the transition matrix $[T]_{\gamma}^{\beta}$. – Gerry Myerson Sep 29 '16 at 13:17
• I thought that transformations had to take $u$ in the original basis (in this case, $\beta$) and convert to another basis ($\gamma$)... I can't understand how you could put $u$ in the other basis (for lack of better wording? This is what I don't get)? Edit: I think of transformations usually in different dimensions. I can see how a basis would span $u$ if it had the same dimension as another basis, but not if they differed? – Asinine Sep 29 '16 at 13:24
• @Asinine do you mean that you're specifically confused by the proof? – Omnomnomnom Sep 29 '16 at 13:27
• I'm really not sure about the theorem or the proof. Definitely lost with the proof; usually those help me understand what the theorem is saying if I don't get it immediately. – Asinine Sep 29 '16 at 13:35
• I think I have narrowed down my issue: I apparently have no clue how bases are applied. All the examples in my notes and in my book have been "standard ordered" bases... for $P_n(\Real)$ they're always $\{[x^0,]x^1,x^2,... x^n\}$, or the "e" ones $\{(1,0,...0), (0,1,...0), (0,0,... 1)\}$. Now I'm looking at a problem with a basis of $\{(1,1),(-2,1)\}$ and I'm completely clueless. But I think this has pinpointed the area causing issues. I will ask my professor about this in class today. – Asinine Sep 29 '16 at 14:03

The theorem says that you can use $[T]_\gamma^\beta$ in precisely the way we should expect. If $[u]_\beta$ is the coordinate vector of $u$ with respect to $\beta$, then the coordinate vector of the output , that is $[T(u)]_\gamma$, is precisely what you get from the matrix product $[T]_\beta^\gamma [u]_\beta$.
In other words, the matrix $[T]_\beta^\gamma$, when left-multiplied to a vector, "computes" the output of the transformation $T$. This is precisely why care about "the matrix of a linear transformation" in the first place.
• Hmm, I think I almost understand. I keep forgetting to think of transformations in terms of "plotting" them out, I guess... of sorts, how $f(x)$ relates to $x$. So basically, you're saying that $[u]_\beta$ is the "$x$" in old-school ~7-8th grade algebra terms, and $[T(u)]_\beta^\gamma$ is the "$f(x)$" (or "$y$")? If this is the case, it's becoming a bit clearer but I still don't understand the $[T(u)]_\gamma$ bit. – Asinine Sep 29 '16 at 13:31
• Not quite: $[T]_\beta^\gamma$ is the "function itself", $f$. On the other hand, $[T(u)]_\gamma$ is the "output" of the $f$. So, for example: if $f(x) = 2x$, then $f(1) = 2$. In this case, $1$ plays the role of $[u]_\beta$, $2$ plays the role of $[T(u)]_\gamma$, and the function $f(x)$ itself plays the role of $[T]_\beta^\gamma$. – Omnomnomnom Sep 29 '16 at 16:53