Of interest: Theorem 8.ii
In the proof, Kunze writes the change of basis matrix relationship in sigma sum notation. I am confused by the steps in which Kunze left multiplies both sides by Q, the inverse matrix of P (the original change of basis matrix). I am confused at the following points:
-
I know that $I = PP^{-1}$; however, I do not know how to translate this into sigma sum notation as Kunze presents. In particular, I do not understand how Kunze simplifies the second to last step shown in the attached image to $\alpha_k$. Why does the index letter change?
-
I am also confused as to why the index used matters. Don't i, j, and k all have n elements? I feel like I am missing some fundamental understanding of indices.
-
What does $\alpha_k$ mean? My understanding is that it represents the ordered basis vectors. So, $\alpha_1$ is the first basis vector (of the old basis), for example.
tldr; How do you simplify $\sum_{i=1}^{n}(\sum_{j=1}^{n}P_{ij}Q_{jk})\alpha_i$ where Q is the inverse matrix of P?
Any help would be greatly appreciated!
Best Answer
In that equation, $k$ is a free index - it exists independent of the sigma notation, and it's there to indicate that Kunze is actually representing $k$ separate equations. $\sum_j Q_{jk} \alpha_j'$ isn't (explicitly) a matrix multiplication, it's just an equation where he's writing a linear combination of the vectors $\alpha_j'$ whose coefficients are the $k$-th row of $Q$. In other words, it is just:
$\sum_j Q_{jk} \alpha_j' = Q_{1k} \alpha_1' + Q_{2k} \alpha_2' + \ldots + Q_{nk} \alpha_n'$
Since $\alpha_j'$ is a basis of the vector space $V$, this linear combination is going to be some vector in $V$. By expanding out $\alpha_j'$ vectors in terms of the other basis $[\alpha]$, we get $\alpha_j' = \sum_i P_{ij} \alpha_i = P_{1j} \alpha_1 + P_{2j} \alpha_2 + \ldots + P_{nj} \alpha_n$. Notice that in this equation, $j$ is "free", but then when we substitute that into the original equation, those $j$s get bound up in the sum, so we have (parentheses added to make it clearer how the sums are grouped:
$\begin{eqnarray} \sum_j Q_{jk} \alpha_j' & = & \sum_j \left( Q_{jk} \left(\sum_i P_{ij} \alpha_i \right) \right) \\ & = & \sum_j \left(Q_{jk} \left(P_{1j} \alpha_1 + P_{2j} \alpha_2 + \ldots + P_{nj} \alpha_n \right) \right) \\ & = & Q_{1k}P_{11}\alpha_1 + Q_{1k}P_{21}\alpha_2 + \ldots + Q_{1k}P_{n1} \alpha_n \\ && + Q_{2k}P_{21}\alpha_1 + Q_{2k}P_{22}\alpha_2 + \ldots + Q_{2k}P_{n2}\alpha_n \\ && + \ldots \\ && + Q_{nk}P_{n1}\alpha_n + Q_{nk}P_{n2}\alpha_2 + \ldots + Q_{nk}P_{nn}\alpha_n\end{eqnarray}$
Then, Kunze does some rearrangement courtesy of the associative and distributive rules, which lets him basically move the sigmas around somewhat freely:
$\begin{eqnarray} \sum_j Q_{jk} \alpha_j' & = & \sum_j Q_{jk} \sum_i P_{ij} \alpha_i \\ & = & \sum_j \sum_i \left( Q_{jk} P_{ij} \alpha_i \right) \\ & = & \sum_i \left( \sum_j Q_{jk} P_{ij} \right) \alpha_i \\ & = & \sum_i \left( \sum_j P_{ij} Q_{jk} \right) \alpha_i \end{eqnarray}$
Now, up to this point, we haven't actually been doing any matrix stuff, it's all just been expanding and collecting sums of things that happen to also be coefficients in a couple of matrices. But at this point, we do use the fact that $Q = P^{-1}$. Not as a left multiplication $QP$, but as a right multiplication $PQ$ - in particular, notice that $\sum_j P_{ij} Q_{jk}$ is exactly the expression you use for the $ik$-th element of the matrix $PQ$, i.e. the entry in the $i$-th row and $k$-th column. But since $PQ = I_n$, we know that this will equal 0 if $i \neq k$ and 1 if $i = k$, so that $\sum_j P_{ij} Q_{jk} \alpha_i = \alpha_i$ if $i = k$, and the zero vector otherwise. And if you add up lots of zeros plus one lot of $\alpha_k$, you get the last line.