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I think there's a mistake or a typo in the solutions -- but just to make sure I'm correct I'd appreciate if someone could verify it. Exercise 4 in Ch. 5 of Shilov's Linear Algebra asks to express the coordinates of a vector $x \in \mathbf{K}_2$ after two consecutive bases transformations in terms of the coordinates of the original basis. Let us denote by $A$ the matrix of the operator transforming basis $\{e\}$ to $\{f\}$ and by $B$ the matrix of the operator transforming basis $\{f\}$ to $\{g\}$. Given the coordinates of x wrt $\{e\}$, $(\xi_1, \xi_2)$, the coordinates of x wrt $\{g\}$, $(\tau_1, \tau_2)$, are given by $$ \pmatrix{\tau_1\\\tau_2}=(AB)^{-1}\pmatrix{\xi_1\\\xi_2}=B^{-1}A^{-1}\pmatrix{\xi_1\\\xi_2}$$ Is this right? The solutions say the matrix should be $BA^{-1}$ so indeed it seems to be a typo. I followed the reasoning in the said chapter, where it is shown that $AB$ transforms $\{e\}$ to $\{g\}$ directly (the order is not very intuitive), and hence through its inverse one gets the components in the new basis.

I'm also confused by the issue raised in this question: Vector Component Transformation Matrix in Shilov's Linear Algebra even after reading the answer. I can't comment at the moment because of the low rep, so asking it again here. Why is it that Shilov writes (ch. 5, p. 122) in the conclusion that if $P$ transforms $\{e\}$ to $\{f\}$, then the coordinates of $x$ wrt $\{f\}$ are obtained through $(P^{-1})^T$? Even though it is $P^{-1}$ in equation (12) directly above? I'm clearly missing something.

geoant
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1 Answers1

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There are most likely some typos. Note that in the exercise, it was written that

$\tau_1=b_{11}\xi_1+b_{12}\xi_2,\quad\tau_2=b_{21}\xi_1+b_{22}\xi_2,$

and the last sentence of the exercise is

“Express the components $\tau_1,\tau_2$ in terms of the components $\xi_1,\xi_2$.”

Since the author had already written down the expressions himself, these expressions couldn’t possibly be what he really asked for. What he really wanted, I believe, is to express $\tau_1,\tau_2$ in terms of the components $\color{red}{\eta_1,\eta_2}$ instead. As it was given in the exercise that $\eta = A\xi$ and $\tau = B\xi$, the answer is $\tau = B\xi = BA^{-1}\eta$.

user1551
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