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If I have a linear transformation $T$ and the basis $b$ and $a$, when I have $[T]_{b}^{a}$ It means that I'm doing the transformation starting on the basis $a$ and the result is given with respect of basis $b$?!

If so, it is correct to apply first the transformation with respect to basis $a$ : $[T]_{a}^{a}$ and then apply another transformation, $[Q]_{b}^{a}$ for example, that change the basis from $a$ to $b$?

What I am trying to get is that the relation below is true or not: $$ [T]_{b}^{a} = [Q]_{b}^{a}\cdot [T]_{a}^{a} $$

Thanks.

Bruno Reis
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  • So $[T]{b}^{a} = [Q]{b}^{a}\cdot [T]_{a}^{a}$ it's a valid relation? – Bruno Reis Oct 01 '16 at 18:29
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    Correction to a previous comment: Unfortunately, notations differ across textbooks and authors, so whether you're correct depends on which textbook you're reading. That being said, you seem to be following the opposite of the convention that I'm used to. – Ben Grossmann Oct 01 '16 at 18:34

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Note that if you want $[Q]^a_b$ to take a coordinate-vector with respect to $a$ and produce a coordinate-vector with respect to $b$, then the transformation $Q$ itself "shouldn't do anything". That is, for any vector $x$, we should have $Q(x) = x$. That is, $Q$ should be the identity operator. Typically, $I$ is used to denote this identity operator.

With that being said, it is indeed true that $$ [T]^a_b = [I]^a_b[T]^a_a $$

Ben Grossmann
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  • Yeah, it's a isomorphic transformation that isn't going to change my vector, just gimme another coordinates for it... Right? By the way, thanks a lot for your comment and answer my friend. – Bruno Reis Oct 01 '16 at 18:37
  • "Isomorphic transformation" doesn't mean anything, what you're trying to say is "isomorphism". Note: it's not $Q$ itself that should give you the new coordinates, its the $[\cdot]^a_b$ around the $Q$ which are doing that. For this reason, $Q$ itself needs to "do nothing". – Ben Grossmann Oct 01 '16 at 18:41
  • Yeah, I'm new to linear algebra... You said what I wanted to say in my previous comment. If I had $[Q]_{a}^{a}$ it would've been the identity matrix, right? – Bruno Reis Oct 01 '16 at 18:44
  • @BrunoReis that's right. So, $Q$ is the linear transformation $Q(x) = x$. Or, as I put it, $Q = I$. – Ben Grossmann Oct 01 '16 at 18:45
  • And you're welcome – Ben Grossmann Oct 01 '16 at 18:46
  • Thanks my friend. See you around through the forum :) – Bruno Reis Oct 01 '16 at 18:58