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I've been going over matrix similarity and base changes and I am stuck. I was taught that a change in basis from standard basis to another basis say 'B' the transformation matrix was the inverse of the basis vectors of B. i.e if $$B=\begin{pmatrix} 1&1 &0 \\ 0&1 &0 \\ 1&3 &1 \end{pmatrix}$$ Then $$B^{-1}=\begin{pmatrix} 1&-1 &0 \\ 0&1 &0\\ -1&2 &1 \end{pmatrix}$$was the transformation matrix. However in my textbook there is an example of basis transformations including linear maps that doesn't seem to fit that pattern. Here's the example:

Consider the linear map $T: \mathbb{R}^2\rightarrow\mathbb{R}^2$ such that $T((x,y))=(x+y,y)$. Find the matrices associated to T in the bases $((1,0),(0,1))$ and $((1,2),(3,1))$ and the similarity transformation between them.

I can find the matrices of T w.r.t the two bases. These are: $$T_N=\begin{pmatrix} 1&1\\ 0&1 \end{pmatrix}$$ Calling the non-standard base 'W'. $$T_{W}=\begin{pmatrix} 3/5&-1/5\\ 4/5&7/5 \end{pmatrix}$$ However I am having trouble understanding what to do next. I thought to find the change of basis between $N$ (standard) and $W$, you find $W^{-1}$ and this transforms standard vectors into $W$ basis vectors and the same with the linear maps. However the answer in my book is $W$ i.e $W=\begin{pmatrix} 1&3\\ 2&1 \end{pmatrix}$. This then fits when multiplying out as $T_NW=WT_W$. Surely $W$ is the basis change matrix for $W$ to standard basis?

I would much appreciate any help

George

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Write now each element of the second basis as linear combination of the first one, and take the transpose of the coefficients' matrix, say $\;P\;$ . This is the similarity map between the matrices $\;T_N\;,\;\;T_W\;$ ...

It is not clear what you call $\;W^{-1}\;$ to: I thought this is your notation for the new matrix...

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  • $\begingroup$ $W^{-1}$ refers to the inverse of W. I was under the impression that changing from the standard basis to another basis ($W$) you multiply by the inverse of the other basis? $\endgroup$ – George1811 Jan 3 '14 at 15:09
  • $\begingroup$ There is no "inverse of a basis" ! You can talk of the inverse of a regular map, though. $\endgroup$ – DonAntonio Jan 3 '14 at 15:16
  • $\begingroup$ Yes sorry, I meant arranging the basis vectors into a matrix map and taking the inverse of that. $\endgroup$ – George1811 Jan 3 '14 at 15:40
  • $\begingroup$ Well, yes: since the basis $\;N\;$ is the canonical basis, one can do that. It usually is not that easy as writing one basis as lin. combination of another one can be way more messy. $\endgroup$ – DonAntonio Jan 3 '14 at 15:43

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