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



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...

  • $\begingroup$ $W^{-1}$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.