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Are the transformation matrices the relation between basis $\beta$ and $\beta'$?

We are told to get the coordinates of vector $v = (1,2,1)$ in both basis $\beta$ and $\beta'$, and find what the relation between the two basis is:

$$\beta = \{(1,1,-1),(0,-1,2),(2,0,1)\}$$

$$\beta' = \{(1,0,-1),(1,0,1),(0,1,1)\}$$

I get that the coordinates of the vector are:

$$M_\beta(v)= \left( \begin{matrix} 9\\ 7\\ -4\\ \end{matrix}\right) , M_\beta'(v)= \left( \begin{matrix} 1\\ 0\\ 2\\ \end{matrix}\right)$$ And I get both transformation matrices: $$M_{\beta}^{\beta'}= \left( \begin{matrix} 3/2&-3/2&1/2\\ -1/2&3/2&3/2\\ 1&-1&0\\ \end{matrix}\right) , M_{\beta'}^{\beta}= \left( \begin{matrix} -3&1&6\\ -3&1&5\\ 2&0&-3\\ \end{matrix}\right)$$

I've tested that both transformation matrices work and also that I get $I_3$ when multiplying both of them:

$$M_\beta^{\beta'}.M_{\beta'}^{\beta} = I_3\\ M_\beta^{\beta'}.M_\beta(v) = M_{\beta'}(v)\\ M_{\beta'}^{\beta}.M_{\beta'}(v) = M_\beta(v)$$

Can I say that the transformation matrices are the relation between $\beta$ and $\beta'$?

Hope this is not a stupid question...

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I suppose so.

Note that if you form the matrix consisting of the columns of $\mathcal B$ ($\mathcal B'$) you get the change of basis matrix from $\mathcal B$ (respectively $\mathcal B'$) to the standard basis. Thus we should have:$$[I]_{\mathcal B}^{\mathcal B'}=[I]_S^{\mathcal B'}[I]_{\mathcal B}^S=\begin {pmatrix}1&1&0\\0&0&1\\-1&1&1\end{pmatrix}^{-1}\cdot \begin{pmatrix}1&0&2\\1&-1&0\\-1&2&1\end{pmatrix}$$. This should be your $[M]_{\mathcal B}^{\mathcal B'}$. Indeed it is.

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  • $\begingroup$ Wow I didnt know this way of getting the matrix, seems to be faster this way. $\endgroup$ – user605734 MBS Oct 19 '18 at 15:44
  • $\begingroup$ It's a good trick to know... $\endgroup$ – Chris Custer Oct 19 '18 at 15:45
  • $\begingroup$ The thing is that you suppose that the answer to my question is yes, but you suppose it then I dont know what to do ~ $\endgroup$ – user605734 MBS Oct 19 '18 at 15:49
  • $\begingroup$ I guess you will find out in a matter of time. I'm not sure what else they could be looking for. $\endgroup$ – Chris Custer Oct 19 '18 at 15:57

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