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$T\in\mathcal{L}(V) $ where $\dim(V) <\infty$

Consider $\beta_1, \beta_2, \beta_3, \beta_4$ four bases of $V$ .

Does this implies the matrix $[T]_{\beta_1 }^{\beta_2}$ is similar to the matrix $[T]_{\beta_3}^{ \beta_4}$ ?

If the bases $\beta_1=\beta_2$ and $\beta_2=\beta_3$ , then the proof is known.

Few examples support that the relation similarity is true. But examples are not enough , we need a proof or a counter example.

What are the required the condition on those four bases to make sure that $[T]_{\beta_1 }^{\beta_2}$ and $[T]_{\beta_3}^{ \beta_4}$?

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    $\begingroup$ What is $\mathcal{M}$? A matrix representation of $T$? $\endgroup$
    – paperskilltrees
    May 21 at 16:42
  • $\begingroup$ If so, try to write down the definition of similarity, the relation between bases ($\beta_i$ in terms of $\beta_{i+2}$), and the expression of $T$ through $\mathcal{M}[T]_\cdots$. This should suffice to answer the question. $\endgroup$
    – paperskilltrees
    May 21 at 16:47

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Going with the more standard notation, it's a standard result that: $$[T]^{\beta_4}_{\beta_3} = M_{\beta_2, \beta_4}[T]^{\beta_2}_{\beta_1} M_{\beta_1, \beta_3}^{-1},$$ so you certainly have the equivalence of these two matrices. But unless you have the two change of coordinate matrices are equal: $$ M_{\beta_2, \beta_4} = M_{\beta_1, \beta_3}$$ I do not think you can expect similarity. In fact, it looks like you can just take $\beta_3 = \beta_1$ and $\beta_4 = 2\beta_2,$ then you will have: $$[T]^{\beta_4}_{\beta_3} = \frac 12 [T]^{\beta_2}_{\beta_1},$$ so they certainly won't be similar.

It just dawned on me while writing this actually that you can take determinants from the very first equation and immediately conclude that unless the change of coordinate matrices have the same determinant, you can never have similarity. Even then, that's just a necessary condition, certainly not sufficient.

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  • $\begingroup$ $T:\Bbb{R}^2\to\Bbb{R}^2 $ defined by $T(x, y) =(-y, x) $\\ Can you find $4$ bases of $\Bbb{R^2} $ w. r.to which the above matrices are not similar? $\endgroup$
    – Lost in Space
    May 23 at 12:33
  • $\begingroup$ $\beta_1 = \beta_2 = \beta_3 = \{e_1,e_2\}$ and $\beta_4 = \{2e_1, 2e_2\}$ and you will have the matrix in $\beta_1\to \beta_2$ as: $$\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$ and the other matrix is half of this. So obviously, they are not similar. $\endgroup$
    – dezdichado
    May 23 at 14:59
  • $\begingroup$ Excellent.Deserve upvote. Last question before acceptance .Is there any condition to make those two matrices $[T]_{\beta_1 }^{\beta_2}$ and $[T]_{\beta_3}^{ \beta_4}$ similar? If $\beta_1=\beta_2$ and $\beta_3=\beta_4$ then obvious. Are there any strong assumptions on these basis to make matrices similar? $\endgroup$
    – Lost in Space
    May 23 at 17:11
  • $\begingroup$ @SouravGhosh as written in the answer: $$ M_{\beta_2, \beta_4} = M_{\beta_1, \beta_3}$$ would be a sufficient condition to ensure similarity, which is "weaker" than just assuming the bases are pairwise equal. You would want weaker conditions if you want general. I suspect this is also necessary but I do not see an immediate way to do this. In the linked question, the fault in logic seems to be assuming this general result, which is now seen false. But that question seems purely computational - worst comes to worst, one can name the entries of $\beta_1, \beta_2$ and just bash the answer. $\endgroup$
    – dezdichado
    May 23 at 19:10
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    $\begingroup$ Great. It's now very clear. You are really amazing. $\endgroup$
    – Lost in Space
    May 23 at 19:17

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