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For a matrix $A\in M_n(\mathbb{F})$ consider the linear transformation $T_A:\mathbb{F}^n\rightarrow \mathbb{F}^n$ such that $x\mapsto Ax$. Suppose A is diagonalizable and $B=\{v_1,...,v_n\}$ is a basis of eigenvectors for $T_A$ corresponding to eigenvalues $\lambda_1,...,\lambda_n$. Let $D=[T_A]_B$ and $C\in M_n(\mathbb{F})$ with columns $v_1,...,v_n$. Prove that $$D=C^{-1}AC. $$

I now really sure how to approach this. In a previous problem, I computed $D$ as a diagonal matrix with eigenvalue entries (If I did the problem correctly). But I'm confused about what is going on with the Matrix $C$. It isn't clear if each column are the basis vectors or the diagonal entries for $C$.

I'm also not sure how to compute $C^{-1}$. Could you just try to prove that $CD=AC$? Any help in understanding the problem is much appreciated.

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  • $\begingroup$ Each column of $C$ is one of the $v_{j}$ (eigenvectors from the basis $B$), as the question says. Yes, it is equivalent to show that $CD = AC$ (make sure you can explain why!). Some hints for showing this: if $v_{j}$ is the $j$-th column of $C$, then what is the $j$-th column of $AC$? And what is the $j$-th column of $CD$ in terms of the $\lambda_{i}$'s and the $v_{i}$'s? $\endgroup$ – Minus One-Twelfth Feb 15 at 12:48
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This is just the change of basis formula for linear operators, from the standard basis of $K^n$ to the basis $B$. That the $v_i$ are eigenvectors for $T_A$ has nothing to do with the formula (though it does tell you the form that $D$ will have).

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