# Matrix representation of a linear transformation under change of basis.

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.

• 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? – Minus One-Twelfth Feb 15 at 12:48

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).