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Let $D$:= diag($\lambda_1, \ldots, \lambda_n$), i.e., $D$ is a diagonal matrix in $\mathbb{C}^{n\times n}$ with entries $\lambda_1, \ldots, \lambda_n$$\mathbb{C}$ on its diagonal.

Furter let $U\in\mathbb{C}^{n\times n}$ be an invertible matrix and define $A$ := $U^{−1} $D$ $U$, i.e. A$ is similar to $D$. Find the eigenvalues and eigenvectors of $A$.

As $U$ is invertible and $D$ is diagonal I think that $A$ should also be a diagonal matrix.

And as $A$ is similar to $D$ then I suppose that $A$ has the same eigenvalues as $D$, namely ${\lambda n}^{n}$.

Could someone tell me if I am on the right way?

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If $u_i$ is an eigenvector of $D$: $$ Du_i=\lambda_iu_i $$ than, from: $$ A=U^{-1}DU\Rightarrow AU^{-1}=U^{-1}D$$

we have: $$ A(U^{-1}u_i)=U^{-1}Du_i=U^{-1}\lambda_iu_i=\lambda_i(U^{-1}u_i) $$

that gives the eigenvectors for the same eigenvalues.

Geometrically you can think at $U$ as a change of basis, so $A$ represents the same linear transformation of $D$ in a new basis, so its eigenvectors are the same, but expressed in the new basis.

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