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I need to prove that an $n\times n$ matrix $A$ is diagonalizable only to a matrix $D$ that has its diagonal entries as a permutation of the eigenvalues of A, in order to solve a related question.

The related question is disproving that a matrix A is diagonalizable to a diagonal matrix with one of the entries not being an eigenvalue of A.

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Suppose $A=P^{-1}DP$, where $D$ is diagonal, and let $d_i$ denote the $i$th diagonal entry of $D$. Then $AP^{-1}=P^{-1}D$. Let $\mathbf{e_i}$ be the $i$th standard basis vector. Then $$ AP^{-1}\mathbf{e_i}=P^{-1}D\mathbf{e_i}=P^{-1}d_i\mathbf{e_i}=d_iP^{-1}\mathbf{e_i}. $$ Therefore $P^{-1}\mathbf{e_i}$ is an eigenvector of $A$, with eigenvalue $d_i$.

This proves that all the diagonal entries of $D$ are eigenvalues of $A$.

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The simplest explanation is that similarity transformations preserve eigenvalues, and that the diagonal entries of a diagonal matrix are exactly its eigenvalues (since the standard unit vectors are corresponding eigenvectors).

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Here's a better claim: if $D$ and $E$ are similar diagonal matrices, then there's a permutation matrix $P$ with $P^{-1}DP = E$.

That'd prove your claim, for if $A$ were similar to both $D$ and $E$, then $D$ and $E$ would be similar, and hence similar via some permutation.

Why is it a better claim? Because, for instance, if your $4 \times 4 $ matrix has only $1$ and $-1$ as eigenvalues, and it's similar to

\begin{bmatrix} 1& & & & \\ & 1& & & \\ & & & 1& \\ & & & &-1 \end{bmatrix}

then it cannot be similar to \begin{bmatrix} 1& & & & \\ & 1& & & \\ & & & -1& \\ & & & &-1 \end{bmatrix}

even though both of those diagonal matrices have entries that are all eigenvalues of $A$. This richer observation may not be needed for the thing you want to prove, but since this is no harder to prove than the other one, we might as well do it.

Now...the question is, "How do I prove that claim?"

Suppose that $$ M^{-1} D M = E, i.e.,\\ DM = ME. \newcommand{\be}{\mathbf e} $$ for some pair of diagonal matrices $D$ and $E$. Note that $\be_1$ is an eigenvector corresponding to the first entry $e_{11}$ of $E$. So $$ ME\be_1 = Me_{11}\be_1 = e_{11} M\be_1 $$ But this is the same as $DM\be_1$, by the equality above, thus $M\be_1$ is an eigenvector of $D$, with eigenvalue $e_{11}$ as well. Since $D$ is diagonal, all of its eigenvectors are standard basis vectors, so there must be some value $j$ with $D\be_j = e_{11}\be_j$. So $e_{11}$ is one of the diagonal entries of $D$.

Consider the restriction of $v \mapsto Ev$ to the subspace orthogonal to $\be_1$ (whose matrix is just $E$ with the first row and column deleted) and the restriction of $v \mapsto Dv$ to the subspace orthogonal to $\be_j$ (whose matrix is $D$ with the $j$th row and column deleted). These are still similar, (via a matrix $N$ that is simply $M$ with the first column and $j$th row deleted) and diagonal, and an inductive argument finishes the claim.

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  • $\begingroup$ Gotcha! So disproving D and E being similar will then disprove A and E (whatever one I want to disprove) are similar. Giving it a go! $\endgroup$ – Uq'''12wn1F12u2x3uW31H1JBk9m Apr 5 '16 at 0:13
  • $\begingroup$ Alright I think I'm getting it. Why can't we just have $e$ as any eigenvalue of $E$ instead of just the largest one? $\endgroup$ – Uq'''12wn1F12u2x3uW31H1JBk9m Apr 5 '16 at 0:26
  • $\begingroup$ You can; I've written out an argument now. $\endgroup$ – John Hughes Apr 5 '16 at 0:29

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