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I am getting conflicting answers depending on how I approach the problem and I don't understand why. Let $A=\begin{pmatrix}2&-5&7\\1&-2&0\\0&0&3 \end{pmatrix}$. If I go directly to the characteristic polynomial I get $\det(A-I\lambda)=\det\begin{pmatrix}2-\lambda&-5&7\\1&-2-\lambda&0\\0&0&3-\lambda \end{pmatrix}$ and then calculating the determinant I get that $\lambda_1=3, \lambda_{2,3}=\pm i$.

I then conclude that $A$ is not diagonalizable over $\mathbb{R}$ as we have non-real eigenvalues.

However if I put $A$ into an upper diagonal matrix I have that $A=\begin{pmatrix}2&-5&7\\0&0.5&-3.5\\0&0&3 \end{pmatrix}$

I then see the eigenvalues as the entries of our upper diagonal matrix, i.e. distinct and real. Then isn't this representation of $A$ diagonalizable? But $A$ hasn't changed. What am I misunderstanding?

Thanks

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  • $\begingroup$ There should be an error in your upper diagonal similar matrix. $\endgroup$ – mathcounterexamples.net Dec 8 '20 at 14:01
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    $\begingroup$ "But $A$ hasn't changed": er, $A$ has changed in three elements ! $\endgroup$ – Yves Daoust Dec 8 '20 at 14:05
  • $\begingroup$ Ok, there is my misunderstanding. Performing elementary row operations does effect $A$. I thought they were "fair game." $\endgroup$ – Marco Esquandolas Dec 8 '20 at 14:08
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Your new matrix is equivalent to $A$, but not similar. The latter condition is stronger and required to preserve the Eigenvalues.

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    $\begingroup$ Thanks. This is the clearest path to clear up my confusion $\endgroup$ – Marco Esquandolas Dec 8 '20 at 14:11
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The operation you are performing on the matrix $A$ is equivalent to pre-multiplying by another elementary matrix $E$. Just because $EA$ is diagonalizable, does not mean that $A$ is. In particular, as you just demonstrated, the eigenvalues of $A$ and $EA$ are not necessarily the same.

Perhaps a more elementary example of such "innocent" alterations leading to changing eigenvalues are flipping the rows of $$M = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix},$$ which results in the identity matrix. The original matrix $M$ has eigenvalues $-1,1$ and the resulting identity has only $1$ instead (with multiplicity $2$)...

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That means every matrix of order $n\times n$ with rank $n$ is diagonalizable. But this is certainly not true. The matrix you got after you row reduced your $A$ is not simlar to $A$ since there does not exist any invertible $P$ such that $A=PBP^{-1}$ where $B$ is the row reduced version of $A$

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