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I came across this problem while tutoring. Its trivial if you use Jordan canonical form.

If $A$ is a $3\times 3$ real matrix, show that $A$ is similar (in $M(3\times 3,\mathbb{C})$) to a complex diagonal matrix or a real upper triangular matrix.

Well, ok, $A$ has a real eigenvalue. If it only has one real eigenvalue, the others are distinct, so $A$ is diagonilizable over $\mathbb{C}$. It can't have just two real eigenvalues. What if it has 3? If they're distinct its diagonalizable in $\mathbb{R}$. If 2 are the same, and the other is different, its upper triangular in $\mathbb{R}$.

But what if it has a real eigenvalue of algebraic multiplicity 3? Is there some elementary or geometric argument here?

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Suppose $A$ has three real eigenvalue $a,b,c$. Pick a corresponding and extend it to a basis of $\mathbb R^3$. Then there exists an invertible matrix $P$ such that $P^{-1}AP=\pmatrix{a&\ast\\ 0&B}$, where the two eigenvalues of $B$ are $b$ and $c$. Now it suffices to show that $B$ is similar to a real upper triangular matrix, but that is easy if you play the same trick recursively.

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  • $\begingroup$ Sorry, why are b and c eigenvalues of B? $\endgroup$ – Pliny the ill Apr 10 '18 at 11:35
  • $\begingroup$ @Plinytheill Because $bI-A$ and $cI-A$ are singular. $\endgroup$ – user1551 Apr 10 '18 at 11:38
  • $\begingroup$ niice. thanks.! $\endgroup$ – Pliny the ill Apr 10 '18 at 11:40
  • $\begingroup$ I preferred your previous comment. was there something wrong with it? $\endgroup$ – Pliny the ill Apr 10 '18 at 11:41
  • $\begingroup$ the outstanding case is where a=b=c $\endgroup$ – Pliny the ill Apr 10 '18 at 11:44

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