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?


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.