1

Let $ A $ be a $ n \times n $ matrix. Show that over the complex numbers , there exists an invertible matrix P such that $ P^{-1}AP$ is an upper triangular matrix.

Answer: If the matrix $A $ is diagonalisable then $ P^{-1}AP $ is diagonal and hence it is an upper triangualr matrix. Now if $ A $ is not diagonalisable , then $ P^{-1} AP $ has Jordan canonical form , which is upper triangular. Hence the proof . Is it right approach ? Any help

MAS
  • 10,638
  • Can you justify why $A$ has a Jordan form? – Michael Biro Jun 10 '17 at 00:26
  • since every matrix has jordan form – MAS Jun 10 '17 at 00:27
  • I meant, how do you know that? Otherwise you're just shifting the claim. – Michael Biro Jun 10 '17 at 00:28
  • i can't justify right now , but i know one think that -a matrix may not not be diagonalisable but it is similar to a jordan form . – MAS Jun 10 '17 at 00:30
  • Am I right ? or wrong – MAS Jun 10 '17 at 00:31
  • 1
    While the existence of a Jordan canonical form suffices to show your claim, it does seem a possibly oversimplified approach. It depends on how much you were expected to do for this problem. – hardmath Jun 10 '17 at 00:54
  • 1
    The existence of Jordan form is very hard to prove. Much harder than proving this result directly. If I was your professor I would not allow using Jordan form. – D_S Jun 10 '17 at 01:33

1 Answers1

2

Because $C$ is algebraically closed, $A$ has a nonzero eigenvector $v$. Rewriting $A$ relative to any basis $(v,w_2,\dots,w_n)$, we see that $A$ is similar to a block-matrix of the form $\pmatrix{\lambda & B \\ 0 & A_1}$, where $A_1$ is some $(n-1) \times (n-1)$ matrix. Now, considering the action of $A_1$ on the subspace generated by $w_2, \dots, w_n,$ the desired result can be obtained by induction on $n$.

user49640
  • 2,704