proof $B^{-1}MB$ is triangular

How to prove this? Theorem: Let $M$ be a matrix of complex numbers. There exists a non-singular matrix $B$ such that $B^{-1}MB$ is a triangular matrix.

This is corollary from book Linear Algebra by Serge Lang. And there is this proof: This is standard interpretation of the change of matrices when we change bases..

But I don't think that proves anything. So I need a real proof for that.

Thanks :)

• Jordan's theorem? Or, if you need $B$ to be orthogonal, Schurr's form: en.wikipedia.org/wiki/Schur_decomposition – 5xum Feb 10 '14 at 11:36
• The fastest high level proof is to use the fact that $M$ has an eigenvector $v$, and $M/\mathbb C v$ has smaller dimension, so we can use induction to show that we have a basis in which $M$ is triangular. However, that is probably not what you're looking for. What do you know? What have you tried? What tools are at your disposal? – Aaron Feb 10 '14 at 11:37
• Well, Theorem 1.1. of Lang gives conditions on a base for the matrix being triangular and then Theorem 1.2. some properties about existence that gives corollary 1.3. where it is stated the triangularization of every endomorphism. Finally, collorary 1.4. gives the desired proves since passing from matrices to linear mappings and viceversa is trivial supposing you have read the previous chapters. – Josué Tonelli-Cueto Feb 10 '14 at 12:57

Theorem: Let {$v_1, \dots , v_n$} be a fan basis for $A$. Then the matrix associated with $A$ relative to this basis is an upper triangular matrix.
He then proves that if $V$ is a finite dimensional vector space over $C$ and $A : V \to V$ is a linear map, then there exists a fan of $A$ in $V$. The argument used is exactly what @Aaron has mentioned in his comment above.