Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that if $A\in M_{n}\left(\mathbb{F}\right)$ matrix with a characteristic polynomial that can be written as a product of linear elements (?) ($\left(x-\lambda_{1}\right)^{r_1}\dots\left(x-\lambda_{k}\right)^{r_k}$) that $A$ is similar to an upper triangular matrix.

My attempt: Well first I showed that if $\lambda$ is an eigenvalue of $A$ then it can be written as

$P^{-1}AP=\left(\begin{matrix}\lambda & * & \cdots & *\\ 0 & \ulcorner & & \urcorner\\ \vdots & & B\\ 0 & \llcorner & & \lrcorner \end{matrix}\right)$

for some invertible $P \in M_{n}\left(\mathbb{F}\right)$.

Then by induction on n I showed that for a $2\times 2$ matrix the claim holds using the above lemma. I've assumed correctness for some $n$. And then I tried proving it for $n+1$: I used the lemma for the $\left(n+1\right) \times \left(n+1\right)$ matrix (because we assume that its characteristic polynomial can be reduced).

Then I tried using the induction hypothesis on the inner $B$ of the $\left(n+1\right) \times \left(n+1\right)$ matrix. So I've concluded that there's some matrix $Q\in M_{n}\left(\mathbb{F}\right)$ so that $Q^{-1}BQ$ is upper triangular.

The next step was to use it for my current matrix so I've decided to cunstruct the following matrices:

$ Q^{'}=\left(\begin{matrix}1 & 0 & \cdots & 0\\ 0 & \ulcorner & & \urcorner\\ \vdots & & Q\\ 0 & \llcorner & & \lrcorner \end{matrix}\right) $ and $ Q^{'-1}=\left(\begin{matrix}1 & 0 & \cdots & 0\\ 0 & \ulcorner & & \urcorner\\ \vdots & & Q^{-1}\\ 0 & \llcorner & & \lrcorner \end{matrix}\right) $

So that now I can claim that

$Q^{'-1}\left(\begin{matrix}\lambda & * & \cdots & *\\ 0 & \ulcorner & & \urcorner\\ \vdots & & B\\ 0 & \llcorner & & \lrcorner \end{matrix}\right)Q^{'} = \left(\begin{matrix}\lambda & * & \cdots & *\\ 0 & \ulcorner & & \urcorner\\ \vdots & & Q^{-1}BQ\\ 0 & \llcorner & & \lrcorner \end{matrix}\right) $

which is essentially an upper triangular matrix.

But I'm having hard time determining whether it's correct at all because to be honest a lot of indices confuse me :) and I would really like to know whether my proof is correct and is there a more solid\clear\intuitive way to explain the last conclusion.

Edit 1: I'll try to make my question clearer: We are usually required to be extremely rigorous, so the $Q^{'-1} P^{-1} A P Q^{'}$ part needs to be explained as well but I can't seem to find a better explanation then something like "well it has identity in the upper left corner so that's what happens" but I can't figure a nice way of writing this mathematically without babbling too much.

share|improve this question
It looks good and crystal clear to me. –  user1551 Apr 14 '13 at 2:56
@user1551 thank you, I've tried to clarify my question. –  Scis Apr 14 '13 at 8:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.