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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.

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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

1 Answer 1

up vote 0 down vote accepted

Just so that this question does not remain unanswered, I'll paraphrase the proof given in the question in a less technical way.

A matrix $A$ is similar to an upper triangular matrix if and only if for the linear operator $\phi$ of $V=F^n$ defined by $A$ there exist a chain of $\phi$-stable subspaces $$ \{0\}=V_0\subset V_1\subset V_2\subset\cdots\subset V_n=V $$ where $\dim(V_i)=i$ for all $i$. This is so because if $\def\B{\mathcal B}\B=[b_1,\ldots,b_n]$ is any basis, then $\operatorname{Mat}_\B(\phi)$ is upper triangular if and only if each of the subspaces $\langle b_1,\ldots,b_d\rangle$ is $\phi$-stable, for $d=1,2,\ldots,n$.

Now a proof that if the characteristic polynomial $\chi_A$ splits into a product of linear factors $X-\lambda_i$, then $A$ is similar to an upper triangular matrix (by the way, the converse is obvious). Proceed by induction on $n=\dim V$, the case $n=0$ being obvious. Now let $n>0$, and let $X-\lambda$ be one of the factors of $\chi_A$, then there is some eigenvector $v$ of$~A$ for $\lambda$. Put $V_1=\langle v\rangle$, then since $V_1$ is $\phi$-stable, $\phi$ induces a linear operator $\overline\phi$ of the quotient space $V/V_1$, which is of dimension$~n-1$. The characteristic polynomial of $\overline\phi$ is $\chi_A/(X-\lambda)$, so it also splits into a product of linear factors. Then by induction there is a chain of $\overline\phi$-stable subspaces $$ V_1/V_1=W_0\subset W_1\subset W_2\subset\cdots\subset W_{n-1}=V/V_1 $$ with $\dim W_i=i$ for all $i$. Being a subspace of the quotient space we can write $W_i=V_{i+1}/V_1$ for a uniquely defined subspace $V_{i+1}$ of$~V$, which has dimension $i+1$, and is $\phi$-stable (because $W_i$ is $\overline\phi$-stable). Together with $V_0=\{0\}$ this provides our chain of subspaces that shows that $A$ is similar to an upper triangular matrix, completing the proof.

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