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Suppose $A \in M^{F}_{n \times n}$ is non diagonalizable.

Is there a polynomial $P(t) \in F[t]$ of degree $n-1$ such that $P(A)^2=0$ when:

  • $F=\mathbb{R}$
  • $F=\mathbb{C}$

How do I approach this question?

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If $n\ge2$, yes. Hint: at least one Jordan block is not diagonal. – Did Jun 4 '11 at 9:27
In more elementary language, $A$ must have at least one eigenvalue with algebraic multiplicity $\ge 2$. This is because if $A$ has $n$ distinct eigenvalues, then it is necessarily diagonalizable. – Mark Jun 4 '11 at 9:32
Another approach (though the basic idea is the same) is to use the Cayley-Hamilton theorem to find a polynomial $Q(t)\in F[t]$ such that $Q(A)=0$. If $Q$ has a repeated root, you can find $P$ such that $P|Q|P^2$. – Aaron Jun 4 '11 at 9:42
@Mark: that makes sense, but where do I go from there? – daniel.jackson Jun 4 '11 at 10:19
@Aaron: by Cayley-Hamilton, $Q$ is the characteristic polynomial of $A$, but I'm not sure I followed the rest of your suggestion. – daniel.jackson Jun 4 '11 at 10:24
up vote 2 down vote accepted

The answer is in the comments already made, but perhaps some expansion would not be amiss.

$A$ has $n$ eigenvalues (in $\bf C$). If they are all distinct, then $A$ is diagonalizable, so we may assume there is a repeated eigenvalue, hence at most $n-1$ distinct eigenvalues.

Let $Q$ be the characteristic polynomial of $A$. The ring of polynomials with coefficients in $F$ is a unique factorization domain, so $Q=R^2S$ for some polynomials $R$ and $S$ with coefficients in $F$, where $S$ has no repeated factors. We can't have $R$ of degree zero, since that would mean $Q$ has no repeated factors, which we have already ruled out. Now $P=RS$ is the polynomial you're looking for.

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For $F=\mathbb R$, the solutions doesn't work because repeated eignevalues are not the only thing that stands in the way of diagonalizability. Indeed, you cannot diagonalize unless the eigenvalues are real.

For example the rotation matrices $R_{\theta}=\pmatrix{\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta}$ with $\theta \neq 0,\pi$ give a counterexample to the claim.

In general, if suppose that $A$ is an $n\times n$ real matrix with unique eigenvalues, some of which are complex, and let $D=SAS^{-1}$ be a diagonalization of $A$ over $\mathbb{C}$ (there can't be one over $\mathbb{R}$ because there are complex eigenvalues). Let $\lambda_1, \ldots, \lambda_n$ the the diagonal entries of $D$ (the eigenvalues of $A$). Then for any polynomial $P$, we have $P(D)$ is diagonal with entries $P(\lambda_1), \ldots, P(\lambda_n)$. Because $P(D)$ is conjugate to $P(A)$, one vanishes if and only if the other does.

So suppose that $P^2(A)=P^2(D)=0$. Then each $\lambda_i$ would be a root of $P^2$, and hence a root of $P$. Therefore, $P$ would have to have degree at least $n$.

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I think I understand the example but could you also add a small intuitive explanation as to what's going on here? – daniel.jackson Jun 4 '11 at 21:14
@daniel.jackson I have fleshed out the argument. Hopefully this is clearer. – Aaron Jun 4 '11 at 21:42
Maybe a tiny example would help: $A=\pmatrix{0&1\cr-1&0\cr}$ is not diagonalizable over $\bf R$, and it's not hard to show that there are no real numbers $a$ and $b$, not both zero, such that $(aA+b)^2=0$. But I took "non-diagonalizable" to be meant in an absolute sense, that is, over the algebraic closure, that is, over $\bf C$. – Gerry Myerson Jun 4 '11 at 23:39
@Gerry Myerson: Your example is my example, with $\theta=3\pi/2$! But I would have taken the same interpretation that you did if it weren't for the fact that he specifically mentions $\mathbb R$ and $\mathbb C$ separately. – Aaron Jun 5 '11 at 0:10

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