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I'm trying to go about the proof that any matrix that is nilpotent (i.e. $\exists N \in\Bbb N. A^N = \mathbf{0}$) cannot be diagonalizable. I believe that the best way to go about this is by showing that a given eigenvalue's geometric multiplicity is not the same as its algebraic multiplicity. However, I am having some difficulty figuring out what aspect of nilpotency might help me with this calculation.

I can see that if $A^N = \mathbf{0}$ for some $N \in\Bbb N,$ I think it may be reasonable to prove that the only eigenvectors we get from $A$ are the $\mathbf{0}$ vector, although I am not entirely sure if this is the appropriate way to prove the above statement, given that it doesn't necessarily pertain to multiplicity but more of the mechanics of finding eigenvectors and eigenvalues of a matrix. Any recommendations on this problem?

Dietrich Burde
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Jennifer
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    Depends on what tools you want to use. Easiest proof: the minimal polynomial is $x^k$ where $k$ is the index of nilpotence. But an operator is diagonalizable iff its minimal polynomial splits as a product of distinct linear factors (with multiplicity 1). – symplectomorphic Jun 09 '16 at 23:50

6 Answers6

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First note that the only eigenvalue of a nilpotent matrix $A$ is $0$. Now suppose that $A$ is diagonalizable and nonzero. Since $A$ is diagonalizable, $A$ is similar to a diagonal matrix with the eigenvalues of $A$ on the main diagonal, i.e., $A=PDP^{-1}$ where $D$ is the zero matrix so that $A$ is the zero matrix, a contradiction.

Sri-Amirthan Theivendran
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    Just since the claim is not obvious to me. Say that $ A $ is nilpotent for power $ k $.

    Say $ \lambda $ is an eigenvalue of $ A $ with eigenvector $ x \neq 0 $, so $ Ax = \lambda x $.

    Multiply across by $ A^{k-1} $, so $ A^k x = \lambda A^{k-1} x = \lambda A^{k-2}(Ax) = \lambda^2 A^{k-2}x $. Repeating, we see $A^k x = \lambda^k x $.

    As $ A^k = 0 $, the equation is $ \lambda^k x = 0 $

    But $ x \neq 0 $, which gives $ \lambda = 0 $. As $ \lambda $ is arbitrary, all the eigenvalues of $ A $ must equal zero.

     – isaac Feb 26 '24 at 01:20
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Not true (example: $0$ matrix). Though, the statement is true if $A \neq 0$.

If $A$ is diagonalizable, then there exists an invertible $P$ and a diagonal matrix $D$ such that $A = P^{-1} D P$. Then, $A^N = P^{-1} D^N P$, which gives $P^{-1} D^N P = 0$. Therefore $D^N = 0$ and so $(D_{ii})^N = 0$ for all $i$, hence $D_{ii} = 0$ for all $i$ and $D$ is the zero matrix. This gives $A = P^{-1} 0 P = 0$, which is contradictory to the assumption.

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The claim is true for nonzero matrices. The minimal polynomial is of the form $X^n$ for $n>1$ so it has repeated roots. Then your matrix cannot be diagonalizable. Alternatively all the eigenvalues are zero, so the only diagonalizable nilpotent matrix is the zero matrix.

Pedro
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Suppose that $A$ is a nilpotent and diagonalizable matrix.

$A = PDP^{-1}$ and $\exists n$ such that $A^n = O$

$A^n = PD^{n}P^{-1} = O\\ D^n = O$

Since $D$ is diagonal $D^n = O \implies D = O$

The only nilpotent and diagonalizeable matrix is $O$

Doug M
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For an eigenvector $u$ with eigenvalue $\lambda$ we have $$ A u = \lambda u \Rightarrow A^2 u = A \lambda u = \lambda^2 u \Rightarrow A^n u = \lambda^n u $$ If there is a $n$ with $A^n = 0$, then this means $0 = \lambda^n u$ and for a non-vanishing $u$ this requires $\lambda = 0$.

mvw
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One line: Minimal polynomial divides $z^n=0$ so it must have repeated roots.

*Result used: A complex matrix is diagonalizable iff its minimal polynomial has no repeated roots. If $p(A)=0$, minimal polynomial of $A$ divides $p(A)$. If a real matrix cannot be diagonalizable wrt to $\mathbb{C}$ then it surely cannot wrt to $\mathbb{R}$.

Daniel Li
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