Why are eigenvalues of nilpotent matrices equal to zero? [duplicate]

If $A$ is a $\displaystyle 10 \times 10$ matrix such that $A^{3} = 0$ but $A^{2} \neq 0$ (so A is nilpotent) then I know that $A$ is not invertible, but why does at least one eigenvalue of $A$ have to be equal to zero? How would one show that all eigenvalues of $A$ are equal to zero?

• Conceptually, it's because the matrix is collapsing that dimension down irreversibly. May 4, 2015 at 5:57

If $v$ is a non-zero eigen vector corresponding to an eigenvalues $\lambda$ we have, by definition, $Av=\lambda v$. Then $A^2v= A( Av)= A(\lambda v)= \lambda^2v$. It easily follows that $\lambda^n$ is an eigenvalue for $A^n$ but the latter is the zero matrix, for which all eigenvalues are zero, hence $\lambda=0$.

• Wow! I'm impressed. 5 up-votes in 27 seconds! May 4, 2015 at 2:39
• Travis, you are right. But I wrote the words "all eigenvalues are zero" for the zero matrix, thinking it requires no justification. However disregarding nilpotency if $B=p(A)$, with $p$ a polynomial, then it is true that $\lambda$ is an eigenvalue of $A$ iff $f(\lambda)$ is an eigenvalue for $B$. May 4, 2015 at 2:43
• eigenvectors are by definition nonzero.
– abel
May 4, 2015 at 2:45
• @PVanchinathan Yes, see my comment above. Your latter statement is not true, however: If we take $A := -I$, $p(x) := x^2$, then $B = P(A) = I$ but $1$ is not an eigenvalue of $A$ whereas $p(1) = 1$ is an eigenvalue of $B$. May 4, 2015 at 2:47
• @David: think about it. If it can be diagonalized what can be that diagonal matrix? May 4, 2015 at 3:02

Suppose $\lambda$ is an eigenvalue of the nilpotent matrix $A,$ and $u$ its associated eigenvector. Then $$Au = \lambda u, u \neq 0$$ multiplying by $A$ on the right shows $$A^2u = \lambda Au = \lambda^2 u$$ and by induction $$A^n u = \lambda^n u$$

If $A$ is nilpotent, then $A^k = 0$ for some $k>0.$ that implies $\lambda^k = 0\to \lambda = 0.$

• I stuck with the induction. Nov 22, 2020 at 12:15

Alternatively, do the contrapositive. If $A$ has a non-zero eigenvalue, $\lambda,$ then $A^{k} \neq 0$ for all $k.$

Proof: there exists $v \neq 0,$ such that $$Av = \lambda v,$$ so $$A^{k} v = \lambda^{k} v \neq 0.$$ Done.

Suppose $$n$$ is the smallest integer for which $$A^n=0$$. Since $$A$$ is non-zero $$\implies$$ $$n \gt 1$$

let the characteristic polynomial for $$A$$ be

$$A^m+c_1A^{m-1}+\cdots + c_m = 0$$ here $$m \le n$$. multiplying the equation by $$A^{n-1}$$ gives $$c_mA^{n-1} =0$$ hence $$c_m=0$$. this procedure can be repeated to show that all coefficients except the first are zero, hence the equation is simply: $$A^m=0$$ and we must have $$m=n$$

since the eigenvalues are roots of the equation: $$x^m = 0$$ it follows that they must all be zero

• I think $n$ should be greater than $1$ (first line of your post). How do you assure $m \leq n$, i.e. how do exclude the case $m \gt n$? May 4, 2015 at 6:17