# relation between nilpotent matrix and eigenvalues

• Mar 26th 2011, 08:00 PM
Jskid
relation between nilpotent matrix and eigenvalues
Show that if A is nilpotent, then the only eigenvalue of A is 0.

Here's my work $\displaystyle A^k \vec x = A^{k-1}(A \vec x)=A^{k-1}(A \lambda)=\vec 0$
I thought about trying to find the characteristic polynomial but I've got too little information to get far...
• Mar 26th 2011, 08:58 PM
Tinyboss
Suppose A has a nonzero eigenvalue with corresponding eigenvector v. What is $\displaystyle A^n(v)$?
• Mar 26th 2011, 11:54 PM
Deveno
hint #2: write A^n(v) in two ways, and derive a contradiction.
• Mar 27th 2011, 01:57 AM
FernandoRevilla
An alternative:

If $\displaystyle A$ is nilpotent of order $\displaystyle k$ then, $\displaystyle p(t)=t^k$ is an annihilator polynomial of $\displaystyle A$ so, its minimal polynomial divides to $\displaystyle p(t)$ hence, it has the form $\displaystyle \mu(t)=t^s$ . As a consequence its characteristic polynomial has the form $\displaystyle \chi(t)=t^r$ .
• Mar 27th 2011, 01:50 PM
Jskid
Quote:

Originally Posted by Deveno
hint #2: write A^n(v) in two ways, and derive a contradiction.

Here's what I've got
let n be a positvie integer
suppose A has a non zero eigne value $\displaystyle \lambda$ with corresponding eigenvector $\displaystyle \vec v$
$\displaystyle A^n \vec v=A^{n-1}(\lambda \vec v) = \lambda (A^{n-1} \vec v)=...=\lambda^n \vec v$
This means that $\displaystyle A^k \vec v=\lambda^k \vec v=\vec 0$ but it could be either $\displaystyle \lambda$ or $\displaystyle \vec v$ that is zero?
• Mar 27th 2011, 07:45 PM
tonio
Quote:

Originally Posted by Jskid
Here's what I've got
let n be a positvie integer
suppose A has a non zero eigne value $\displaystyle \lambda$ with corresponding eigenvector $\displaystyle \vec v$
$\displaystyle A^n \vec v=A^{n-1}(\lambda \vec v) = \lambda (A^{n-1} \vec v)=...=\lambda^n \vec v$
This means that $\displaystyle A^k \vec v=\lambda^k \vec v=\vec 0$ but it could be either $\displaystyle \lambda$ or $\displaystyle \vec v$ that is zero?

Nop, since by definition an eigenvector must be different from zero.

Tonio
• Mar 27th 2011, 08:51 PM
Deveno
first off, by supposition you say that λ is non-zero. eigenvectors are by definition non-zero. so you have on the one hand A^n(v) = λ^nv where neither is 0, so A^n(v) is non-zero.

on the other hand A^n(v) = 0v = 0, since A is nilpotent. how can a 0-vector be equal to a non-zero vector? it can't. therefore, λ can't be non-zero (that was our ONLY assumption).

@Fernando: i see that the mimimal polynomial of A has to be of the form x^s (because the only divisors of x^k are of that form), but how do you conclude that the characteristic polynomial has no other roots? unless what you're actually saying is that the minimal polynomial of A actually has to be x^k (which i think it does since A^s , s < k, should not be 0, by the definition of the order k of A). if so, then we can take any polynomial p(A) that A satisfies (including the characteristic polynomial), and knock off the powers of A greater than k to get a polynomial of degree at most k-1 that A also satisfies, and since x^k is minimal, the left-over part must be 0. but then i feel i'm missing something, why can't the characteristic polynomial be of the form: (x^k)^mq(x)? unless you're making a tacit appeal to the fact that q(x) has to split in some extension of the base field (the original poster doesn't indicate what the entries of A actually are, presumably real or complex numbers, but maybe not), so that considered as a matrix in Mn(E) (instead of Mn(F)) q(x) has roots in E, and thus A has eigenvalues in E corresponding to these roots), contradicting the minimality of x^k for A. if so, it seems the long way around. tell me what i'm missing....
• Mar 27th 2011, 09:08 PM
Jskid
Quote:

Originally Posted by Deveno
hint #2: write A^n(v) in two ways, and derive a contradiction.

I don't see any contridiction in my solution, what were you refering to? (just out of curiousity)
• Mar 27th 2011, 11:25 PM
FernandoRevilla
Quote:

Originally Posted by Deveno
but how do you conclude that the characteristic polynomial has no other roots? .

For $\displaystyle A\in \mathbb{K}^{n\times n}$ , the characteristic and minimal polynomials have the same irreducible factors on $\displaystyle \mathbb{K}[t]$ .
• Mar 28th 2011, 01:01 PM
Deveno
Quote:

Originally Posted by FernandoRevilla
For $\displaystyle A\in \mathbb{K}^{n\times n}$ , the characteristic and minimal polynomials have the same irreducible factors on $\displaystyle \mathbb{K}[t]$ .

i see the implication one way (since the minimal polynomial divides the characteristic polynomial), what is the implication the other way?
• Mar 28th 2011, 11:58 PM
FernandoRevilla
Quote:

Originally Posted by Deveno
i see the implication one way (since the minimal polynomial divides the characteristic polynomial), what is the implication the other way?

It is not easy to prove it in an autocontent way. Some previous properties are needed. For example, if $\displaystyle \mu (t)$ is the minimal polynomial of $\displaystyle A\in \mathbb{K}^{n\times n}$ then, the characteristic polynomial of $\displaystyle A$ divides to $\displaystyle (\mu (t) )^n$ .