relation between nilpotent matrix and eigenvalues

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March 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 $A^k \vec x = A^{k-1}(A \vec x)=A^{k-1}(A \lambda)=\vec 0$
I thought about trying to find the characteristic polynomial but I've got too little information to get far...
March 26th 2011, 08:58 PM
Tinyboss

Suppose A has a nonzero eigenvalue with corresponding eigenvector v. What is $A^n(v)$ ?
March 26th 2011, 11:54 PM
Deveno

hint #2: write A^n(v) in two ways, and derive a contradiction.
March 27th 2011, 01:57 AM
FernandoRevilla

An alternative:

If $A$ is nilpotent of order $k$ then, $p(t)=t^k$ is an annihilator polynomial of $A$ so, its minimal polynomial divides to $p(t)$ hence, it has the form $\mu(t)=t^s$ . As a consequence its characteristic polynomial has the form $\chi(t)=t^r$ .
March 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 $\lambda$ with corresponding eigenvector $\vec v$
$A^n \vec v=A^{n-1}(\lambda \vec v) = \lambda (A^{n-1} \vec v)=...=\lambda^n \vec v$
This means that $A^k \vec v=\lambda^k \vec v=\vec 0$ but it could be either $\lambda$ or $\vec v$ that is zero?
March 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 $\lambda$ with corresponding eigenvector $\vec v$
$A^n \vec v=A^{n-1}(\lambda \vec v) = \lambda (A^{n-1} \vec v)=...=\lambda^n \vec v$
This means that $A^k \vec v=\lambda^k \vec v=\vec 0$ but it could be either $\lambda$ or $\vec v$ that is zero?

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

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

Quote:

Originally Posted by FernandoRevilla

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

i see the implication one way (since the minimal polynomial divides the characteristic polynomial), what is the implication the other way?
March 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 $\mu (t)$ is the minimal polynomial of $A\in \mathbb{K}^{n\times n}$ then, the characteristic polynomial of $A$ divides to $(\mu (t) )^n$ .