1. eigenvector

Let V be an n-dimensional complex vector space where $n\geq 1$ and let $x:V\rightarrow V$ be a linear map.
show that $x$ has an eigenvector $v \in V$.

i can't solve this question . how can ?

2. Re: eigenvector

The fact that the space is complex is important, since you now that the characteristic polynomial has a root.

3. Re: eigenvector

Yes. I know that the roots of the characteristic polynomial are eigenvalue . but i show x has an eigenvector v\in V

4. Re: eigenvector

Originally Posted by vernal
Yes. I know that the roots of the characteristic polynomial are eigenvalue . but i show x has an eigenvector v\in V
Fix a basis $B=\{v_1,\ldots,v_n\}$ of $V$ , if $\lambda$ is a root of the characteristic polynomial and $A=[T]_B$ then, the system $(A-\lambda I)x=0$ has at least a non zero solution $(\alpha_1,\ldots,\alpha_n)\in\mathbb{C}^n$ . Choose $v=\alpha_1v_1+\ldots+\alpha_nv_n$ .

5. Re: eigenvector

Originally Posted by vernal
Yes. I know that the roots of the characteristic polynomial are eigenvalue . but i show x has an eigenvector v\in V
If you already know that the roots of the characteristic polynomial are eigenvalues, then you are done. The characteristic polynomial does have a complex root and therefore $x$ does have an eigenvalue. By definition, an eigenvalue of a linear map $x$ is a scalar $\lambda$ such that there exists a vector $v$ in the space $V$ such that $xv=\lambda v.$ This is also the definition of the eigenvector $v.$

There is a nice proof of this without the use of determinants/charcteristic polynomials. It's here, page 3.

6. Re: eigenvector

Originally Posted by ymar
If you already know that the roots of the characteristic polynomial are eigenvalues, then you are done. The characteristic polynomial does have a complex root and therefore $x$ does have an eigenvalue. By definition, an eigenvalue of a linear map $x$ is a scalar $\lambda$ such that there exists a vector $v$ in the space $V$ such that $xv=\lambda v.$ This is also the definition of the eigenvector $v.$

There is a nice proof of this without the use of determinants/charcteristic polynomials. It's here, page 3.
thanks..but in page 3 :Theorem 2.1 Every linear operator on a finite-dimensional complex vector space
has an eigenvalue., but my question eiguenvector.

7. Re: eigenvector

Originally Posted by FernandoRevilla
Fix a basis $B=\{v_1,\ldots,v_n\}$ of $V$ , if $\lambda$ is a root of the characteristic polynomial and $A=[T]_B$ then, the system $(A-\lambda I)x=0$ has at least a non zero solution $(\alpha_1,\ldots,\alpha_n)\in\mathbb{C}^n$ . Choose $v=\alpha_1v_1+\ldots+\alpha_nv_n$ .
thanks fernandorevilla..
you said the system (A-\lambda I)x=0 has at least a non zero solution (\alpha_1,\ldots,\alpha_n)\in\mathbb{C}^n , why?

8. Re: eigenvector

Originally Posted by vernal
thanks fernandorevilla.. you said the system (A-\lambda I)x=0 has at least a non zero solution (\alpha_1,\ldots,\alpha_n)\in\mathbb{C}^n , why?
$\textrm{rank}\;(A-\lambda I) . Now, apply Rouche Fröbenius Theorem.

9. Re: eigenvector

Originally Posted by FernandoRevilla
Fix a basis $B=\{v_1,\ldots,v_n\}$ of $V$ , if $\lambda$ is a root of the characteristic polynomial and $A=[T]_B$ then, the system $(A-\lambda I)x=0$ has at least a non zero solution $(\alpha_1,\ldots,\alpha_n)\in\mathbb{C}^n$ . Choose $v=\alpha_1v_1+\ldots+\alpha_nv_n$ .
excuse me

$A=[T]_B$ what is T?

10. Re: eigenvector

Originally Posted by vernal
thanks..but in page 3 :Theorem 2.1 Every linear operator on a finite-dimensional complex vector space
has an eigenvalue., but my question eiguenvector.
I made a small but significant mistake in my earlier post. The vector $v$ is supposed to be non-zero.

The existence of an eigenvalue implies the existence of an eigenvector. With the definitions from my earlier post it's immediate. What does it mean that $x$ has an eigenvalue? This:

1) $\left (\exists \lambda\in\mathbb{C}\right)\left( \exists v\in V\setminus\{0\}\right ) xv=\lambda v.$

What does it mean that $x$ has an eigenvector? This:

2) $\left( \exists v\in V\setminus\{0\}\right )\left (\exists \lambda\in\mathbb{C}\right) xv=\lambda v.$

These formulas are equivalent. The order of the existential quantifiers doesn't matter.

However, Sheldon Axler uses a different definition of an eigenvalue. For him, that $x$ has an eigenvalue means this:

3) There exists $\lambda\in \mathbb{C}$ such that $x-\lambda I$ is not injective.

We want to see that 3) implies 2). What does it mean that $x-\lambda I$ is not injective? It means that we have two vectors $v_1,v_2\in V,$ such that $v_1\neq v_2$ and

$(x-\lambda I)v_1=(x-\lambda I)v_2,$

the last formula being equvalent to

$xv_1 -\lambda v_1 = xv_2 -\lambda v_2,$

and further equivalent to

$x(v_1-v_2)=\lambda (v_1-v_2).$

Since $v_1\neq v_2,$ we have that the vector $v_1-v_2$ is non-zero. Therefore, we have found a non-zero vector $v$ such that

$xv=\lambda v,$

just as we wanted.

11. Re: eigenvector

Originally Posted by vernal
excuse me $A=[T]_B$ what is T?
Sorry, I didn't notice that you call the linear map $x$ instead of $T$ . Of course, it does not matter .

12. Re: eigenvector

Originally Posted by FernandoRevilla
Sorry, I didn't notice that you call the linear map $x$ instead of $T$ . Of course, it does not matter .
thanks:-D

13. Re: eigenvector

Originally Posted by ymar
I made a small but significant mistake in my earlier post. The vector $v$ is supposed to be non-zero.

The existence of an eigenvalue implies the existence of an eigenvector. With the definitions from my earlier post it's immediate. What does it mean that $x$ has an eigenvalue? This:

1) $\left (\exists \lambda\in\mathbb{C}\right)\left( \exists v\in V\setminus\{0\}\right ) xv=\lambda v.$

What does it mean that $x$ has an eigenvector? This:

2) $\left( \exists v\in V\setminus\{0\}\right )\left (\exists \lambda\in\mathbb{C}\right) xv=\lambda v.$

These formulas are equivalent. The order of the existential quantifiers doesn't matter.

However, Sheldon Axler uses a different definition of an eigenvalue. For him, that $x$ has an eigenvalue means this:

3) There exists $\lambda\in \mathbb{C}$ such that $x-\lambda I$ is not injective.

We want to see that 3) implies 2). What does it mean that $x-\lambda I$ is not injective? It means that we have two vectors $v_1,v_2\in V,$ such that $v_1\neq v_2$ and

$(x-\lambda I)v_1=(x-\lambda I)v_2,$

the last formula being equvalent to

$xv_1 -\lambda v_1 = xv_2 -\lambda v_2,$

and further equivalent to

$x(v_1-v_2)=\lambda (v_1-v_2).$

Since $v_1\neq v_2,$ we have that the vector $v_1-v_2$ is non-zero. Therefore, we have found a non-zero vector $v$ such that

$xv=\lambda v,$

just as we wanted.
thanks for your help ymar