# Eigenvectors and eigenvalues - how to find the column vector

1. Nov 19, 2012

### Tala.S

Hi

We have a matrix A (picture), the eigenvalues are λ1 = 4 and λ2 = 1 and the eigenvectors are

λ1 : t(1,0,1)
λ2 : t1(1,0,2) + t2(0,1,0)

I have to examine if there's a column vector v that satifies :

A*v = 2 v

I would say no there doesn't exist such a column vector v because 2 isn't an eigenvalue:

Av = λv

so

Av = 2v

but we know that λ is 4 or 1 and not 2

Am I wrong ?

I would be nice if someone could give me their opinion :)

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Last edited: Nov 19, 2012
2. Nov 19, 2012

### AlephZero

You seem to understand what eigenvalnes and vectors are, which is good.

But you aren't quite right. There IS a vector that satisfies A*v = 2 v, but you might think it's not a very interesting vector.

Try solving (A - 2I)v = 0, and see what you get.

3. Nov 19, 2012

### Tala.S

When I solve it I get

v =

(1 0)
(0 0)
(0 0)

?

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4. Nov 19, 2012

### Erland

You cannot use Maple in this way. You should use linsolve(A1,b) or LinearSolve(A1,b). Or better, do it by hand.

5. Nov 19, 2012

### HallsofIvy

Staff Emeritus
I don't understand what you mean by that. Your vectors before were single columns of three numbers. How could v be two columns?

You were close to right when you said before "doesn't exist such a column vector v because 2 isn't an eigenvalue". What is true is that $\lambda$ is an eigenvalue for matrix A if and only if there exist a non-trivial (i.e. non-zero) vector v such that $Av= \lambda v$.

Since two is not an eigenvalue, there cannot exist a non-trivial vector but, as
Aleph-zero said, "There IS a vector that satisfies A*v = 2 v, but you might think it's not a very interesting vector.".

6. Nov 19, 2012

### Tala.S

So the vector is (0,0,0) ?

Or have I completely misunderstood this ?

7. Nov 19, 2012

### Erland

Right. But it is not counted as an eigenvector.

8. Nov 19, 2012

### AlephZero

Correct. The question asked you to find a column vector, not a non-zero column vector!

You only get non-zero solutions of Av = λv when λ is an eigenvalue, but v = 0 is a solution for any value of λ.