# Tricky one - solve for k - linear algebra

1. Mar 25, 2009

### sciencegirl1

1. The problem statement, all variables and given/known data

A=
1 5 2
1 3 k
2 1 1

Solve for k if A has the eigenvector

3
2
5

3. The attempt at a solution
I first tried to put in eigenvalues L1, L2 and L3

2-L1 [3-L1, k ; -5 , -2-L1] and so on...

and was going to isolate k but it doesnt´make any sense to me.

Can you help?

2. Mar 25, 2009

### tiny-tim

Hi sciencegirl1!
doesn't look right

are you sure you've copied the question correctly?

3. Mar 25, 2009

### Wretchosoft

If x is an eigenvector, Ax=qx, for some constant q, which when multiplied out, will give you three equations in two variables. Unfortunately, when I tried to work it out, I got 23=3q and 13=5q, which leads me to believe that [3, 5, 2] isn't an eigenvector of this matrix.

4. Mar 25, 2009

### sciencegirl1

2 1 -1
1 3 k
-1 -5 -2

-5
-2
3

You´re right. This wasn´t the right A. Sorry. This one is correct :)

5. Mar 25, 2009

### tiny-tim

D'uh!

ok … now just do the obvious

multiply the vector by the matrix … what do you get?

6. Mar 25, 2009

### sciencegirl1

thanks tiny tim you´re a live saver

when I multiply it I get
-15
-11+3k
9

Then I don´t know what to do :(

7. Mar 25, 2009

### Staff: Mentor

What's the definition of eigenvector? Knowing that will help you understand the results you got.

8. Mar 25, 2009

### sciencegirl1

i don´t know...
can you help me solve this?
I don´t have that much time left.

9. Mar 26, 2009

### Staff: Mentor

Right. It doesn't make any sense.
You have a matrix A, and you're told that a given vector is an eigenvector. An eigenvectors is a special vector with regard to a particular matrix because the product Ax is a vector that is parallel to x. Any other vector makes a product, Ax that is not the same direction as x. For an eigenvector x, which is not allowed to be the zero vector, Ax = $\lambda$.

Historically the Greek letter $\lambda$ (lambda) has been used to represent an eigenvalue.

What you need to do is to solve for k in the equation A[3 2 5]^T = $\lambda$[3 2 5]^T. The exponent T means "transpose."

That's equivalent to (A - $\lambda$I)[3 2 5]^T = [0 0 0]^T.

Surely you can do that...

10. Mar 26, 2009

### tiny-tim

Hi sciencegirl1!

I can't help thinking that you try to make everything too difficult

go back to basics

Q: what is the definition of an eigenvector? A: if you multiply it by the matrix, the result is an exact multiple of the original

sooooo … what value of k makes -15 -11+3k 9 an exact multiple of -5 -2 3 ?

11. Mar 26, 2009

### sciencegirl1

So I do -15/3=-5
9/3=3
and -11+5k=-2 and therefore k=5/3 ??

12. Mar 26, 2009

### tiny-tim

oh, its 5k now, is it? :rofl:

well, that does make it a lot easier

nooo, it's not -2 (and how do you get 5/3 out of that anyway? )

Hint: you'll make less mistakes if you do it in stages …

so: stage 1: what is the eigenvalue?

13. Mar 26, 2009

### sciencegirl1

can you show me how you do it?

14. Mar 26, 2009

### tiny-tim

Nope!

You try it … what is the eigenvalue?

15. Mar 26, 2009

### Staff: Mentor

To repeat what I said in post 9,
We could show you how we do it, but we aren't going to. We're trying to help you do your work, and understand it.

16. Mar 26, 2009

### sciencegirl1

3 2 5]^T
why do you use this vector. The vector is -5 -2 3

and why use T?

Can you just show me how to do this simply?

17. Mar 26, 2009

### Avodyne

Some are using the vector and/or matrix you gave originally, not the corrected ones which are now buried in the middle of many posts.

You are given

$$A= \begin{pmatrix} 2 & 1 & -1 \\ 1 & 3 & k \\ -1 & -5 & -2 \end{pmatrix} \ \ \ \ \ \ x= \begin{pmatrix} -5 \\ -2 \\ 3 \end{pmatrix}$$

and are told that $x$ is an eigenvalue of $A$, which means that

$$Ax = \lambda x$$

where on the left, we matrix-multiply the matrix $A$ times the column-vector $x$ to get a new column vector; some of the components of this column vector will depend on $k$. On the right, we multiply each component of $x$ by the number $\lambda$ (which we don't know yet). We equate each component of the left side with the corresponding component of the right side. This gives us three equations in two unknowns ($k$ and $\lambda$). Presumably, the equations will be compatible; that is, there will be a solution. So, pick any two equations, solve them for $k$ and $\lambda$, and then, if you like, check that the third equation is also satisfied.

18. Mar 26, 2009

### sciencegirl1

I do know that Ax= [-15, -11+3k, 9]

Then I have to find \\lamda * [-5, -2, 3] = [-15, -11+3k, 9]

therefore

-5 \\lamda -2\\lamda + 3\\lamda =-15
and = 9
and = -11+3k

???

19. Mar 26, 2009

### lalligagger

You are trying to make the individual vector components match up, the sum of the components is fairly meaningless here.

20. Mar 26, 2009

### Avodyne

No, each component of each side is equal.

First of all,

$$\lambda\times \begin{pmatrix} -5 \\ -2 \\ 3 \end{pmatrix} = \begin{pmatrix} -5\lambda \\ -2\lambda \\ 3\lambda \end{pmatrix}$$

Then, the left side equals the right side:

$$\begin{pmatrix} -15 \\ -11+3k \\ 9 \end{pmatrix} = \begin{pmatrix} -5\lambda \\ -2\lambda \\ 3\lambda \end{pmatrix}$$

This gives us three equations, one for each component:

$$-15 = -5\lambda$$

$$-11+3k = -2\lambda$$

$$9 = 3\lambda$$