# Homework Help: Matrix with one eigenvalue

1. Mar 7, 2010

### snakebite

Hi, this is my 1st post here and i was wondering if I could get some help

Suppose wehave a 2x2 matrix A with one eigenvalue $$\lambda$$, but it is not a scalar matrix. Suppose $$\vec{v2}$$ is a nonzero vector which is not an eigenvector of A; show that $$\vec{v1}$$ = (A-$$\lambda$$)$$\vec{v2}$$ is an eigenvector of A. Also show that if P is the matrix with columns $$\vec{v1}$$ and $$\vec{v2}$$ then P^(-1)AP = [$$\lambda$$ 1
0 $$\lambda$$]

3. The attempt at a solution

I tried calculating (A-$$\lambda$$)$$\vec{v1}$$ to try and proove that it is equal to 0, however i end up with it being equal to (A-$$\lambda$$ I)^2$$\vec{v2}$$

Thank you very much

2. Mar 7, 2010

### Dick

You know that a matrix solves it's characteristic polynomial, right? What's the characteristic polynomial of a 2x2 matrix with one eigenvalue?

3. Mar 7, 2010

### snakebite

well if A [ a b the the characteristic polynomial is
c d]
$$\lambda$$^2 + $$\lambda$$(-a-d) + (ad -bc) but since we have 1 eigenvalue then the dicriminant must be 0 and hence (a+d)^2-4(ad - bc) = 0

I already tried doing this and calculating (A-$$\lambda$$I)^2 but that leads to nowhere and certainly doesn't equal 0

4. Mar 7, 2010

### Dick

If the discriminant is zero that quadratic must factor into a square. The characteristic polynomial must be (x-lambda)^2.

5. Mar 7, 2010

### snakebite

So in that case can we say that the minimal polynomial is therefore (x-$$\lambda$$)^2 and hence we have (A-$$\lambda$$)^2 is equal to 0,
therefore we get from the original equation that
(A-$$\lambda$$)$$\vec{v1}$$=0 and therefore v1 is an eigenvector.
correct?

6. Mar 7, 2010

Correct.

7. Mar 7, 2010

### snakebite

Ok, but what does this tell us about the matrix A, for example in the second part if we take
P=[v1 v2] = [v11 v21 then we have
v12 v22]
P^-1 = v22/det(P) -v21/det(P)
-v12/det(P) v11/det(P)
So how does P-1AP = $$\lambda$$ 1
0 $$\lambda$$

8. Mar 7, 2010

### Dick

You are thinking of this on too detailed a level. P^(-1)AP is just the matrix of A in the basis {v1,v2}. Write down Av1 and Av2 and deduce what the matrix must look like.

9. Mar 7, 2010

### snakebite

ok so we have v1 is eigenvector
hence Av1 = $$\lambda$$v1
hence (A-$$\lambda$$)v1 = 0 but we also have v1 = (A-$$\lambda$$I)v2

hence Av1 = A(A-$$\lambda$$I)v2 and -$$\lambda$$v1 = -$$\lambda$$(A-$$\lambda$$I)v2

hence by calculating it out i find (A-$$\lambda$$I)v1 = (A-$$\lambda$$)^2 v2
To get v1 =(A-$$\lambda$$)v2 which is equivalent to A = v1 + $$\lambda$$v2

but then i m stuck, I dont quite understand how to get A from this because we'll always have it in terms of v1 and v2

Last edited: Mar 7, 2010
10. Mar 7, 2010

### Dick

You WANT to have it in terms of v1 and v2, P^(-1)AP is the matrix of the linear transformation A IN THE BASIS {v1,v2}. You have Av1=lambda*v1 and Av2=v1+lambda*v2, right? Suppose e1=(1,0) and e2=(0,1) were the usual basis and I told you Be1=lambda*e1 and Be2=e1+lambda*e2. Then you could probably tell me the entries of the matrix B, right? What would they be?

11. Mar 7, 2010

### snakebite

yea in ur case we would have
B = $$\lambda$$ 1
0 $$\lambda$$

but in our case we don't have the components of v1 and v2 we can just right A in terms of v11,v12,v21,v22?
so would P-1AP cancel out and we're sort of left in the basis e1 e2 to get B?

12. Mar 7, 2010

### Dick

Hmm. Not sure I'm getting through. Since you KNOW how A acts in the basis {v1,v2} you KNOW what the matrix of A is in the basis {v1,v2}. It's [[lambda,1],[0,lambda]]. In contrast, you DON'T know what the original entries of A were. The thing you KNOW is called P^(-1)AP.

13. Mar 7, 2010

### snakebite

yea i think I get it now, after doing the change of basis on A with v1 and v2, then doing P(-1)AP is equivalent to the matrix B in your earlier statement since A and B are the same but in different bases
correct?

14. Mar 7, 2010

### Dick

Yes, A and P^(-1)AP are the same transformation in two different bases.

15. Mar 7, 2010

### snakebite

great THANK YOU very much, i really appreciated your help :D