Matrix with One Eigenvalue and Non-Eigenvector Eigenvectors

[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]

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

 

[Dick]

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

 

[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-\lambdaI)^2 but that leads to nowhere and certainly doesn't equal 0

(thanks for the quick reply)

 

[Dick]

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

 

[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?

 

[Dick]

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?

Correct.

 

[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

 

[Dick]

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

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.

 

[snakebite]

ok so we have v1 is eigenvector
hence Av1 = \lambdav1
hence (A-\lambda)v1 = 0 but we also have v1 = (A-\lambdaI)v2

hence Av1 = A(A-\lambdaI)v2 and -\lambdav1 = -\lambda(A-\lambdaI)v2

hence by calculating it out i find (A-\lambdaI)v1 = (A-\lambda)^2 v2
To get v1 =(A-\lambda)v2 which is equivalent to A = v1 + \lambdav2

but then i m stuck, I don't quite understand how to get A from this because we'll always have it in terms of v1 and v2

 

[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?

 

[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?

 

[Dick]

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?

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.

 

[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?

 

[Dick]

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?

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

 

[snakebite]

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