Matrix with One Eigenvalue and Non-Eigenvector Eigenvectors

In summary, the matrix A has 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 \lambda
  • #1
snakebite
16
0
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 [tex]\lambda[/tex], but it is not a scalar matrix. Suppose [tex]\vec{v2}[/tex] is a nonzero vector which is not an eigenvector of A; show that [tex]\vec{v1}[/tex] = (A-[tex]\lambda[/tex])[tex]\vec{v2}[/tex] is an eigenvector of A. Also show that if P is the matrix with columns [tex]\vec{v1}[/tex] and [tex]\vec{v2}[/tex] then P^(-1)AP = [[tex]\lambda[/tex] 1
0 [tex]\lambda[/tex]]

The Attempt at a Solution



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

Thank you very much
 
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  • #2
You know that a matrix solves it's characteristic polynomial, right? What's the characteristic polynomial of a 2x2 matrix with one eigenvalue?
 
  • #3
well if A [ a b the the characteristic polynomial is
c d]
[tex]\lambda[/tex]^2 + [tex]\lambda[/tex](-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-[tex]\lambda[/tex]I)^2 but that leads to nowhere and certainly doesn't equal 0

(thanks for the quick reply)
 
  • #4
If the discriminant is zero that quadratic must factor into a square. The characteristic polynomial must be (x-lambda)^2.
 
  • #5
So in that case can we say that the minimal polynomial is therefore (x-[tex]\lambda[/tex])^2 and hence we have (A-[tex]\lambda[/tex])^2 is equal to 0,
therefore we get from the original equation that
(A-[tex]\lambda[/tex])[tex]\vec{v1}[/tex]=0 and therefore v1 is an eigenvector.
correct?
 
  • #6
snakebite said:
So in that case can we say that the minimal polynomial is therefore (x-[tex]\lambda[/tex])^2 and hence we have (A-[tex]\lambda[/tex])^2 is equal to 0,
therefore we get from the original equation that
(A-[tex]\lambda[/tex])[tex]\vec{v1}[/tex]=0 and therefore v1 is an eigenvector.
correct?

Correct.
 
  • #7
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 = [tex]\lambda[/tex] 1
0 [tex]\lambda[/tex]
 
  • #8
snakebite said:
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 = [tex]\lambda[/tex] 1
0 [tex]\lambda[/tex]

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

hence Av1 = A(A-[tex]\lambda[/tex]I)v2 and -[tex]\lambda[/tex]v1 = -[tex]\lambda[/tex](A-[tex]\lambda[/tex]I)v2

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

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
 
Last edited:
  • #10
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
yea in ur case we would have
B = [tex]\lambda[/tex] 1
0 [tex]\lambda[/tex]

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
snakebite said:
yea in ur case we would have
B = [tex]\lambda[/tex] 1
0 [tex]\lambda[/tex]

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.
 
  • #13
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
snakebite said:
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.
 
  • #15
great THANK YOU very much, i really appreciated your help :D
 
1.

What is an eigenvalue in a matrix?

An eigenvalue in a matrix is a scalar value that represents the scale at which a vector is multiplied by the matrix. In other words, it is a number that is associated with a specific vector in the matrix, such that when the vector is multiplied by the matrix, the resulting vector is parallel to the original vector.

2.

What does it mean when a matrix has only one eigenvalue?

When a matrix only has one eigenvalue, it means that all of the eigenvectors of the matrix are multiples of each other. This also means that the matrix is not diagonalizable, as there is only one distinct eigenvalue.

3.

How do you find the eigenvalue of a matrix?

To find the eigenvalue of a matrix, you must first find the characteristic polynomial of the matrix. Then, set the polynomial equal to zero and solve for the values of lambda that satisfy the equation. These values of lambda are the eigenvalues of the matrix.

4.

What is the significance of a matrix with only one eigenvalue?

A matrix with only one eigenvalue has the property that all of its eigenvectors are multiples of each other. This can be useful in certain applications, such as in physics and engineering, where symmetry and uniformity are important concepts.

5.

Can a matrix with one eigenvalue be diagonalized?

No, a matrix with one eigenvalue cannot be diagonalized because diagonalization requires the existence of distinct eigenvalues. Since a matrix with one eigenvalue only has multiples of the same eigenvector, it cannot be decomposed into a diagonal matrix and a transformation matrix.

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