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Matrix with one eigenvalue

  • Thread starter snakebite
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  • #1
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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
 

Answers and Replies

  • #2
Dick
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You know that a matrix solves it's characteristic polynomial, right? What's the characteristic polynomial of a 2x2 matrix with one eigenvalue?
 
  • #3
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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
Dick
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If the discriminant is zero that quadratic must factor into a square. The characteristic polynomial must be (x-lambda)^2.
 
  • #5
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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
Dick
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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
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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
Dick
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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
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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 dont quite understand how to get A from this because we'll always have it in terms of v1 and v2
 
Last edited:
  • #10
Dick
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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
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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
Dick
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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
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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
Dick
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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
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great THANK YOU very much, i really appreciated your help :D
 

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