Eigenvalue Proof

  • #1

Homework Statement



Prove that similar square matricies have the same eigenvalues with the same algebraic multiplicities.

Homework Equations


C^-1PC=Q



The Attempt at a Solution


Am I supposed to show that (P-[tex]\lambda[/tex]I)x=(C^-1PC-[tex]\lambda[/tex]I)x?
 

Answers and Replies

  • #2
Dick
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The eigenvalues are the roots of the characteristic polynomial. Can you show Q=C^(-1)PC and P have the same characteristic polynomial? I.e. det(Q-xI)=det(P-xI)?
 
Last edited:
  • #3
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Am I supposed to show that (P-[tex]\lambda[/tex]I)x=(C^-1PC-[tex]\lambda[/tex]I)x?
Well, that's equivalent to C-1PC=Q, isn't it?



solution:
What you want to do is to show, that they have the same char polynomial (up to scalarmultiplication). But that's easy since [tex]det(\mathbf{P} -\lambda \mathbf{I})=det(\mathbf{C^{-1}QC} -\lambda \mathbf{C^{-1}C})=det(\mathbf{C^{-1}}(\mathbf{Q} -\lambda\mathbf{I} )\mathbf{C})[/tex]
Now use det(AB)=det(A)*det(B) and your done.
cheers
 
  • #4
Dick
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Well, that's equivalent to C-1PC=Q, isn't it?



solution:
What you want to do is to show, that they have the same char polynomial (up to scalarmultiplication). But that's easy since [tex]det(\mathbf{P} -\lambda \mathbf{I})=det(\mathbf{C^{-1}QC} -\lambda \mathbf{C^{-1}C})=det(\mathbf{C^{-1}}(\mathbf{Q} -\lambda\mathbf{I} )\mathbf{C})[/tex]
Now use det(AB)=det(A)*det(B) and your done.
cheers
Hi, 1777-1855. The forum rules prohibit posting complete solutions (i.e. doing the problem for the poster). That's awfully close. Try and give a hint first and see if they can figure out the rest themselves, ok?? It's more fun that way and they learn more.
 
  • #5
The eigenvalues are the roots of the characteristic polynomial. Can you show Q=C^(-1)PC and P have the same characteristic polynomial? I.e. det(Q-xI)=det(P-xI)?
So if I prove those two determinants are equal, it necessarily follows that the two matricies have the same eigenvalues and algebraic multiplicities, because the characteristic polynomials are the same?
 
  • #6
Dick
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So if I prove those two determinants are equal, it necessarily follows that the two matricies have the same eigenvalues and algebraic multiplicities, because the characteristic polynomials are the same?
Yes, those two determinants are actually polynomials in x, they are the characteristic polynomials. Their roots (and the multiplicities of them) are the eigenvalues and their algebraic multiplicities.
 

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