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 June 8th, 2012, 11:07 AM #1 Member   Joined: Jan 2012 Posts: 51 Thanks: 1 Eigenvalues and Eigenvectors Let A be an nxn invertible matrix and let ? be an eigenvalue of A. Show that 1/? is an eigenvalue of A^-1. Can someone help me with this proof ? Sorry about my english, I'm brazilian.
 June 8th, 2012, 11:38 AM #2 Member   Joined: Jan 2012 Posts: 51 Thanks: 1 Re: Eigenvalues and Eigenvectors Well, i think i got the idead of the exercise above, but I'll let it here to anyone who wants to solve. Now I'm having trouble with this one: Let A be an nxn matriz with the propertie of A=A². Show that if ? is an eigenvalue, then it is 0 or 1.
June 8th, 2012, 03:01 PM   #3
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Re: Eigenvalues and Eigenvectors

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 Originally Posted by bonildo Let A be an nxn invertible matrix and let ? be an eigenvalue of A. Show that 1/? is an eigenvalue of A^-1. Can someone help me with this proof ? Sorry about my english, I'm brazilian.
I am using | | to mean determinant.
? is an eigenvalue means |A - ?I| = 0. Let B = A^(-1).

|I - ?B| = |BA - ?B| = |B(A - ?I)| = |B||A - ?I| = 0.

Therefore 1/? is an eigenvalue for B.

June 8th, 2012, 04:41 PM   #4
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Re: Eigenvalues and Eigenvectors

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Quote:
Originally Posted by mathman
Quote:
 Originally Posted by bonildo Let A be an nxn invertible matrix and let ? be an eigenvalue of A. Show that 1/? is an eigenvalue of A^-1. Can someone help me with this proof ? Sorry about my english, I'm brazilian.
I am using | | to mean determinant.
? is an eigenvalue means |A - ?I| = 0. Let B = A^(-1).

|I - ?B| = |BA - ?B| = |B(A - ?I)| = |B||A - ?I| = 0.

Therefore 1/? is an eigenvalue for B.
Sry, but i did'nt got your proof. Why are you evaluating |I - ?B| ? Should'nt it be |I? - B| ? And why |B||A - ?I| = 0 implies 1/? is an eigenvalue for B ??? Sry, I've just initialized in the world of demonstrations .
? is an eigenvalue of A. |I? - B| = 0 is the same as |I - (1/?)B| = 0.
The proof shows 1/? is an eigenvalue of B.

 June 8th, 2012, 07:02 PM #5 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Eigenvalues and Eigenvectors I wouldn't bother with the determinant. Simply note that if v is an eigenvector of A with eigenvalue $\lambda$, then $Av= \lambda v$. Now multiply both sides by $A^{-1}$.

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### let be an eigenvalue of an invertible matrix a show that

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