Showing A^-1 has eigenvalues reciprocal to A's eigenvalues

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SUMMARY

The discussion centers on proving that the eigenvalues of a nonsingular matrix A^-1 are the reciprocals of the eigenvalues of A using the concept of similar matrices. The user Ian initially establishes that A is nonsingular if 0 is not an eigenvalue of A, leveraging the determinant relationship det(A^-1) = 1/det(A). However, Ian struggles to apply the concept of similar matrices to complete the proof, questioning how to relate A and A^-1 effectively. The conversation highlights the need for a clear understanding of eigenvalue properties and matrix similarity in linear algebra.

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Homework Statement


If A is nonsingular, prove that the eigenvalues of A-1 are the reciprocals of the eigenvalues of A.

*Use the idea of similar matrices to prove this.

Homework Equations


det(I[tex]\lambda[/tex] - A) = 0
B = C-1AC (B and A are similar, and thus have the same determinants)


The Attempt at a Solution


At first I showed that A is nonsingular iff 0 is not an eigenvalue of A. To do this I just used the fact that det(A-1) = 1/det(A) and that if lambda was 0, then we'd have det(A) = 0, which would mean A-1 is undefined. If lambda isn't 0, then we have det(I[tex]\lambda[/tex] - A) = 0, which tells us A is nonsingular.

As for the other proof, I'm convinced I have to use the idea of similar matrices, because the book I am using is focusing on them right now, and these exercises are relating to them.
But A-1 and A are not the same linear transformation (unless A = I), so they are not similar. Therefore I cannot use the formula: B = C-1AC

So, I am lost on how to do this proof using the idea of similar matrices. Anyone know how I could accomplish this?

-Ian
 
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Are you required to use "similar matrices"? If [itex]\lambda[/itex] is an eigenvalue of A, with eigenvector v, then [itex]Av= \lambda v[/itex]. Now take [itex]A^{-1}[/itex] of both sides of that equation.
 
Well, if I didn't then I wouldn't be learning anything about them. I mean, most of these proofs I don't see how to use similar matricies to figure them out... There's a huge number:

a) If A can be diagonalized, then its eigenvalues are distinct. I said this is false. I just used a counter example in the form of the 3x3 identity matrix. How would I use similar matricies for this??

b) If all eigenvalues of A are equal to 2, then B-1AB = 2I for some nonsingular B. I see how this one relates to similar matricies, but have no clue on how to prove it.

c) if all eigenvalues of A are zero, then A is the zero matrix. (No clue how to prove)

d) If all eigenvalues of A are zero, then A is similar to the zero matrix.

e) If B = C-1AC then B3 = C-1A3C (I think I can prove this using determinants)

f) If A is nonsingular, then all its eigenvalues are nonzero. - This is true, but it easily follows from my proof I presented earlier. I did not use similar matricies.

g) There exist 3x3 real matricies with no real eigenvalue (I don't see how similar matrices comes into play here)


I have a test today, so that is why I am trying to figure this stuff out. Any idea on how similar values would apply to any of these proofs? Any idea how I would prove some of them?
-Ian
 

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