anonymous
  • anonymous
Suppose that delta is an eigenvalue of an invertible matrix A. Show that 1/delta is an eigenvalue of A inverse.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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TuringTest
  • TuringTest
write the definition of an eigenvalue
anonymous
  • anonymous
So I know what an eigenvalue is, but how would you write the definition of it, what do u mean by that?
TuringTest
  • TuringTest
Start with showing what an eigenvalue is: a scalar, \(\delta\) in this case, such that\[AI=\delta A\]Now, since we can assume \(A\) has an inverse \(A^{-1}\), multiply both sides by the inverse. What do you get?

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TuringTest
  • TuringTest
*I meant of course\[AI=\delta I\]multiply both sides by \(A^{-1}\) and what do you get?
anonymous
  • anonymous
Thanks for the help
anonymous
  • anonymous
So I do what u said and that should be the answer? That should prove what we need to prove?
TuringTest
  • TuringTest
Not quite, you have not demonstrated what we have set out to prove. why don't you write out what you get by multiplying the definition of the eigenvalue\[AI=\delta I\]by the inverse of the matrix, \(A^{-1}\) and show me what you get? I will help you from there.
anonymous
  • anonymous
So this is what I did, so here's my work: A^-1 AI = delta*I*A^-1 I = delta*I*A^-1 So I that's what I got, I'm pretty sure the right side can't be simplified any further, so if you could help me with the rest, that'd be great.
TuringTest
  • TuringTest
divide both sides by delta and you're done check it out, you get the definition of the eigenvalue, with the eigenvalue of A^-1 as 1/delta
anonymous
  • anonymous
So on the right side, we then get I * A^-1, but that's just still A^-1 right?
TuringTest
  • TuringTest
yes\[AI=\delta I\]\[A^{-1}AI=II=I^2=I=A^{-1}\delta I\]remember that scalars are commutative so we can move delta around\[\delta A^{-1}I=I\]\[A^{-1}I=\frac1\delta I\]which is the definition of the eigenvalue
anonymous
  • anonymous
And one more thing, we also divide I from both sides too right, because that's how we get the one right?
TuringTest
  • TuringTest
I is the identity matrix, multiplying or dividing by it changes nothing, just like the scalar number 1
anonymous
  • anonymous
Right, so that'll just give us that one.
anonymous
  • anonymous
Because it's just I / I.
TuringTest
  • TuringTest
matrices do not become scalars matrix division is defined as multiplication by its inverse the inverse of the identity matrix is still the identity matrix, so\[II^{-1}=II=I^2=I\]still the identity matrix, not the scalar number 1 big difference, gotta get that straight in linear algebra
anonymous
  • anonymous
Ok, so I think I see, ok so in the end we get this: A^-1 * I = 1/delta * I, so then the I's just cancel out since they're on both sides and that is what gives us A^-1 = 1/delta right?
TuringTest
  • TuringTest
you don't need to cancel the I's, it's just like times 1 if I had 1x=1y would I need to cancel the 1's ? a matrix times the identity matrix is itself, so\[AI=AI^{-1}=A\] in the definition of the eigenvalue, I is written explicitly to show that the scalar eigenvalue lambda is multiplying into a matrix\[A=\lambda I\]. if the I's cancel you get\[A=\lambda\]that's wrong because the thing on the right is a scalar and the thing on the left is not Just leave your last line as\[A^{-1}=\frac1\delta I\]and that makes clear the properties of the eigenvalue we were looking for
anonymous
  • anonymous
All right that makes sense, and can u help out with one more problem related to this same stuff?
TuringTest
  • TuringTest
I can try, but my connection is horrible right now, I may only be able to pm you
anonymous
  • anonymous
Ok, so here's the question: If matrix A has an inverse A^-1, use equation 2: Av = lambda*v, to show that A^-1 has the same eigenvectors as A. Determine a relationship between the eigenvalues of A and A^-1. Illustrate with a suitable example.

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