Question

Suppose that delta is an eigenvalue of an invertible matrix A. Show that 1/delta is an eigenvalue of A inverse.

Answer 1

write the definition of an eigenvalue

Answer 2

So I know what an eigenvalue is, but how would you write the definition of it, what do u mean by that?

Answer 3

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?

Answer 4

*I meant of course\[AI=\delta I\]multiply both sides by \(A^{-1}\) and what do you get?

Answer 5

Thanks for the help

Answer 6

So I do what u said and that should be the answer? That should prove what we need to prove?

Answer 7

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.

Answer 8

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.

Answer 9

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

Answer 10

So on the right side, we then get I * A^-1, but that's just still A^-1 right?

Answer 11

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

Answer 12

And one more thing, we also divide I from both sides too right, because that's how we get the one right?

Answer 13

I is the identity matrix, multiplying or dividing by it changes nothing, just like the scalar number 1

Answer 14

Right, so that'll just give us that one.

Answer 15

Because it's just I / I.

Answer 16

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

Answer 17

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?

Answer 18

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

Answer 19

All right that makes sense, and can u help out with one more problem related to this same stuff?

Answer 20

I can try, but my connection is horrible right now, I may only be able to pm you

Answer 21

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.