## Cali_Native559 Group Title Suppose that delta is an eigenvalue of an invertible matrix A. Show that 1/delta is an eigenvalue of A inverse. one year ago one year ago

1. TuringTest Group Title

write the definition of an eigenvalue

2. Cali_Native559 Group Title

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

3. TuringTest Group Title

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?

4. TuringTest Group Title

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

5. Cali_Native559 Group Title

Thanks for the help

6. Cali_Native559 Group Title

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

7. TuringTest Group Title

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.

8. Cali_Native559 Group Title

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.

9. TuringTest Group Title

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

10. Cali_Native559 Group Title

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

11. TuringTest Group Title

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

12. Cali_Native559 Group Title

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

13. TuringTest Group Title

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

14. Cali_Native559 Group Title

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

15. Cali_Native559 Group Title

Because it's just I / I.

16. TuringTest Group Title

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

17. Cali_Native559 Group Title

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?

18. TuringTest Group Title

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

19. Cali_Native559 Group Title

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

20. TuringTest Group Title

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

21. Cali_Native559 Group Title

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.