Question

If I have a 3 by 3 matrix (B) with eigenvalues 0, 1, and 2, can someone tell me how you can find the eigenvalues of (B^2 + I)^-1 where I is the identity matrix.

Answer 1

Ok B has a eigendecomposition
B=P D Inv(P), where D is a diagonal matrix with the eigenvalues of B.
B^2 = P D^2 Inv(P). (Check this).
B^2 + I = P D^ Inv(P) + I
Now you can express I as P*I*Inv(P).

I think you can figure out the rest yourself.

Answer 2

Just in case you want the rest of the answer since i won't be awake (its 5am here muahahaha), or will be in class: Then

B^2+ I = P D^2 Inv(P) + P I Inv(P)
= P (D^2+I) Inv(P)
Using
Inv(ABC)(ABC) = I => Inv(C)Inv(B)Inv(A)ABC //derivation of 3 matrix inverse

Inv(B^2+I) =P Inv(D^2+I) Inv(A)

Now the eigenvalues of this new matrix will be on the diagonal matrix Inv(D^2-I). To find the inverse of a diagonal, if i'm not wrong you simply take the reciprocal of each term. Don't quote me on that one because my memory is terrible.

D^2+I has the diagonals 1,4,9 + 1,1,1 = 1,5,10. Then the eigenvalues of the inverse of this will be 1, 1/5, 1/10.

Answer 3

Actually it'd be 0, 1, 4 + 1 + 1 +1 for D^2 +I. Then the eigenvalues of the inverse would 1, 1/2, 1/5. The rest of your logic is sound. Thanks!

Answer 4

Sorry I meant, 0, 1, 4 + 1, 1, 1

Answer 5

Oh, and you're right about the inverse of a diagonal matrix. It's what you said it would be since you get the Identity matrix after multiplying them together.

Answer 6

OOPS lol. I read the eigenvalues to be 1,2 and 3. X.X

Answer 7

We know that if \[\lambda\] is an eigenvalue for the matrix A then \[\lambda ^{n}\] is an eigenvalue for A^n so for B^2 we have \[\lambda\]^2 as my eigenvalue.
for B^2 + I we have \[\lambda\]^2 + 1 as my eigenvalue and for the inverse of it we jst inverse the eigenvalue i.e. \[1\div (\lambda ^{2}+1)\]
therefore the values are 1, 1/2,1/5