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

i can explain geometrically, but i havent done this in a while so the algebra may be slightly dodgy

let B have eigenvalues \[\lambda _a \text{ , } \lambda _b \text{ , } \lambda _c\]

eigenvalues give us the scale factor for the associated eigenvector under the transformation B , ei
\[Bv_a = \lambda _av_a\]
so

\[B^2v_a = BBv_a = B\lambda _av_a = \lambda_a^2 v_a\]

so eigenvalues of B^2 are \[\lambda _a^2 \text{ , } \lambda _b^2 \text{ , } \lambda _c^2\]

(to be continued)

Answer 2

ok i typed the rest of my answer but then openstudy died on me

Answer 3

basically start with eigenvalues of B and then use that to find eigenvalues of B^2

then use the property:

\[\det(B^2 - \lambda I) = 0\]
substituting in the eigenvalues of B^2 as lambda

then with B^2 + 1 , and the eigenvalues as unknowns
\[\det((B^2 +I)- \lambda I) = 0\]\[\det(B^2 -(\lambda -1)I) = 0\]

can you see eigenvalues of B^2 + 1 are one more than eigenvalues of B^2 ?

Answer 4

now again consider the kinda definition of eigenvalues, given matrix M and its respective eigenvector and eigenvalue
\[v_m \text{ , } \lambda _ m\]
\[Mv_m = \lambda_m v_m\]

now lets multiply by inverse M

\[M^{-1}Mv_m = v_m = \lambda_mM^{-1} v_m \]

using that last bit
\[\frac{1}{\lambda_m}v_m = M^{-1}v_m\]


so eigenvalue of inverse M is reciprocal of eigenvalue of M

therefore eigenvalue of (B^2 + I)^-1 = reciprocal of eigenvalue of B^2 + 1