Question

How to prove this theorem? Let `A` be an `n×n` matrix and `lambda` an eigenvalue of `A`. Then `lambda+mu` is an eigenvalue of the matrix `M = A+muI`, where `I` is the `n × n` unit matrix?

Answer 1

If `lamda` is an eigenvalue of `bb(A)` with corresponding eigenvector `bb(ulv)` then we know that (by definition)

`bb(A) bb(ulv) = lamda bb(ulv)` ..... [A]

Now, let us determine if `lamda+mu` is an eigenvalue of the given matrix:

`bb(M) =bb(A)+mu bb(I)`

Consider:

`bb(M) bb(ul v) = (bb(A)+mu bb(I)) bb(ul v)`

`\ \ \ \ \ \ \ = bb(A)bb(ul v) + mu bb(I)bb(ul v) \ \ \` (by linearity)

`\ \ \ \ \ \ \ = lamdabb(ul v) + mu bb(ul v) \ \ \` (by [A] and identity properties)

`\ \ \ \ \ \ \ = (lamda + mu )bb(ul v) \ \ \` (by linearity)

Thus we have shown that `lamda+mu` is an eigenvalue of the matrix `bb(M)` with corresponding eigenvector `bb(ulv)`. QED.