Prove: If λ is an eigenvalue of A, x is a corresponding eigen - vector, and s is a scalar, then λ - s is an eigenvalue of A - sI and x is a corresponding eigenvector.

If λ is an eigenvalue of A and x is the corresponding eigenvector, then Ax=λx. Now, notice that (A-sI) x=Ax-sx=λx-sx = (λ-s) x. As x is different for zero we can affirm that (λ-s) is an eigenvalue of A-sI and x is the corresponding eigenvector

Step-by-step explanation:

Recall from the definition of eigenvalue: we say that the number (real or complex) λ is an eigenvalue of the matrix A if and only if there is a vector x, different from zero such that Ax=λx.

So, we only need to show that (A-sI) x = (λ-s) x.