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.

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