Prove: If λ is an eigenvalue of A, x is a corresponding eigenvector, and s is a scalar, then λ- s is an eigenvalue of A- sI and x is a corresponding eigenvector.
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Suppose λ is an eigenvalue of A with eigenvector x. Then
Ax = λx.
Note then that for any scalar s
(A - sI)x = Ax - sIx = λx - sx = (λ - s)x.
Hence λ - s is an eigenvalue of A - sI with eigenvector x.
Ax = λx.
Note then that for any scalar s
(A - sI)x = Ax - sIx = λx - sx = (λ - s)x.
Hence λ - s is an eigenvalue of A - sI with eigenvector x.