Suppose that A is an eigenvalue of A that is Ax Ax for some

Suppose that A is an eigenvalue of A, that is, Ax = Ax, for some nonzero vector x. Show that this x is an eigenvector of B = A - 7I and find the eigenvalue.

Solution

If ₁ is an eigenvalue of A , then the ₁ satisfies the equation Ax = ₁ x, or, equivalently, the equation (A – ₁I)x = 0 has a solution for some non-zero x, say x₁ . Now, if x₁ is an eigenvector of B, then x₁ is a non-zero solution of (B – I)x₁ = 0 for some scalar . Since B = A – 7I, we must have ( A – 7I –I)x₁ = 0 or, { A – ( + 7)I}x₁ = 0, which means that + 7 is an eigenvalue of A. We know that if ₁ is an eigenvalue of A , then (A – ₁ I)x₁ = 0 for some non-zero x₁ . Now, if { A – ( ₁+ 7)I}x₁ is also = 0, then ₁+ 7 is also an eigenvalue of A. Conversely, if ₁+ 7 is an eigenvalue of A, and x₁ is an eigenvector of A corresponding to the eigenvalue ₁ , then x₁ is an eigenvector of B = A – 7I corresponding to the eigenvalue ₁ + 7.