Prove that if X #0 is an eigenvalue of an invertible matrix A, then is an eigenvalue of A-¹ Proof: Suppose is an eigenvector of eigenvalue A, then Au = Xu. Since A is invertible, we can multiply both sides of Au = Xu by Au = λύ. so This implies that Since A #0 we obtain that Thus is an eigenvalue of 4-¹. Q.E.D.

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