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Suppose $\mu$ is an eigenvalue of the $B$ in Exercise $15,$ and that $\mathbf{x}$ is a corresponding eigenvector, so that $(A-\alpha I)^{-1} \mathbf{x}=\mu \mathbf{x}$ Use this equation to find an eigenvalue of $A$ in terms of $\mu$ and $\alpha .[\text {Note} : \mu \neq 0 \text { because } B \text { is invertible. }]$

$\lambda=\alpha+(1 / \mu)$ is an eigenvalue of $A$ with corresponding eigenvector $\mathbf{x}$

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 8

Iterative Estimates for Eigenvalues

Vectors

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So in this question, given that we know this, we basically want to find on item value off a in terms of you and so physical. Next, multiply that invest on this line So it's a left hand multiplication. So this is equal to a minus. Alpha identity, new X expanding brackets You have i a new X last. Why Alfa Identity has a new X. He was a constant second. Just threw that out and get new a X minus Phi Well, identity times about anything is just the same thing. So it's just how far new X So now we move a extra itself. So we have Yes, you could see one of them. You'll and ex Ellen plus Alfa New X factory out X. We have one ever knew what's one plus by Alfa you Thanks. Which can be written as I hope, plus one over New X so excellent. So what we've just done. We're just showing. Is that a X zika too Alfa plus buy one from you. Thanks. And so therefore, this is your wagon value

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