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Show that $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ is an eigenvalue of $A^{T} .\left[\text { Hint: Find out how } A-\lambda I \text { and } A^{T}-\lambda I\right.$ are related.

see the proof

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 1

Eigenvectors and Eigenvalues

Vectors

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Hello. One strand is given question. We have to show that Lambda is an Eigen value off on Lee. And only if, if and only if lambda is an icon. Value off, I suppose. Sophie. Right. So it is enough to show that the equation that they question a X is closed toe lamb decks has a non driver solution if and only if the equation a transpose X equals two Lambda X has a non travel solution. Right? So we can live. Right? This equation in the form off identity metrics that is 80 minus slammed. I yes, excessive cost, Ojeda. Okay, since that the dominant off this is request toe the determinant off transport C minus Lambda I is request to be dominate off a transports minus lambda making. Right here it I identity medics transposed right. So if the sequester lambda So the a minus slammed I don't suppose right as you know, it is a questo Determinate off a minor. Slammed I Yeah, so the bad. This is in both the equations a simultaneously in vegetable or simultaneously not invaluable. It follows that the both equations or simultaneously have known travel solution or simultaneously have on Lee the driver solution, which means that Lambda is an ideal Eigen value off a if and only if it is an Eigen value off transpose off, correct as it is.

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