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an eigenvalue of A, then 1/λ is an eigenvalue of

A^−1.

det((A-λI))

det((A-λI)^-1)

=det(A^-1 - λ^-1 * I)

=det(A

^{-1}-1/λ*I)

Is this enough to show that?

Another question I have is:

Let A be an n × n matrix. Show that A is not

invertible if and only if λ = 0 is an eigenvalue

of A.

Not sure how to approach this prob.

det(A-λI)=0

det(A-0I)=0

det(A)=0

But idk how to show what the problem is asking.