Prove if lambda is an eigenvalue of a matrix A, then lambda + 1 is an eigenvalue of A + I.
Solution 1
Lets suppose $v $ is eigenvector of $A$ with eigenvalue $λ$, then, $Av=λv$, now take the same $v$ and apply $ A+I $ we have $(A+I)v=Av+v=λv+v=(λ+1)v $ so we found λ+1 is eigenvalue of the linear transformation $A+I$.
Solution 2
Hints:
There exists $\;0\neq v\in V\;$ s.t. $\;Av=\lambda v\;$ , and thus
$$(A+I)v=Av+v=\ldots$$
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TheOne15
Updated on March 03, 2020Comments
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TheOne15 over 3 years
Any advice on solving this question, I realize that in this case Ax = lambda(x) and that the determinant of the identity matrix is equal to 1. I don't know where to go from here though. I initially thought I could just replace A with lambda and take the determinant of I, but I don't feel like that works.
Took another attempt at it, is this correct?: A + I = lambda + 1 [substitute lambda for A] lambda + I = lambda + 1 [subtract lambda from both sides] I = 1 [substitute 1 for I into the original equation] A + 1 = lambda + 1 A = lambda
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Artem over 6 yearsStart with writing the definition of something being an eigenvalues.
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imranfat over 6 yearsI somehow feel that this question has been here before on MSE
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