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Linear Algebra - Diagonalizable and Eigenvalue Proof

by B_Phoenix
Tags: algebra, diagonalizable, eigenvalue, linear, proof
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B_Phoenix
#1
Apr7-09, 07:50 PM
P: 1
1. The problem statement, all variables and given/known data

"Let A be a diagonalizable n by n matrix. Show that if the multiplicity of an eigenvalue lambda is n, then A = lambda i"

2. Relevant equations



3. The attempt at a solution

I had no idea where to start.
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VKint
#2
Apr7-09, 08:10 PM
P: 111
Since [tex] A [/tex] is diagonalizable, we can choose some invertible matrix [tex] S [/tex] such that [tex] A = S D S^{-1} [/tex], where [tex] D [/tex] is diagonal and the diagonal entries of [tex] D [/tex] are the eigenvalues of [tex] A [/tex]. We can translate the assumption regarding the multiplicity of [tex] \lambda [/tex] into a statement about [tex] D [/tex], after which the result follows by using [tex] A = S D S^{-1} [/tex].


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