Linear Algebra  Diagonalizable and Eigenvalue Proofby B_Phoenix Tags: algebra, diagonalizable, eigenvalue, linear, proof 

#1
Apr709, 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. 



#2
Apr709, 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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