If A has n linearly independent eigenvectors relating to the eigenvalue 1, then A is the identity matrix.
I'm little puzzled with this - any ideas? Is the above statement true?
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The matrix can be diagonalized in the form $A=PIP^{-1}$. – Li Chun Min Apr 30 '17 at 01:44
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Hint: you are given that the geometric multiplicity of eigenvalue $1$ is $n\,$. The algebraic multiplicity must then be at least $n\,$, therefore must be exactly $n\,$, so the characteristic polynomial is $(x-1)^n$.
dxiv
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Could you tell me why algebraic multiplicity is at least $n$?If $A$ is $n\times n$ matrix can't its algebraic multiplicity be less than $n$? Please clarify my doubt? Thank you for the nice answer. – mathscrazy May 01 '17 at 05:59
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@mathscrazy The geometric multiplicity of an eigenvalue can never be larger than the algebraic multiplicity, see Examples for proof of geometric vs. algebraic multiplicity for example. Putting that the other way around, the algebraic multiplicity must be at least equal to the geometric multiplicity (or possibly larger). Here, the geometric multiplicity is given to be $n$, so the algebraic one must be $\ge n,$, and it cannot be $\gt n$ since it's (implied to be) an $n$-by-$n$ matrix. – dxiv May 01 '17 at 06:08
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1Thanks for your reply. I completely understood and up voted your answer. :) – mathscrazy May 01 '17 at 06:14
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1This is the problem with the most of the downvoters. At least mention the reason or ask for any queries before downvoting. – mathscrazy May 16 '17 at 01:56
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This means that $Av_i=v_i$ for all $i$, where your basis is $v_1,\ldots,v_n$. As all vectors $u$ are linear combinations of the $v_i$ then $Au=u$ for all $u$.
Angina Seng
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