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Wikipedia claims that

"Given an n×n matrix A.... both algebraic and geometric multiplicity are integers between (including) 1 and n."

But how can the geometric multiplicity possibly be n? Since $(A-\lambda I)$ is a square matrix (as opposed to a matrix with more columns than rows), each of A's eigenspaces $Nul (A-\lambda I)$ has at most $(n-1)$ dimensions, isn't it?

I.e. The geometric multiplicity of an eigenvalue must be a number between between $1$ and $(n-1)$, right?

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up vote 7 down vote accepted

If $A=\lambda I$, then $A-\lambda I=0$, and the null matrix obviously has kernel dimension $n$.

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And, of course, that's the only example where the geometric multiplicity is $n$. – Robert Israel Aug 7 '12 at 7:33
@celtschk Oh yes, of course! Thanks!! Thanks, Robert, too. – Ryan Aug 7 '12 at 7:42

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