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Linear Algebra
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Is the geometric multiplicity of an eigenvalue of the matrix A equal to the number of free variables the matrix (A-e*I) has, where e is the eigenvalue and I is the identity matrix?
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I don't think so, consider \[A=\begin{bmatrix}1&2\\0&1\end{bmatrix} \implies \lambda_{1,2} = 1\] with multi 2. But \[(A-\lambda_1\mathbb{I})=\begin{bmatrix}0&2\\0&0\end{bmatrix}\] has only one free variable.
Good counter example.
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