Let $V$ be a finite dimensional vector space over $\mathbb{C}$ and suppose the linear transformation $T:V\rightarrow V$ has eigenvalue $\lambda_0$ with algebraic multiplicity $n_0=3$.

(a) Show that the geometric multiplicity $d_0$ of the eigenvalue $\lambda_0$ satisfies $d_0\leq 3$.

(b) Suppose $\lambda_0$ is the only eigenvalue of $T$. Is it possible that $dimV=4$?

(a) I know that the geometric multiplicity can't exceed the algebraic multiplicity however I do not know how to show this for a $n\times n$ matrix.

(b) If there is only one eigenvalue then the dimension of the matrix is equal to the algebraic multiplicity. Therefore $dim V$ can not be equal to $4$.

Could any shed some light on this for me. Any help would be greatly appreciated.


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