Let $T$ be an operator on a complex Vector space $V$. Then, the algebraic multiplicity of an eigen value is equal to $\dim ~null~ (T - \lambda I)^{\dim V}$

Which means, if we obtain the upper triangular matrix form for this operator, the number of times , the eigen value $\lambda$ appears on the diagonal is $\dim ~null~ (T - \lambda I)^{\dim V}$.

Attempt for a Proof:

Let us say the eigen value $\lambda$ appears $k$ times on the diagonal. Now, the matrix of $T-\lambda I$ will have $k$ zeroes on it's main diagonal.

Also, $T-\lambda I$ will be a nilpotent matrix and $(T-\lambda I)^{\dim V}$ is definitely a null matrix.

How do I proceed and make a link from here to prove the result above? ( Please don't try using induction )

Thank you very much for your help in this regard.

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    $\begingroup$ @MatthewLeingang Sorry and Yes. I have corrected it now. $\endgroup$ – MathMan Jan 30 '15 at 20:18
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    $\begingroup$ Algebraic multiplicity is just the degree of the linear term corresponding to $\lambda$ in the characteristic polynomial. Every matrix is similar to an upper triangular matrix and the determinant is invariant under similarity. Is this not enough to prove the claim? $\endgroup$ – user113529 Jan 30 '15 at 20:23

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