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The question is: Suppose that T is a self-adjoint operator on a finie-dimensional inner product space and that 2 and 3 are the only eigenvalues of T. Prove that $T^2-5T+6I = 0$.

To prove this question, can I take the inner product of $(T^2-5T+6I)v$ and $Tv$, where $Tv = 2v$ and $v \neq 0$, and then show that the inner product of those two vectors is equal to $0$? Since $v \neq 0$, would that imply that $T^2-5T+6I = 0$? Or am I coming at this question completely wrong?

Thanks.

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    Must have 2 roots, they must factor as $(x-\lambda_1)(x-\lambda_2) = x^2 -x(\lambda_1+\lambda_2) + \lambda_1 \cdot \lambda_2$. $2 \cdot 3=6$ and $2+3=5$. – 2017-02-19
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    One possible way to do this is to check that it holds on a basis of eigenvectors for $T$. – 2017-02-19

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As $T$ is self-adjoint, it is diagonizable, so the minimal polynomial splits into linear factors and has only simple roots. This together with the fact that the only eigenvalues are $2$ and $3$ implies that the minimal polynomial is $(X-2)(X-3) = X^2-5X+6$. By the definition of the minimal polynomial, we have $T^2-5T+6I=0$.

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    Very nice and complete answer. +1 – 2017-02-19