Proof of $f(z)=(z-\lambda_{2})\cdot\cdot\cdot(z-\lambda_{m})$ is equal to zero when applied to operator $T$ with $n$ distinct eigenvalues.

Question

$\mathbb{F}=\mathbb{C}$, Dim$V=n$ and $T\in\mathcal{L}(V)$ has $n$ distinct eigenvalues $\lambda_{1},..., \lambda_{n}$. Let $f$ be the polynomial defined by $f(z)=(z-\lambda_{2})\cdot\cdot\cdot(z-\lambda_{m})$.

Prove that $f(T)=0$

My proof:

First, notice that $f$ is defined by the standard factorization of a polynomial with complex coefficients.

$ \begin{align} \longrightarrow f(T)\vec{v}&=(T\vec{v}-\lambda_{1}\vec{v})\cdot\cdot\cdot(T\vec{v}-\lambda_{n}\vec{v})\\ &=a_{o}I\vec{v} + a_{1}T\vec{v}+\cdot\cdot\cdot+a_nT^n\vec{v}\\ &=(a_{o}I + a_{1}T+\cdot\cdot\cdot+a_nT^n)\vec{v}\\ &=(T-\lambda_1I)\cdot\cdot\cdot(T-\lambda_nI)\vec{v} \end{align}$

Isolating the last term, $(T-\lambda_nI)\vec{v}$ which is equal to $0$ by the definition of eigenvalues/eigenvectors we see that $f(T)=0$.

I am concerned with the lambdas being incorrectly placed/distributed. I'm really bad at the algebra part of linear algebra. Thanks for the check on that and anything else.

Do you ask about the m? Otherwise, that is the problem statement as given to me. — rocksNwaves, May 02 '17 at 14:46
Yes, the $m$. There's an $n$ in the rest of the question, not $n$. Also do you really want to start the product with $z-\lambda_1$? — Angina Seng, May 02 '17 at 17:15

Proof of $f(z)=(z-\lambda_{2})\cdot\cdot\cdot(z-\lambda_{m})$ is equal to zero when applied to operator $T$ with $n$ distinct eigenvalues.

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