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On page 186 there is a lemma that says:

Suppose that $T\alpha=c\alpha$. If f is any polynomial, then $f(T)\alpha=f(c)\alpha$

My proof was the following:

I was trying to do induction over $deg(f)=k$.

On k=1 I have that: $(fT)\alpha=(a_0I_d+a_1T)\alpha=a_0\alpha+a_1c\alpha=f(c)\alpha$

Then I assume true for k, i.e $(a_0I_d+a_1T+...+a_nt^n)\alpha=(a_0+a_1c+...+a_nc^n)\alpha$

I have to prove it holds for k+1

Here is what I don't know if I am getting right: $f(T)(\alpha)=a_0I_d\alpha+a_1T\alpha+...+a_nT^n\alpha+a_{n+1}T^{n+1}\alpha=a_0\alpha+a_1c\alpha+...+a_nc\alpha+a_{n+1}c^{n+1}\alpha$

And I don't know what else to do, I found another question related to this one but the solution is not quite completed.

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  • $\begingroup$ It is called the "spectral mapping theorem". $\endgroup$
    – Jean Marie
    Nov 1 '19 at 19:05
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Let's assume the proposition holds for all polynomials with degree less than $n$. Then we have

$$f(T)(\alpha)=a_0I_d\alpha+a_1T\alpha+...+a_nT^n\alpha+a_{n+1}T^{n+1}\alpha=\\ (a_0I_d\alpha+a_1T\alpha+...+a_nT^n\alpha)+(a_{n+1}T^{n+1}\alpha)=\\ (a_0\alpha+a_1c\alpha+...+a_nc^n\alpha)+a_{n+1}T(T^{n}\alpha)=\\ (a_0\alpha+a_1c\alpha+...+a_nc^n\alpha) +a_{n+1}T(c^{n}\alpha)=\\ (a_0\alpha+a_1c\alpha+...+a_nc^n\alpha) +a_{n+1}(c^{n+1}\alpha) $$

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Your $T$ is a matrix, $\alpha$ a vector and $c$ a scalar (+eigenvalue)? If yes, then you don't need this induction over degree of $f$. You have only to observe that

  1. if $A,B$ are arbitrary matrices and $a,b$ scalars then $(aA+bB)\alpha= aA\alpha +bB \alpha$ (so the action is linear) This statement can be inductively exitended to finite sum $\sum_{i=1}^na_i A_i$

  2. and if $\alpha$ is an eigenvector of $A$ with eigenvalue $c$ then $A^n\alpha=c^n\alpha$.

Combine 1 and 2 gives the desired result

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The proof uses only the linearity of $T.$ First observe that $T^k\alpha=c^k\alpha.$ This can be proved by induction: $$T^k\alpha = T(T^{k-1}\alpha)=T(c^{k-1}\alpha)=c^{k-1}T\alpha = c^{k-1}c\alpha = c^k\alpha. $$ Now let $f(x)=\sum_{k=0}^n a_kx^k$ be a polynomial. Then $$f(T)\alpha = \sum_{k=0}^n a_kT^k\alpha = \sum_{k=0}^n a_kc^k\alpha = f(c)\alpha.$$

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Try induction to prove that $T^n (\alpha)=c^n(\alpha)$. It can be proved very easily(note that $T^n (\alpha)=T (T^{n-1} (\alpha)$).

Now let $f (x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0$. Then \begin{align} f (T)(\alpha )&=(a_nT^n+a_{n-1}T^{n-1}+\cdots +a_0I)(\alpha )\\ &=a_nT^n(\alpha )+a_{n-1}T^{n-1}(\alpha )+\cdots +a_0I(\alpha )\\ &=a_nc^n(\alpha )+a_{n-1}c^{n-1}(\alpha )+\cdots +a_0(\alpha )\\ &=(a_nc^n+a_{n-1}c^{n-1}+\cdots +a_0)(\alpha )=f (c)(\alpha) \end{align}

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