# Doubt about a lemma in Hoffman and Kunze's *Linear Algebra* section 6.2

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.

• It is called the "spectral mapping theorem". Nov 1 '19 at 19:05

Let's assume the proposition holds for all polynomials with degree less than $$n$$. Then we have

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

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

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 $$f(T)\alpha = \sum_{k=0}^n a_kT^k\alpha = \sum_{k=0}^n a_kc^k\alpha = f(c)\alpha.$$

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}