Vtyjkyuk

For each positive integer $k$ we write $T^k = Tcircdotscirc T$, the $k$-fold composition of the map $T$. With this notation, we can define $p(T) := a_0mathrmId_V + a_1T +dots + a_nT^n$ whenever $p(x) = a_0 + a_1x +dots + a_nx^n$ is a polynomial with complex coefficients
$a_0,dots, a_n$.

a) Show that if f $λ$ is any eigenvalue of $T$ and $p$ is any polynomial, $p(λ)$ is an eigenvalue of $p(T)$

b) Show that if $β$ is a basis of $V$ then $p([T]^β_β)$ is diagonalizable for any polynomial $p(x)$ with complex coefficients. Deduce that $p(T)$ is diagonalizable.

c) Suppose that All of the eigenvalues s λ of T satisfy |λ| < 1. Show that for
any basis α of V , the matrix limk→∞[T^k]α
α = O, in the sense that the components of
the matrix [T^k]α
α
all tend to 0 as k → ∞.

For (a), let $v$ be an eigenvector of $lambda$ for $T$, i.e. $Tv=lambda v$, $p$ a polynomial as in your question. Then you have

$$p(T)v=(sum_i=0^na_iT^i)v=sum_i=0^n(a_iT^i)v=sum_i=0^na_iT^iv=sum_i=0^na_ilambda^iv=(sum_i=0^na_ilambda^i)v$$

The first two equalities just follow from basic properties of linear maps and their sums like distribution over addition, etc. The most important property used, is that $T^iv=lambda^iv$ for this eigenvector $v$.

You can see this, by considering an induction on $i$:

1. The base case for $i=0$ just amounts to: $T^0v=id_Vv=v=1cdot
   v=lambda^0v$.



2. 
   

   Now, for the induction step, assume that $T^iv=lambda^iv$ for some
   $i$, then $T^i+1v=TT^iv=Tlambda^iv=lambda^i Tv=lambda^i+1v$.

   
   
   

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For (b), consider what the conditions for a matrix are to be diagonalizable in the complex numbers. Especially, to deduce diagonalizability of $p(T)$, how is $p(T)$ connected with $p([T]^beta_beta)$?

Observe that $T^n(v) = T^n-1(T(v)) = T^n-1(lambda v) = lambda^nv$.

We are looking for a $v$ such that $p(T)(v) = p(lambda)(v)$.

Assume that $v$ is an eigenvector of $T$, i.e. $T(v) = lambda v$.

Then

$p(T)(v) = (a_nT^n + ... +a_1T + a_0)(v) =$

$= a_nT^n(v) + ... + a_1T(v) + a_0v =$

$=a_nlambda^nv + ... + a_1lambda v + a_0v$

$= p(lambda)(v)$

For a), use the hint in the comments, just to write out $p(T)$ and apply it to an eigenvector.

For b), it is easy to prove that any power of a diagonalizable matrix is diagonalizable (and is similar to $D^n$, where $PAP^-1=D$).