Show that $T:{P_{10}} \to {P_{10}}$ defined by $T(p(t)) = p(t+1)$ is NOT diagonalizable.
So the standard basis for $P_{10}$ is {$1,t,t^2,t^3,...$} and we know that $Rank(T)+ Nullity(T) = dim(T)$
I have all the pieces I need I think. How can I put them together?
Let $T$ be a diagonalizable linear transformation on a finite-dimensional vector space $V$ which has a single eigenvalue $\lambda$. Let $\{x_1,x_2,\dots, x_n\}$ be an eigenbasis for $V$, so that $Tx_i=\lambda x_i$. We can therefore write any $x\in V$ as $\sum a_ix_i$, and thus
$$Tx=T\left(\sum_{i=1}^n a_ix_i\right) = \sum_{i=1}^n a_i T(x_i) = \lambda x.$$
Hence $T=\lambda I$.
Can you see how to get from this result to the answer?
Consider the matrix of T with respect to the standard basis. We have
$T(1) = 1, T(t) = t + 1, T(t 2) = (t + 1)^2 = t^2 + 2t + 1$, and so on.
We see that in general, $p(t)$ and $T(p(t))$ have the same degree. This forces the matrix of T with respect to the standard basis to be upper triangular. Moreover, $$T(t^n) = (t + 1)^n = t^n + nt^(n−1) + (other stuff)$$ so the diagonal entries of this matrix are all 1’s. This means $T$ has a single eigenvalue, 1. Since the rank of $[T]_S − I$ is $n − 1$ (or at least, since the rank is nonzero) we see that there will not be enough eigenvectors. As an alternative, if we wish to find an eigenvector, call it $p(t)$, then we need $T(p(t)) = 1 · p(t)$, or $p(t) = p(t + 1)$, and this is only true for constant polynomials. Either way we look at it, $T$ is not diagonalizable
You’ve already determined that $T$ has only one eigenvalue, $1$, so its characteristic polynomial is $(x-1)^{11}$. For $T$ to be diagonalizable, this means that its minimal polynomial must be $x-1$. However, $T(p(t))-p(t)=p(t+1)-p(t)=0$ only holds for all $t$ when $p$ is a constant polynomial, hence $T$’s minimal polynomial is not $x-1$.
