# Diagonalizablity of a transformation from ${P_{10}} \to {P_{10}}$

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?

• You may have all the right pieces, but those don't seem to be the right pieces. Diagonalizability is usually thought of as a statement about eigenvectors, or minimal polynomials. On the other hand, that basis gives you an upper-triangular matrix, which might help. Commented May 5, 2017 at 2:10
• I can see that the diagonal of the transformation of the matrix is going to be all 1. Wouldn't that means there won't be enough eigenvectors to form a basis to diagonalize the matrix? Commented May 5, 2017 at 2:15
• Do $P_2$ first, $P_3$ if that is too easy Commented May 5, 2017 at 2:28

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?

• It is a start. I am working on it Commented May 5, 2017 at 2:30
• Sounds good! :) Commented May 5, 2017 at 2:30
• can you check my answer below? Commented May 5, 2017 at 2:38

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

• Yeah, looks good. Commented May 5, 2017 at 2:42

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 $x-1$.

• Ahh, nice! That second sentence is what I was grappling for before I had the other idea. Thanks for the effort on an 'answered' question :) Commented May 5, 2017 at 2:40
• @EricStucky It’s pretty much the same idea that the OP had—you run out of eigenvectors (very quickly, in fact).
– amd
Commented May 5, 2017 at 2:43