Stack Exchange
Dimension of quotient of vector space of polynomials
Let $A = \begin{pmatrix}1 & 1 & 0 & 0 \\ 0 &1 & 1 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$
Let $\mathcal{P}$ be the vector space of polynomials with real coefficients and $\mathcal{P}(A) =\{p(t) \in \mathcal{P}: P(A) =0\}$. Compute $\text{dim}\mathcal{P}/\mathcal{P}(A)$. 
An attempt: $I - A$ is nilpotent of index 4. Hence if $p(t) = \sum_{k=0}^{4}a_{k}t^{k}$. Then $0 = p(A)$ if $a_{k} = \binom{4}{k}(-1)^{k}$. I claimed that $\mathcal{P}(A)$ is generated by $p(t) = t^{4} - 4t^{3} + 6t^{2} - 4t + 1$. I then concluded that $\mathcal{P}/\mathcal{P}(A) \cong \mathbb{R}[t]/(p(t))$ and I concluded that the dimension being searched was 4. Is this correct?

As $A$ is currently written, $I-A$ is nilpotent with index 3.

hahaha. I messed up there then.

Definition of $\mathcal P(A)$ has $3$ types of $p$ in it. Typo?

I corrected it, Jack's wasted life.

The characteristic polynomial of $A$ is $\;\chi_A(t)=(t-1)^4$, hence the minimal polynomial of $A$ is $p(t)=(t-1)^r,\enspace r\le 4$. It's easy to check $r=3$.  Hence $\mathcal P/((p(t)$ is isomorphic to $\mathcal P_2$, the subspace of polynomials of degree $\le 3$, and its dimension is $3$.
