Nilpotent matrices proof 
Prove that there does not exist a $7 \times 7$ matrix $B$ with real entries such that $B^2 + B + I$ is nilpotent.

Since I need to show this statement for any $7 \times 7$ matrix $B$, I know that it isn't enough to give a counterexample, so I need to proxy it generally. I know the definition of nilpotent: the existence of $k \geq 0$ such that $(B^2 + B + I)^k = 0$. But I cannot fiugre out how to start a rigorous argument, other than that it may be appropriate to proceed by contradiction. I would appreciate any help on this problem.
Updated attempt based on a helpful answer:

Seeking a contradiction, suppose that $B^2 + B + I$ is nilpotent, so there is a $k > 0$ such that $(B^2 + B + I)^k = 0$. Let $B$ be a $7 \times 7$ matrix with real entries. Define its characteristic polynomial
\begin{align*}
q_B (\lambda) = \det(\lambda I - B). 
\end{align*}
This is a monic polynomial of degree $7$, and every odd polynomial has a real root, so there exists some $\lambda \in \mathbb{R}$ such that $\det(\lambda I - B) = 0 $. This $\lambda$ is an eigenvalue of $B$, so there exists a non-zero eigenvector $v$ such that $Bv = \lambda v$.

Here is my thought as to how to proceed. I plugged in values of $k$ and multiplied out $(B^2 + B + I)^k$ and ended up with a polynomial in $\lambda$. For example:
\begin{align*}
(B^2 + B + I)v & = B^2 v + Bv + Iv = \lambda^2 v + \lambda v + v \\
(B^2 + B + I)^2 v & = (B^4 + 2B^3 + 3B^2 + 2B + I)v = B^4 v + 2B^3 v + 3B^2 v + 2B + v.
\end{align*}
It must be the case that $(B^2 + B + I)^k v \neq 0$ because $v \neq 0$ and we're always adding $v$ at the end. But I can't prove rigorously that the terms don't "cancel out."
 A: Hint: Since $B$ is a real $7 \times 7$ matrix, the characteristic polynomial $p(\lambda) = \det(\lambda I-B)$  is a real polynomial with degree $7$. Hence, $p(\lambda)$ has at least one real root, and thus, $B$ has a real eigenvalue.
Now, let $\lambda$ be this real eigenvalue and $v \neq 0$ be the corresponding eigenvector, i.e. $Bv = \lambda v$. Can you express $(B^2+B+I)^kv$ in terms of $\lambda$, $k$, and $v$? Is it possible for $(B^2+B+I)^kv = 0$ for some integer $k \ge 0$?
A: The solution to the problem is very simple as long as you bear in mind the important fact the previous answer highlights, namely that any real polynomial (by which I mean a polynomial in the ring $\mathbb{R}[X]$) of odd degree has at least one real root together with the following general:

Lemma. Let $K$ be an arbitrary commutative field, $n \in \mathbb{N}^{\times}=\mathbb{N} \setminus \{0\}$ a nonzero natural, $f \in K[X]$ an arbitrary polynomial and $M \in \mathscr{M}_n(K)$ a square matrix of order $n$ over $K$. If $\lambda \in K$ is an eigenvalue of $M$ then $f(\lambda)$ is an eigenvalue of $f(M)$.

In your particular instance, since $B^2+B+\mathrm{I}_7=(X^2+X+1)(B)$ is required to be nilpotent, all its eigenvalues must be $0$. Fixing $\alpha \in \mathbb{R}$ to be a certain eigenvalue of $B$ -- which exists by virtue of the order $7$ being odd -- the lemma above would entail that $\alpha$ is also a root of the polynomial $X^2+X+1$. However, this polynomial is irreducible over $\mathbb{R}$ and has no real roots.

The general claim that any eigenvalue of a nilpotent matrix $M \in \mathscr{M}_n(K)$ must be $0_K$ -- where $K$ is once again an arbitrary commutative field -- is easily proved by remarking that relation $MX=\lambda X$ -- where $\lambda$ is an arbitrary eigenvalue of $M$ and $X \in \mathscr{M}_{n, 1}(K)$ a corresponding eigenvector -- entails $M^r=\lambda^rX$ for any $r \in \mathbb{N}$. If in particular $M^r=\mathrm{O}_n$ for a certain $r \neq 0$, it then follows that $\lambda^r=0_K$ or $X=\mathrm{O}_{n1}$. Since eigenvectors are by definition non-zero vectors, it can only be that $\lambda^r=0_K$ which further entails $\lambda=0_K$, as $r \neq 0$ and since in any field (be it even noncommutative) a product is zero only if one of the factors is zero.
It can be furthermore shown that the characteristic polynomial of a nilpotent matrix, square of order $n$ over a commutative field $K$ is $X^n \in K[X]$, however your particular problem does not require a result of this depth.
