Let $A \in M_n (\mathbb{R})$ be a real matrix of odd rank $n$. Prove that there exists $\lambda \in \mathbb{R}$ such that $$\mbox{Rank} (A - \lambda I) < n$$
I am trying to prove this statement, yet I don't seem to find any information I can deduct from the given information. All I could deduct from the information that $n$ is odd is that the characteristic polynomial is of rank $n$, therefore has an odd (maximal) rank.
Every polynomial with odd degree has a real root, as Alan pointed out. In other words, the matrix $A$ has an eigenvalue $\lambda \in \mathbb{R}$. Let $v$ be a corresponding eigenvector, then for this eigenvector we have $(A-\lambda I) v =0$. By the definition of an eigenvector, $v \neq 0$. Hence $ \dim \text{Null} (A-\lambda I) >0$.
What does this tell you about $\text{Rank} (A-\lambda I)$? Hint: you can use the Fundamental Theorem of Linear Maps.
Additional information
For a matrix $A \in \mathbb{R^{n \times n}}$ it holds that
$$A \text{ nonsingular} \iff \text{rank} A=n \iff \text{Null }A= \{0\}$$
In the present case, we have shown for the matrix $B=A-\lambda I$, that it has a nontrivial solution (there exists a vector $v \neq 0$ such that $Bv=0$), that is, $\text{Null} B\neq \{ 0\}$ . Thus from the above equivalence we can conclude $\text{rank} B \neq n$, since $\text{rank} B \leq n$, we conclude $\text{rank} B<n$.
And in addition we can conclude that $B$ is singular (it is not non-singular).