If $A$ is a self-adjoint operator on an $n$-dimensional vector space where $n$ is even and $det(A) < 0$, then there is a $v$ such that $Av \bot v$ "If $A: E \rightarrow E$ is a self-adjoint operator on a real $n$-dimensional vector space (with inner product) where $n$ is even and $det(A) < 0$, then there is a non-zero vector $v$ such that $v$ is orthogonal to $Av$."
Is that true? The question in the book actually phrases it in the language of matrices:
"If $A$ is a symmetric $n × n$ with $n$ even and $det(A) < 0$, then using the canonical inner product, there is a $v \in \mathbb{R}^n$ where $v$ is orthogonal to $Av$"
I can prove there are vectors $u,w \in E$ where $\langle u,Au \rangle > 0$ and $\langle w,Aw \rangle < 0$. Then, my mind instantly went to the Intermediate Value Theorem, but I guess this would be a function from $\mathbb{R}^n$ to $\mathbb{R}$ and I've only studied single-variable analysis, so I'm not sure. I'd love a hint!
 A: This may be resolved in a (more or less) purely algebraic manner, without using the Intermediate Value Theorem:
Since $A$ is symmetric, i.e. $A = A^T$, all $n$ of its eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$ are real, and 
$\prod_1^n \lambda_i = \lambda_1 \lambda_2 \ldots \lambda_n = \det(A) < 0. \tag{1}$
It is clear from (1) that $A$ must have an odd number of negative eigenvalues, and also an odd number of positive eigenvalues, since $n$ is even.  Otherwise, we could not have $\det(A) < 0$, since the product of an even number of negative reals is positive
This means there exists at least one eigenvalue $\lambda_n < 0$, and at least one $\lambda_p > 0$.  Let $e_n$, $e_p$ be the corresponding normalized eigenvectors:
$Ae_n = \lambda_n e_n, \tag{2}$
$Ae_p = \lambda_p e_p, \tag{3}$
with $\langle e_n, e_n \rangle = \langle e_p, e_p \rangle = 1$.
For $a, b \in \Bbb R$ we compute
$\langle A(ae_p + be_n), ae_p + be_n \rangle = \langle aAe_p + bAe_n, ae_p + be_n \rangle = \langle a\lambda_p e_p + b\lambda_n e_n, ae_p + be_n \rangle$
$= a^2 \lambda_p + b^2 \lambda_n, \tag{4}$
since $\langle e_n, e_p \rangle = 0$, being eigenvectors associated with the distinct eigenvalues $\lambda_n < 0 < \lambda_p$.  It is now possible to choose $a, b \ne 0$ such that
$\dfrac{a^2}{b^2} = -\dfrac{\lambda_n}{\lambda_p} > 0, \tag{5}$
and if we set
$v = ae_p + be_n \tag{6}$
for such $a, b$, the result follows.  QED.
Of course, the answers of Will Jagy (in the comment) and daw are also very nice; lauds to each of them.
A: Consider the scalar function
$$
\phi(t) := \langle A(u+t(w-u)), u+t(w-u)\rangle.
$$
This function $\phi$ is continuous from $\mathbb R$ to $\mathbb R$, $\phi(0)>0$, $\phi(1)<0$. Then there is $t_0\in(0,1)$ such that 
$$
\phi(t_0)=0.
$$
Now it remains to check that $v:=u+t_0(u-w)$ is non-zero. You would not want the zero-vector.
If $v$ would be zero, then $w=(1+t_0)u$, moreover,
$$
0>\langle Aw,w\rangle = (1+t_0)^2 \langle Au,u\rangle >0,
$$
a contradiction.
