Let $H$ be a $n \times n$ real symmetric matrix that has eigenvalues with absolute value less than 1. Define the matrix $M = \prod_{i=1}^n (I - e_ie_i^{\top}H)$ where $e_i$ denotes the $i^\text{th}$ canonical basis vector of $\mathbb{R}^n$.
Assuming that $H$ has at least one negative eigenvalue, is it true that $M$ has an eigenvalue with absolute value greater than 1?
OK, looks like I've got it. Since it is pretty late here now, it would be nice if someone could check that I haven't written some nonsense (I apologize in advance if I have).
Let $N(H)$ be the kernel of $H$, let $Q$ be the orthogonal projector to $N(H)$ and let $P$ be the orthogonal projector to $N(H)^\perp$. Note that since $H$ is real symmetric, we have $QH=HQ=0$ and $PHP=H$.
Now let $e$ be a unit vector and let $X=x-\langle Hx,e\rangle e$. Then $PX=Px-\langle PHP(Px),Pe\rangle Pe$. Thus, if $x_j\in \mathbb R^n$ satisfy the recursion $x_{j+1}=x_j-\langle Hx_j,e_j\rangle e_j$ where $e_j$ cycle through an orthonormal basis in $\mathbb R^n$, then $y_j=Px_j\in P(\mathbb R^n)$ satisfy the recursion $y_{j+1}=y_j-\langle \widetilde H y_j,v_j\rangle v_j$, where $v_j=Pe_j$ cycle through some complete system of vectors whose norms are bounded by $1$. Moreover, if $H$ has a negative eigenvalue, so does $\widetilde H$, but the advantage is that $\widetilde H$ is invertible in $P(\mathbb R^n)$.
Thus, we reduced the problem to the following:
Let $E$ be a real Euclidean space, let $v_j\in E$ be a sequence cycling through a fixed complete system of vectors with norm at most $1$, let $H:E\to E$ be any invertible symmetric linear operator of norm at most $1$, let $x_0\in E$ be any vector with $\langle Hx_0,x_0\rangle<0$. Then the sequence of vectors defined by the recursion $x_{j+1}=x_j-\langle Hx_j,v_j\rangle v_j$ grows at least geometrically in norm.
The key inequality is $$ \langle Hx_{j+1},x_{j+1}\rangle = \langle Hx_{j},x_{j}\rangle-\langle Hx_{j},v_{j}\rangle^2(2-\langle Hv_{j},v_{j}\rangle) \\ \le \langle Hx_{j},x_{j}\rangle-\langle Hx_{j},v_{j}\rangle^2\,. $$ Since $H$ is invertible, we have $|HX|\ge c|X|$ for all $X\in E$ with some $c>0$. Now, if $|\langle Hx_j,v_j\rangle|<\delta|x_j|$ throughout the whole cycle with very small $\delta$, then all vectors in the cycle differ from the first vector $X$ in the cycle by at most $O(\delta)|X|$ and we derive that $|\langle HX,v_j\rangle|=O(\delta)|X|$, which, due to the completeness of the system of vectors $v_j$ over which we cycle, results in $|HX|=O(\delta)|X|$, contradicting the lower bound if $\delta>0$ is small enough. Thus, in each cycle we never increase $\langle Hx_j,x_j\rangle$ and have the inequality $$ \langle Hx_{j+1},x_{j+1}\rangle \le \langle Hx_j,x_j\rangle-\delta^2|x_j|^2 \\ \le (1+\delta^2)\langle Hx_j,x_j\rangle $$ at least once. That is the desired geometric growth. The end.
