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? 



Bound the eigenvalue of product of matrices?
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.


