Solve matrix vector equation

Question

Let $A$ be a real $n\times n$ matrix and $w,x$ real $n\times 1$ vectors.

For fixed $A$ and $w$ solve the following for $x$:

$(x^\top A x)w - (x^\top w) (A+A^\top) x = 0$

Any hints? I do not really feel confident on what kind of rearrangements are legit since multiplication is generally not commutative and I do not know how to deal with the scalar products in there.

Answer 1

Firstly, if $w=0$ then any $x$ is a solution. We now consider the nontrivial case $w\neq 0$.

Let us assume that $x$ is a solution, and let $\mathcal N:= {\rm Ker}(A+A^T)$. Notice that any $x\in\mathcal N$ is a solution (to see this you may need to observe that $x^TAx = \frac12x^T(A+A^T)x$).

Let us now look for solutions $x\notin\mathcal N$, and let us decompose $x=x_\parallel + x_\perp$ with $x_\parallel\in \mathcal N$ and $x_\perp \in \mathcal N^\perp$. Then, $$ (x^TAx)w - (x^Tw)(A+A^T)x = (x^TAx)w - (x^Tw)(A+A^T)x_\perp $$ We have two options now:

• either $x^TAx = 0$, in which case necessarily $x\perp w$ (since $x\notin\mathcal N$ we have $x_\perp\neq 0$);
• or $x^TAx\neq 0$ and $w=\frac{x^Tw}{x^TAx}(A+A^T)x$.

The first case tells us that any vector in the set $\{x:~x^TAx = 0,~x\perp w\}$ is a solution.

In the second case, we may recursively replace $w$ on the rhs, yielding \begin{align} \color{blue}w ={} & \frac{x^T\color{red}w}{x^TAx}(A+A^T)x = \frac{x^T\color{red}{\left( \frac{x^Tw}{x^TAx}(A+A^T)x \right)}}{x^TAx}(A+A^T)x = \frac{(x^T(A+A^T)x) \frac{x^Tw}{x^TAx} } {x^TAx} (A+A^T)x \\ ={} & \frac{2x^TAx \frac{x^Tw}{x^TAx} } {x^TAx} (A+A^T)x = 2 \frac{x^Tw} {x^TAx} (A+A^T)x = \color{blue}{2w} \end{align} So necessarily $w=2w$ and hence $w=0$, which we excluded at the beginning.

It follows that the set of solutions is $$ X^\star = \begin{cases} \mathbb R^n & \text{if } w=0 \\ \{x:x^TAx=0\} \cap \{w\}^\perp & \text{otherwise} \end{cases} $$

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Extra: how to determine the set $\mathcal N_0:=\{x:~x^TAx=0\}$ If $(A+A^T)$ is semidefinite (either positive or negative), then it is trivial to observe that such set is simply ${\rm Ker}(A+A^T)$.

In general, $\mathcal N\subseteq N_0$. To determine it you can compute the eigenvalues $\lambda_1\ldots\lambda_n$ and a corresponding orthonormal basis of eigenvectors $e^1\ldots e^n$ of $A+A^T$.

(note: in order to avoid confusion I use superscripts for vectors and subscripts for scalars, and it should be obvious that when both appear, then the superscript is a power...).

Then each $x$ decomposes as $x=\sum_{i=1}^nx_ie^i$ and $$ x^TAx = \sum_{i=1}^n\lambda_ix_i^2 $$ Let $(p,q,r)$ be the signature of $A+A^T$, and reorder eigenvectors/eigenvalues according to the sign of the eigenvalues as follows: $$ \lambda_1\ldots\lambda_p >0 \quad \lambda_{p+1}\ldots\lambda_{p+q} <0 \quad \lambda_{p+q+1}\ldots\lambda_n = 0 $$ ($r=n-p-q$). For clarity, let $\mu_1=-\lambda_{p+1}\cdots \mu_q=-\lambda_{p+q}>0$ so that $$ x^TAx = \sum_{i=1}^p\lambda_i x_i^2 - \sum_{j=1}^q\mu_j x_{p+j}^2 $$ From the expression above it is clear that in order for it to vanish, the two sums must cancel out while the components $x_{p+q+1}\ldots x_n$ can be anything (indeed $\mathcal N\subseteq\mathcal N_0$). Furthermore, notice that, trivially, $\mathcal N_0$ is a vector space and hence it suffices to find a set of generators. Considering couples $w^{i,j} = a_{i,j}e^i + b_{i,j}e^{p+j}$ for $i=1\ldots p$ and $j=1\ldots q$ and for some $a_{i,j},b_{i,j}$ to be determined, we have $$ {w^{i,j}}^TAw^{i,j} = a_{i,j}^2\lambda_i - b_{i,j}^2\mu_j $$ In order for it to be zero we may consider $a_{i,j} = \sqrt{\mu_j}$ and $b_{i,j} = \sqrt{\lambda_i}$.

We hence have that $$ \mathcal N \oplus {\rm span} \bigl\{ \sqrt{\mu_j}e^i+\sqrt{\lambda_i}e^{p+j} \,:~ i=1\ldots p,~ j=1\ldots q \bigr\} \leq\mathcal N_0 $$ I guess that with a small extra effort it can be proved that the opposite inclusion works as well. I encourage you to try and let me know :)