Solve an equation with an unknown vector

Question

Let's assume that $a$ is an $n\times1$ vector and $A_1$ and $A_2$ are $n\times n$ hermitian matrix how can I find unknown vector $a$ in equation ${a^HA_1a}={a^HA_2a}$ ?

Answer 1

First of all note that $a = 0$ solves the problem for all matrices $A_1$ and $A_2$. As I believe that you are looking for nontrivial solutions, let me propose the following:

Note that both $A_1$ and $A_2$ are hermitian, that means that $A = A_1 - A_2$ is hermitian, too. Now note that finding your vector is equivalent to finding a (non-trivial) vector that satisfies $a^H A a = 0$. Thanks to the hermiticity of $A$, we may (pseudo) diagonalize $A$ using a diagonalization of the form $A = P D P^H$ with $D$ diagonal (containing the eigenvalues in the right order corresponding to $P$). If we introduce $b = P^H a$, the problem reduces to $b^H D b = 0$. Let $b = \left(b_1, b_2, \ldots, b_n \right)^T$, if $D$ contains $\lambda_i$ on the $i$th diagonal entry, the problem reduces to finding $b_1,\ldots,b_n$ such that $\lambda_1 \vert b_1 \vert^2 + \ldots + \lambda_n \vert b_n \vert^2 = 0$. Note that if $\lambda_i$ is zero, one solution is a vector containing zeros and a one on the $i$th spot. This would correspong to $A_1$ and $A_2$ having a simultaneous zero eigenvector. If $\lambda_i$ is positive and $\lambda_j$ is negative, another solution would be a vector containing a one in the $i$-th position and $\pm \sqrt{\frac{\lambda_i}{\lambda_j}}$ in the $j$-th position. You can also see that there may not be a solution if $A$ is positive definite and that if a solution exists, it may not be a unique.

After you have found a $b$, you can transform it back via $a = P b$.