My question is as follows:
Let the equation $V^{\top}MV=F$.
Such as $V^{\top}=(x_1,x_2,...,x_n)$ a line vector of n unknown coefficients, M a known diagonal matrix (of size n) and F a real number also known.
So there are one or more solutions for this problem? Theoretically and mathematically what can we say about this problem?
thank you in advance
On
Since $$M=\left(\begin{matrix}m_1&0&\cdots&0\\0&m_2&\cdots&0\\\vdots&\vdots&&\vdots\\0&0&\cdots&m_n\end{matrix}\right)$$ is a diagonal matrix, we can express the matrix equation in another form after doing matrix multiplications:
$$m_1x_1^2+m_2x_2^2+\dots+m_nx_n^2=F$$
If the equation has infinitely many solutions where $n\geq2$, then it means that $m_1, m_2\dots m_n,F$ are all non-negative or all non-positive (with some $m_i\neq 0$), or $\exists i,j$ such that $m_i>0,m_j<0$.
The only case for a unique solution for $n\geq2$ is when all $m_i$ are positive (or all are negative) and $F=0$. The solution is trivial ($x_i=0$). This is due to the fact that $x^2\geq0$ for real $x$, and $x^2=0$ only when $x=0$.
The case for no solution is when all $m_i=0$ and $F\neq0$, or all $m_i\geq0$ and $F<0$ (or $m_i\leq0$ and $F>0$).
In the case where all $m_i$ are non-negative (non-positive case is similar, just multiply everything by $-1$), arrange the terms in increasing $|m_i|$. We can start generating a solution iteratively by picking $x_1$ in the range $\left[-\sqrt{\frac{F}{m_1}}, \sqrt{\frac{F}{m_1}}\right]$ (if $m_i$ is 0, there is no restriction). Once $x_1$ is picked, we can change $F$ to $F^\prime=F-m_1x_1^2$, and we can generate the rest of the solution with this equation:
$$m_2x_2^2+m_3x_3^2+\dots+m_nx_n^2=(F-x_1^2m_1)$$
In the case where not all $m_i$ are the same sign, we can let $F=G+H$, where $G\geq0$ and $H\leq0$, such that we can move all the positive $m_i$ to an equation adding to $G$ and all the negative $m_i$ to an equation adding to $H$.
The answer depends on real $F$.Set $n=2$ and $M=I_2$