I have looked at various answers on hamadamard products, but none of them seem to answer this.
Given a matrix $M=F\circ G$ where $F=UDU^{-1}$ is diagonalizable and $G=xx^T$ is rank 1. Is there way I can diagonalize $M$ or know the eigen values of $M$? Even computing the square root of $M$ would be interesting for me.
Any help is appreciated.
It is possible to re-formulate the problem without reference to the Hadamard product (whose drawback is that it has few properties by itself).
Let $$\tag{1}x=(x_1,x_2,... x_n) \ \ \text{and} \ \ \Delta:=diag(x_1,x_2,... x_n).$$
Then $$\tag{2}F \circ G = \Delta F \Delta.$$
Straightforward proof: left-multiplying (resp right-multiplying) by diag$(x_1,x_2,... x_n)$ amounts to multiply the k-th row (resp. th k-th column)by $x_k$.
Example with $n=2$:
$$\pmatrix{a&b\\c&d} \circ \left(\pmatrix{x_1\\x_2}*\pmatrix{x_1&x_2}\right)=\pmatrix{x_1&0\\0&x_2} *\pmatrix{a&b\\c&d} *\pmatrix{x_1&0\\0&x_2}$$
(where * denotes the usual matrix product).
The question is now: are there relationships between the eigenvalues of $F$ and those of $\Delta F \Delta$ ?
I am tempted to say that an answer is that the more $\Delta$ is away from $I_n$ (Identity matrix) (otherwise the more $X$ is far away from $(1,1,\cdots 1)$), the less connection one has with the initial eigenvalues. Here is an example that illustrates that with the $3 \times 3$ matrix:
$$F=\pmatrix{0&-1&0\\0&0&-1\\3&3&1}$$
whose eigenvalues are $1$ and $\pm\sqrt{3}$ (red points on graphics below).
Edit: I have taken 5000 realizations of $F \circ G$, where $G=XX^T$ with $X=(1+\epsilon_1,1+\epsilon_2,1+\epsilon_3)^T$ where the $\epsilon_k$s have (small) uniform variations in $[-1/4,1/4]$. Here are the 5000 results ($3 \times 5000$ points representing the eigenvalues. One can observe rather important fluctuations around the initial eigenvalues: