Curious how to retrieve original vector $x$ from an $x^Tx$ operation I am doing some analysis on a signal $x(t)$ which is a vector of length $m$. After performing the operation $x^Tx$ of the vector on itself, I get a matrix $Y$ of size $m\times m$. Now, how can I retrieve the original $x(t)$?
Should this be the elements in the diagonal of $Y$? I did some basic plotting and while the diagonal looks similar to original signal vector, they are not equal. Looks straightforward but I am missing something, probably need to normalize it somehow.
In summary,

*

*Is it always possible to retrieve the original signal without information loss?


*If it is possible, how is it done?

(I used your notation but you should be aware that the standard is to consider vectors as columns)
If you take a row-vector $z$ and you do $Yz^T$,
you'll get
$$
Yz=(x^Tx)z^T=x^T(xz^T)=(xz^T)\,x^T.
$$
So as long as the result is nonzero, you are getting a scalar multiple of the original vector. If you use each of the element $e_k$ of the canonical basis, you have that the $k^{\rm th}$ column of $Y$ is
$$
Ye_k=(xe_k^T)\,x^T=x_k\,x^T.
$$
In other words, the columns of $Y$ are multiples of $x^T$, with the coefficients being precisely the entries of $x$. So, if $Y_{kk}=x_k^2\ne0$, you recover $x^T$ up to sign as
$$
x^T=\pm\frac1{\sqrt{Y_{kk}}}\,Y_k,
$$
where $Y_k=Ye_k$ is the $k^{\rm th}$ column of $Y$.
