Curious how to retrieve original vector $x$ from an $x^Tx$ operation

Question

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,

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

2. If it is possible, how is it done?

Answer 1

The diagonal holds the squares of the original vector. With it, and the signs of the other entries, you can find $x$ or $-x$ but can't decide between the two.

Answer 2

(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 $Y$.