I have a question about the proof of Hurwitz's 1-2-4-8 theorem about the sum of squares.
I have consulted Chapter 1 of Rajwade's "Squares" book, notes by Keith Conrad, and notes by Daniel Shapiro. They all feature the same step that I am confused about.
Namely, they derive identity $$ (Y_1,Y_2,\dotsc,Y_n)[(X_1^2+X_2^2+\dotsc+X_n^2)I_n-A^TA]\begin{pmatrix}Y_1\\Y_2\\\vdots\\Y_n\end{pmatrix}=0 $$ where $Y_1,\dotsc,Y_n,X_1,\dotsc,X_n$ are indeterminates, and $A$ is a matrix whose entries are linear forms in $X_1,\dotsc,X_n$. From that they somehow deduce that $$ A^T A=(X_1^2+X_2^2+\dotsc+X_n^2)I_n. $$
Question 1:Is this a correct deduction? If so, why?
As far as I can tell, we can derive only a weaker conclusion. Namely, writing $A=X_1A_1+X_2A_2+\dotsb+X_nA_n$, and expanding everything as a sum of monomials in $X_1,X_2,\dotsc,X_n$ we obtain that $$ (Y_1,Y_2,\dotsc,Y_n)(A_i^TA_i-I_n)\begin{pmatrix}Y_1\\Y_2\\\vdots\\Y_n\end{pmatrix}=0\qquad\text{for each }i=1,\dotsc,n. $$ and there certainly exist non-zero matrices $M$ such that $Y^TMY=0$. For example, $$ (Y_1,Y_2)\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}Y_1\\Y_2\end{pmatrix}=0. $$
I thought that I could rescue this by showing that $A_i^T A_i$ is same for all $i$ (and then one can just change coordinates, and run the proof of Hurwitz's theorem unchanged), but my fix was wrong (as called out by a student in my class today).
Question 2: If this step is indeed in error, how does a correct proof go?
Sadly, I do not have training in German, and my attempt to use a dictionary to understand how Hurwitz made this step in his original paper was unsuccessful. I can read English, Russian and French. I would thus welcome references in those languages.
In Conrad's notes he writes "as $\bf y$ varies" just before (2.4)
If you have a symmetric matrix $M$ and column vectors $y$ such that $y^T M y = 0$ for every $y,$ this means that $M=0.$ This assumes that the characteristic of the field is not $2.$
First, we get the resulting bilinear form from $$ x^T M y = (1/2) \left( (x+y)^T M (x+y) - x^T M x - y^T M y \right). $$ The right hand side is always $0,$ so $x^T My$ is always $0.$
Now take $x$ as the vector with entry $i$ set to $1,$ all other entries $0.$ Take $y$ as the vector with entry $j$ set to $1,$ all other entries $0.$ Then $x^T M y = M_{ij}.$ Therefore $M_{ij}=0.$ We do this for each $ij$ ordered pair, with the result that $M=0.$