Let $K$ be a field with $\text{char}(K) \neq 2$ and $Q$ be a quadric in the projective space $P(K^n)$. Let $M$ be a symmetric $n×n$-matrix over $K$ such that $Q \leftrightarrow x^T M x$.
In my course notes we defined that $Q$ is degenerate if $\text{det}(M) = 0$. Now my question is why this is well defined.
In other words, let $M'$ be another symmetric $n×n$-matrix over $K$ such that $Q \leftrightarrow x^T M' x$. Then why is it impossible that $\text{det}(M) = 0$ and $\text{det}(M') \neq 0$?
Thanks a lot in advance!
Suppose the following: $\newcommand{\ra}[1]{\stackrel{#1}{\Rightarrow}}$
i. $x^\top M x = 0$ and $x^\top M' x = 0$ have the same solution set
ii. $det(M)=0$
Then we can choose a nonzero element $a\in ker(M)$ and prove that $a\in Ker(M')$ as well, as shown below. It follows that $det(M') = 0$.
Lemma: $\forall x\in K^n: a^\top M' x = 0$
Suppose that there exists an $x\in K^n$ such that $a^\top M' x \neq 0$. Then choose $\lambda=-\frac{x^\top M' x}{2a^\top M'x}$.
Using the facts $a^\top Ma=0 \ra{} a^\top M'a=0$ and $a^\top M'x = x^\top M'a$, we can prove that the equivalence class of $x+\lambda a$ is on the quadric Q:
$$ \begin{align} (x+\lambda a)^\top M' (x+\lambda a) &= x^\top M'x + \lambda\cdot\left[ a^\top M'x + x^\top M'a \right] + \lambda^2 \cdot a^\top M'a \\ &= x^\top M'x + 2\lambda\cdot a^\top M'x \\ &= 0 \end{align} $$
From $(i)$ follows that $(x+\lambda a)^\top M (x+\lambda a) = 0$ as well. Because $a\in ker(M)$, it follows that $x^\top Mx=0 \wedge (x+a)^\top M(x+a)=0$, so $x^\top M'x=0 \wedge (x+a)^\top M'(x+a)=0$.
Subtracting both equations gives $$ \begin{align} 0 &= x^\top M' a + a^\top M' x + a^\top M' a \\ &= 2a^\top M' x \end{align} $$ From which follows that $a^\top M' x = 0$, as $char(K)\neq2$. This is a contradiction, so clearly the lemma must hold.
Conclusion: $det(M')=0$
The lemma is equivalent to $\forall x\in K^n: x^\top(M' a) = 0$. Clearly this implies $$ M'a=\vec{0} \ra{} a\in ker(M') \ra{} det(M')=0 $$
Note: This is a very ad hoc proof. It would provide greater insight to study which symmetric matrices describe the same quadric, but that deserves its own question. The intuitive statement "M and M' describe the same quadric iff M and M' differ by a scalar factor" is wrong (shown by OP in the comments), but perhaps a modified statement could work.