Theorem: If $Q(\mathbf{x}) = \mathbf{x}^TA\mathbf{x}$ is a real quadratic form with $A$ symmetric, then the numbers $(p, q, r)$ in any canonical form $\sum_{i=1}^p y_i^2 - \sum_{i=p+1}^{p+q} y_i^2$ are uniquely determined by $A$. They are called the signature.

Proof: Suppose we have two diagonal representations via changes of coordinates $\mathbf{x} = P\mathbf{y}$ and $\mathbf{x} = R\mathbf{z}$: $Q = \sum_{i=1}^p y_i^2 - \sum_{i=p+1}^{p+q} y_i^2 = \sum_{i=1}^{p'} z_i^2 - \sum_{i=p'+1}^{p'+q'} z_i^2$

We must show $p = p'$ and $q = q'$.

Step 1: Define positive and negative spaces.

Let $V^+ = \{\mathbf{x} : Q(\mathbf{x}) > 0 \text{ or } \mathbf{x} = \mathbf{0}\}$ using the first representation. This is the span of the first $p$ coordinate directions in $\mathbf{y}$-coordinates, so $\dim(V^+) = p$.

Let $W^- = \{\mathbf{x} : Q(\mathbf{x}) \leq 0\}$ using the second representation. This contains the span of the last $n - p'$ coordinate directions in $\mathbf{z}$-coordinates, so $\dim(W^-) \geq n - p'$.

Step 2: Intersection argument.

If $\mathbf{x} \in V^+ \cap W^-$, then $Q(\mathbf{x}) > 0$ (from $V^+$) and $Q(\mathbf{x}) \leq 0$ (from $W^-$), so $\mathbf{x} = \mathbf{0}$.

Thus $V^+ \cap W^- = \{\mathbf{0}\}$.

Step 3: Dimension counting.

By linear algebra, for subspaces with trivial intersection: $\dim(V^+) + \dim(W^-) \leq n$

Therefore: $p + (n - p') \leq n$, which gives $p \leq p'$.

Step 4: Symmetry.

By the same argument with roles reversed (using $V^-$ from first representation and $W^+$ from second), we get $p' \leq p$.

Combining: $p = p'$.

Step 5: Determine $q$.

Since $p + q$ equals the rank of $A$ (which is well-defined), and we've shown $p$ is unique, $q$ must also be unique.

Alternatively, apply the same intersection argument to negative spaces: define $V^-$ (dimension $q$) and $W^+$ (codimension $q'$), proving $q = q'$ similarly.

Conclusion: The signature $(p, q, r)$ is uniquely determined by $A$, independent of how we diagonalize. ∎