What is the maximal dimension of a subspace of \(\RR^4\) on which the quadratic form \[ x_1x_2-3x_2^2+x_3^2+2x_2x_4+x_4^2 \] is positive definite?
Solution
Let \[ q(x_1,x_2,x_3,x_4) \coloneqq x_1x_2-3x_2^2+x_3^2+2x_2x_4+x_4^2. \]
1. The quadratic form satisfies\[ q = (x_4+x_2)^2 - \left(2x_2-\frac{x_1}{4}\right)^2 + \left(\frac{x_1}{4}\right)^2 + x_3^2. \]
Expanding the right-hand side gives \[ \begin{aligned} &(x_4^2+2x_2x_4+x_2^2) - \left(4x_2^2-x_1x_2+\frac{x_1^2}{16}\right) + \frac{x_1^2}{16} + x_3^2\\ &\qquad= x_1x_2-3x_2^2+x_3^2+2x_2x_4+x_4^2. \end{aligned} \]
2. The linear coordinates\[ y_1=x_3, \qquad y_2=x_4+x_2, \qquad y_3=2x_2-\frac{x_1}{4}, \qquad y_4=\frac{x_1}{4} \]form an invertible change of coordinates, and in these coordinates\[ q=y_1^2+y_2^2-y_3^2+y_4^2. \]
Step 1 gives the displayed diagonal form. The coordinate change is invertible because the original coordinates can be recovered by \[ x_1=4y_4, \qquad x_2=\frac{y_3+y_4}{2}, \qquad x_4=y_2-\frac{y_3+y_4}{2}, \qquad x_3=y_1. \]
3. There is a \(3\)-dimensional subspace on which \(q\) is positive definite.
In the \(y\)-coordinates, take \[ W \coloneqq \{(y_1,y_2,y_3,y_4):y_3=0\}. \] Then \(\dim W=3\) and, for every nonzero vector in \(W\), \[ q=y_1^2+y_2^2+y_4^2>0. \] Thus \(q|_W\) is positive definite.
4. No \(4\)-dimensional subspace of \(\RR^4\) can have positive-definite restriction of \(q\).
The only \(4\)-dimensional subspace of \(\RR^4\) is \(\RR^4\) itself. In the \(y\)-coordinates, the nonzero vector \[ (0,0,1,0) \] satisfies \[ q(0,0,1,0)=-1<0. \] Hence \(q\) is not positive definite on the whole space.
5. Therefore the maximal dimension is\[ \boxed{3}. \]
Step 3 gives a positive-definite subspace of dimension \(3\), and step 4 rules out dimension \(4\).
6. Q.E.D.
Step 5 is the requested maximal dimension.