Let u be a unit vector in n space with q(u) positive. Now q(2u) is 4 times as large as q(u). The same holds for q(-2u). It follows that q(u) is positive along the entire line determined by u, except for q(0), which equals 0.
The sign of q(x) depends only on the direction of x. It is enough to establish the positive and negative regions on the unit sphere; these extend to all of Rn.
Assume q(x) never changes sign. If it is positive along one vector, it is never negative along another. An example is the ellipse x2/4 + y2/9. Set this equal to a positive constant c and find an ellipse, i.e. the curve x2/4+y2/9 = c. As c increases the ellipse grows larger, but it doesn't change shape. The shape of the conic section is captured by the quadratic form q(x). We don't worry too much about the constant c, because that only changes the size.
When c = 1 the ellipse surrounds the origin, and that means q(x) is positive along every vector. Thus q is a quadratic form that does not change sign.
We know q is never negative, but is it ever zero? We would like to show that q vanishes precisely on the kernel of M. One direction is easy. If x is in the kernel of M, then x*M*xT is obviously zero. We just need to show the converse.
Let q(x) = 0 for some unit vector x. Let y be any nonzero vector. Compute q(x+sy) for an arbitrary scaling factor s. Expand q(x+sy) using matrix notation.
q(x+sy) =
(x+sy) * M * (x+sy)T =
x*M*xT + sy*M*xT + sx*M*yT + s2y*M*yT =
q(x) + 2sy*M*xT + s2*q(y) =
0 + 2sy*M*xT + s2*q(y)
Let s range from -∞ to +∞. We have a quadratic expression in s that is nowhere negative, since q is nowhere negative. Furthermore, when s = 0 q(x+sy) = 0. So the parabola attains its minimum at s = 0. The first derivative at 0 must be 0, implying y*M*xT = 0. The choice of y was arbitrary, so when y = M*xT, we have (M*xT).(M*xT) = 0. The length of M*xT is 0, hence M*xT is the zero vector. The quadratic form vanishes on the vector x iff x is in the kernel of M.
If you review the above proof, you'll see that it can be generalized just a bit. Let q(x) = 0, and let q have the same sign around x. This is a local property; it holds in a neighborhood of the unit sphere about x. Let y be any vector and evaluate q(x+sy). for small s, q(x+sy) is always nonnegative, and q(x+0) is still 0. The parabola in s still has its minimum at 0, the first derivative is 0, and x is in the kernel of M. If this local property holds everywhere, i.e. q has unchanging sign throughout the unit sphere, then q vanishes on the kernel of M.
If q(x) has unchanging sign, what does the surface look like? Rotate coordinates so that q is in standard form. In other words, M has become a diagonal matrix. Some of the diagonal entries may be zero. The corresponding coordinates span the kernel of M, and represent the subspace where q(x) is zero. The remaining entries on the main diagonal are all positive (or all negative). Set q(x) equal to a constant of the same sign and find an ellipsoid in n space. The entries of M establish the shape of the ellipsoid. If they are all equal, for instance, the ellipsoid becomes a sphere.