According to Wikipedia, any quadratic function can be written as $z^TMz$, where $z$ is a column vector and $M$ is a symmetric real matrix. However, this quadratic function is strictly convex only when $M$ is symmetric positive definite. Why? I thought any quadratic function should be convex?
Doesn't $z^TMz>0$ shows only that the range of this function is greater than zero?
Why isn't any symmetric matrix $M$ (which represents a quadratic function) convex?
Why is it only the case that when $z^TMz > 0$ the function is strictly convex?
UPDATE: As pointed out in the comments by @Erik, positive definiteness is a sufficient condition for strict convexity. However in these case, positive definiteness is indeed directly implied since the second derivative is a positive definitive matrix.
PREVIOUS ANSWER: For any twice differentiable function, it is strictly convex if and only if, the Hessian matrix is positive definite. You can find it from any standard textbook on convex optimization. Now here the function at hand is $z^TMz$ which is clearly twice differentiable (by virtue of being quadratic). Now the Hessian of this function is $M$ (please verify yourself, it helped me a lot to memorize it). So $M$ should be positive definite for that quadratic function to be convex.