Signature of a quadratic form?

Question

I think it's clear enough that if we have $f(x)$ = $x^2$ , then the function is positive definite, as for each value different from 0, $x^2$ assumes a positive value. However, if I have $f(x, y, z, t)$ = $x^2$ , is it still positive definite or is it only positive? Because if I calculate the associated matrix I have three eigenvalues = 0, and the textbook says that a matrix is definite positive only if all its eigenvalues are positive. However, what's the difference? According to the definition, a function is definite positive if, whatever input apart from 0, the output is positive. Isn't it still the case? Thanks

Answer 1

If $f(x_1,\dots,x_n)$ is positive definite if it takes positive values whenever $(x_1,\dots,x_n)\neq(0,\dots,0)$.

That is the case for $f(x)=x^2$ but not for $f(x,y,z,t)=x^2$.

Note that e.g. $f(0,1,1,1)=0$ while $(0,1,1,1)\neq(0,0,0,0)$.

Answer 2

For your second function to be positive definite, $f(x,y,z,t)=0$ must hold for only $(x,y,z,t)=(0,0,0,0)$, which is not the case. So your function is positive semi-definite.

It's true that the matrix corresponding to the quadratic form (corresponding meaning $\mathbf{x^T}A\mathbf{x}=f(x,y,z,t)=x^2$) has three eigenvalues as zero, which is fine for a positive semi-definite matrix since it's eigenvalues only need to be non-negative.