# Surjectivity and Orthogonal Complements with Quadratic Forms

My question is asking to clarify the final part of a proof proposed in the following link:

Guarantees of a Quadratic Form for Two Related Matrices

The question being:

Suppose I have a matrix: $\mathbf{Q} =\mathbf{Z}^T\mathbf{B}\mathbf{Z}$, where $\mathbf{Z} \in \mathbb{R}^{n\times m},\mathbf{B}\in\mathbb{R}^{n\times n}, m \leq n.$

Can I say that if $\mathbf{Q}$ is positive definite, then $\mathbf{B}$ is also positive definite?

Now the responder has helped met a lot, but I cannot fathom why "$B$ is not definite positive" in the final part of the proof. Namely in the paragraph:

Now, if $Z$ is not surjective, it's false. Choose B which is definite positive in the range of $Z$ and definite negative in the orthogonal complement. Then for all $X\neq0$, $X^T Q X > 0$ but B is not definite positive.

Simply expanding $X^T Q X$ I arrive at $X^T Z^T B ZX$ but I cannot see how we use orthogonal complement properties from here, or the fact that $Z$ is now non-surjective.

Something clicked and I think I understand it now. I am posting my interpretation here for others to comment on:

Suppose $Z$ is not surjective. This means dim(Im($Z$)) < n.

Everything that Im($Z$) cannot map to is by definition the orthogonal compliment by the following statement for finite dimensional inner product spaces:

$V=Im(Z) \bigoplus Im(Z)^{\perp}$

So let's assume we construct a B such that there exists $x \in Im(Z) : x^TBx > 0$ and there exists $y \in Im(Z)^{\perp} : y^TBy \leq 0$. That is the $B$ is positive definite in the image, and negative definite in the orthogonal compliment.

Then $u^TQu > 0,$ $\forall u \in \mathbb{R}^m$ but $u^T B u$ not necessarily always $> 0$ due to its construction. That is $u \in \mathbb{R}^m$ but sometimes $u \in Im(Z)$ and other times $u \in Im(Z)^{\perp}$.

(I think in fact if $B$ is symmetric then $Im(Z)^{\perp} = ker(Z^T) = ker(Z) \Rightarrow u^T B u = 0$).

Let me know how others feel about the precision of this argument.