what are the conditions on $X$ such that $X^TAX$ is positive definite?

Question

Suppose I have a matrix $A$ which is positive definite and $X$ is a matrix with spectral radius less than $1$ (I think this information is irrelevant). under what conditions we get that $X^TAX$ is also PD. My guess is that $X$ has to also be PD. Also, forgot to mentions that all matrices are square here.

Answer 1

$X$ only has to be invertible, no other condition necessary. Write $\langle v, X^T AX v \rangle = \langle Xv, AXv \rangle = \langle w, Aw \rangle \ge 0$ where $w = Xv$, with equality iff $w = 0$ since $A$ is PD. This condition is equivalent to $v = 0$ iff $X$ is invertible.

Answer 2

@Qiaochu Yuan provided already a complete answer. Nevertheless, I want to add some "intuition", which is too long for a comment:

Note that $A$ is p.d iff $\langle w, Aw\rangle > 0$ for all $w\in\mathbb R^n$ with $w\neq 0_n$. Let $f:\mathbb R^n\rightarrow\mathbb R^n$ be a function. If $f$ is a surjective function $f:\mathbb R^n\rightarrow\mathbb R^n$ then for every $w\in\mathbb R^n$ there exists a $v\in\mathbb R^n$ such that $w = f(v)$. Hence we have that $\langle w, Aw\rangle = \langle f(v), Af(v)\rangle$. However, there may the problem that there exists a $v^*\in\mathbb R^n$ with $v^*\neq 0_n$ such that $f(v^*) = 0_n$. Consequently, we cannot ensure that $\langle f(v),Af(v)\rangle > 0$ for all $v\in\mathbb R^n$. To ensure that this does not happen, we have to assume, in addition, that $f$ is an injective function. Because then there exists only one $v^*\in\mathbb R^n$ such that $f(v^*) = 0_n$, namely $v^* = 0_n$. If $f$ is surjective and injective, then the inverse function $f^{-1}$ exists and is unique, and we have that $\langle w,Aw\rangle > 0$ for all $w\in\mathbb R^n$ iff $\langle f(v), Af(v)\rangle > 0$ for all $v\in\mathbb R^n$.

In the special case where $f$ is a linear function, i. e. $f(v) = Xv$ for some matrix $X$, we have that $f$ is injective iff $f$ is surjective. So it suffices to show that $f$ is either injective or surjective. In most cases it's easier to show that $f$ is injective, because $f$ is injective iff $f(v) = 0_n$ if and only if $v = 0_n$. This is what Qiaochu Yuan suggests.