Why is $R(X'X)^{-1}R'$ in the $F$-test for OLS coefficients invertible? Let $b$ denote the OLS estimator, and assuming normality of the errors we get $b \sim \mathop{\mathcal N}\left(\beta, \sigma^2\left(X'X\right)^{-1}\right)$, where $X\in\mathbb R^{N\times K}$ is of full column rank $K$. Now, for $R\in\mathbb R^{J\times K}$ with $J\leq K$ it must be$$Rb\sim \mathop{\mathcal N}\left(R\beta,\sigma^2R\left(X'X\right)^{-1}R'\right)$$which then implies $\left(Rb-R\beta\right)'\left[\sigma^2 R\left(X'X\right)^{-1}R'\right]^{-1}\left(R\beta-R\beta\right)\sim\chi^2_J$.
My question is, how do we know that $R\left(X'X\right)^{-1}R'$ is invertible? $R$ is assumed to be full row rank $J$, and $X'X$ is a symmetric invertible matrix, but I don't see how this implies that the $J\times J$ product is of full rank/invertible. I have seen the derivation from different sources including SE and nobody's mentioned this.
 A: Lemma $1.$ If $\mathbf A$ is an $n\times n$ p. d. matrix and $\mathbf C$ is $p\times n$ matrix of rank $p,$ then $\mathbf C\mathbf A\mathbf C^\top$ is p. d.
The proof is straight forward by noticing $\mathbf x^\top\mathbf C\mathbf A\mathbf C^\top\mathbf x\geq 0$ and
$$\mathbf x^\top\mathbf C\mathbf A\mathbf C^\top\mathbf x= 0\iff\mathbf C^\top\mathbf x =\mathbf 0 \iff \mathbf x= \mathbf 0.\tag 1$$
$\blacksquare$
Observation $1.$ A p. d. matrix is nonsingular.

Reference:
$\rm[ I]$ Linear Regression Analysis, George A. F. Seber, Alan J. Lee, John Wiley & Sons, $2003, $ sec. $\text{A}.4, $ p. $461.$
A: We can write the real positive-definite matrix $\left(X^\top X\right)^{-1}$ as its eigendecomposition $\left(X^\top X\right)^{-1} = Q\Lambda Q^\top$. With $\Lambda = \Lambda^{1/2}\left(\Lambda^{1/2}\right)^\top$ and $Y\mathrel{:=}\left(\Lambda^{1/2}\right)^\top Q^\top R^\top$ we have $R\left(X^\top X\right)^{-1}R^\top = Y^\top Y$, which is positive-definite (and hence invertible) iff $Y$ has full column rank. But since $\left(\Lambda^{1/2}\right)^\top Q^\top$ is invertible, the column rank of $Y$ is the column rank of $R^\top$, which is the row rank of $R$.
