Why is this matrix invertible?(nonsingular, full column rank) Assume that $X = [X1 : X2]$ is of full column rank matrix (X is not necessarily square)
then
$$X_2^T(I-X_1(X_1^TX_1)^{-1}X_1^T)X_2$$
is it nonsingular(invertible)?
$P_1=X_1(X_1^TX_1)^{-1}X_1^T$(the projection matrix of X1)
rewrite above matrix
$$X_2^T(I-P_1)X_2$$
Why this matrix is invertible?
What property is it related to?
(I saw this question but didn't understand it clearly
When is Block-Partitioned Matrix Invertible?)
 A: The OP seems to tacitly be working over reals without saying so, otherwise e.g. it isn't clear that $(X_1^T X_1)^{-1}$ exists.
0.) $Q:= I -X_1(X_1^TX_1)^{-1}X_1^T$
1.)  over reals we have
$\text{rank}\Big(A^TA\Big) =\text{rank}\Big(A\Big)$
and since $X_2^T Q X_2 =X_2^T Q^2 X_2 = X_2^TQ^T Q X_2$  it suffices to compute
$\text{rank}\Big(X_2^T Q X_2\Big) = \text{rank}\Big(Q X_2\Big)$ and show that the RHS has full column rank.  Equivalently we want to prove
$\text{rank}\big(Q X_2\big) =\text{rank}\big(X_2\big)$
2.)  Since the original matrix $\mathbf X$ has all columns linearly independent but may not be square, it becomes convenient to extend this to a basis, resulting in the $\text{n x n}$ matrix
$\mathbf X' := \bigg[\begin{array}{c|c|c}X_1 & X_2 & X_3\end{array}\bigg] =\bigg[\begin{array}{c|c}\mathbf X & X_3\end{array}\bigg]$
such that $\det\big(\mathbf X'\big) \neq 0$
suppose $X_1$ has r columns, then
$n-r= \text{rank}\Big(Q\mathbf X'\Big) = \text{rank}\Big(Q\Big) = \text{trace}\Big(Q\Big)$
And
$Q\mathbf X' = \bigg[\begin{array}{c|c|c}Q X_1 & Q X_2 & Q X_3\end{array}\bigg]= \bigg[\begin{array}{c|c|c}\mathbf 0 & Q X_2 & Q X_3\end{array}\bigg] $
where the Right Hand Side is an $\text{n x n}$ matrix with the first $r$ columns zero'd out and has rank $n-r$ i.e. this implies $\text{rank}\big(QX_2\big) = \text{rank}\big(X_2\big)$ which completes the proof.
