TITLE: Do the maximal eigenvalues of $(X^TX)^{-1}$ matrix increase when number of columns of $X$ increases?
QUESTION [7 upvotes]: Suppose we have a square real $n\times n$ matrix $X=[x_1,...,x_n]$, where $x_i$ is $i$-th column of the matrix. 
Now define  $X_k=[x_1,..,x_k]$, i.e. matrix $X_k$ columns are the first $k$ columns of the matrix $X$. Define $\lambda_k$ as the maximal eigenvalue of $(X_k^TX_k)^{-1}$. 
Is it possible to prove that $\lambda_1\le \lambda_2\le ... \le \lambda_n$? 
If $X$ is orthogonal, then the answer is yes. Maybe this holds only for certain matrices? Any pointers would be greatly appreciated.

REPLY [5 votes]: The answer is Yes:
Assume that $X_{k+1}^TX_{k+1}$ is invertible. One can check that its lowest absolute eigenvalue is given by
$$\left\vert{\frac1{\lambda_{k+1}}}\right\vert=\min\limits_{y\in{\mathbb R}^{k+1},\,\left\lVert y\right\lVert_2=1}\left\lVert{X_{k+1}y}\right\lVert_2^2\,,$$
It holds
$$\min\limits_{y\in{\mathbb R}^{k+1},\,\left\lVert y\right\lVert_2=1}\left\lVert{X_{k+1}y}\right\lVert_2^2\leq\min\limits_{y\in{\mathbb R}^{k},\,\left\lVert y\right\lVert_2=1}\left\lVert{X_{k}y}\right\lVert_2^2\,.$$
This completes the proof.