Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
"matrix compression" and "cauchy's interlacing theorem" might be helpful here for you. – user1709 Dec 22 '10 at 21:23
I guess you are assuming that $\mathbf X$ has full rank... there isn't anything special about the columns of $\mathbf X$? – J. M. Dec 23 '10 at 1:32
@J. M., yes $X$ has full rank. Concerning columns, we can assume that limit $\lim_{n\to\infty}1/n X^TX$ exists and is a full rank matrix. Not sure that it helps. – mpiktas Dec 23 '10 at 3:37
up vote 5 down vote accepted

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.

share|cite|improve this answer
thanks. Who could have thought that bounties can be so effective :) I'll award you one when it is possible (you can award one only after 24 hours), but please fix the formatting ;) – mpiktas Jan 14 '11 at 14:36
@mpiktas, you welcome. I did my best to fix the formatting ;) – Nabyl Bod Jan 14 '11 at 15:52
could you provide a reference or explain how you got the first equality? Pardon the ignorance.. – user1736 Jan 14 '11 at 17:49
@user1736. The spectral theorem shows that the lowest eigenvalue of $X^TX$ is the min of the euclidean norm of $X$ on the unit sphere. – Nabyl Bod Jan 14 '11 at 23:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.