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I was looking at UCLA's Department of Mathematics's Fall 2015 Basic Qualifying Exam, and question #12 is the following.

Show that the following matrix is positive definite.

$$M=\begin{pmatrix} 2 & 1 & 1 & \ldots & 1\\ 1 & 3 & 1 & \ldots & 1\\ 1 & 1 & 4 & \ldots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 1& 1& \ldots &n+1 \end{pmatrix}$$

I wasn't super sure how one should proceed; showing $x^TMx>0$ for all $x \neq 0$ seems like a pain, and I was curious if there was a slick way to do this. Other ways would be to write $M = AA^T$ or go through the characteristic polynomial, but I'm not super sure about those either. If anyone can give some hints, that would be great!

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2 Answers 2

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Write $$M=\begin{pmatrix} 1 & 0 & 0 & \ldots & 0\\ 0 & 2 & 0 & \ldots & 0\\ 0 & 0 & 3 & \ldots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0& 0& \ldots &n \end{pmatrix}+ \begin{pmatrix} 1 & 1 & 1 & \ldots & 1\\ 1 & 1 & 1 & \ldots & 1\\ 1 & 1 & 1 & \ldots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 1& 1& \ldots &1 \end{pmatrix}.$$

Now it's easy to see that the first of these matrices is positive-definite, while the second is positive-semidefinite (in particular, its eigenvalues are $n$ and $0$).

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    $\begingroup$ +1 Nice solution! If you don't immediately see that the matrix of $1$ is positive definite, you can also leave the lower part of $1$'s in the first matrix and the upper part in the second matrix. Then, since you can read the eigenvalues of triangular matrix on the diagonal, you immediately see that both matrices are positive definite and you're done :) $\endgroup$
    – Ant
    Oct 1, 2017 at 11:24
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Maybe you can consult https://en.wikipedia.org/wiki/Sylvester%27s_criterion. Proceed by induction.

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