# If $A=(a_{ij})$ is positive semidefinite, prove that $a_{ij}^2 \leq a_{ii}a_{jj}$ for all $i \neq j$

If $A=(a_{ij})$ is positive semidefinite, prove that $a_{ij}^2 \leq a_{ii}a_{jj}$ for all $i \neq j$.
I don't even know how to get started, any hint is appreciated, thanks a lot.

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see related math.stackexchange.com/questions/235170/… –  Will Jagy Nov 20 '12 at 18:06

Hint: By Cauchy-Schwarz you have $(x^tAy)^2 \le (x^tAx)(y^t Ay)$ for any $x,y \in \mathbb R^d$.

You can of course try to mimic the proof of Cauchy-Schwartz and start with $$0 \le (e_i + se_j)^tA(e_i + se_j) = a_{ii} + 2sa_{ij} + s^2a_{jj}$$ Now let $s = -\frac{a_{ij}}{a_{jj}}$. You of course have to consider the case $a_{jj} = 0$ seperately.

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Is there any alternative approach that does not require the use of this inequality? –  drawar Nov 20 '12 at 11:04
@drawar Why do you not want to use Cauchy-Schwartz? –  martini Nov 20 '12 at 11:08
Actually I'm practicing for my exam and Cauchy-Schwarz Inequality is not covered in my course so I'm not supposed to use it. –  drawar Nov 20 '12 at 11:14
@user1551: My bad. Thanks! –  drawar Nov 20 '12 at 11:15
You need $(x^tAx)\geq0$ for all $x$ and $x$. Now if you set $x=e_i+te_j$, you get a quadratic polynomial in $t$ which must be non-negative for all $t$. So this polynomial cannot have any non-repeated roots, as polynomials must be positive on one side of these and negative on the other.
So the discriminant of this polynomial (i.e. $b^2-4ac$ if your polynomial is $at^2+bt+c$) must be non-positive - when you work out what the coefficients are in terms of the matrix entries, this will be the inequality you want.
Your first line should be You need $(x^t A x) \geq 0$ for all $x.$ –  Will Jagy Nov 20 '12 at 18:09