Linear Algebra: Nullity of Matrix Product Question

workerant
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Prove that for any m*n matrix A and any n*n matrix B, nullity A is less than or equal to nullity AB.

Sorry Dick: I know nullity of A is n-rank A and this is always greater than or equal to 0 and I figured nullity AB is like nullity A * nullity B, which are both greater than or equal to 0. Is this in the right direction?
 
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Any thoughts on this? Even if you have no idea how to solve it you are supposed to at least thrash around and show an attempt. Such are the forum rules.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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