Linear Algebra: Proving AB Not Invertible for mXn Matrix

gotmilk04
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Homework Statement


If A is an mXn matrix, B is an nXm matirx, and n<m, then AB is not invertible.


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The Attempt at a Solution


By doing A is a 2X1 and B is a 1X2, I find that AB is not linearly independent, so it cannot be invertible, but I'm not sure how to show that for all matrices of this nature.
 
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Can you show there is a nonzero vector x such that Bx=0? That would make big problems for AB being invertible. And don't PM people about problems, ok? Just post it on the forums and wait a bit.
 
Since n<m, there will be a free variable in the nXm matrix B when reduced to echelon form, correct? So then there is obviously more than the trivial solution.
I'm still confused as to why that creates a problem for AB being invertible.
 
If AB has an inverse (AB)^(-1), then for every x, (AB)^(-1)*ABx=x. What happens if ABx=0 and x is not zero?
 
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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