Prove help. rank of inverse matrix

pcming
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I can't find out how to prove this question. Can anyone help?

Let A be an n x m matrix of rank m, n>m. Prove that (A^t)A has the same rank m as A.

Where A^t = the transpose of A.

I seen someone else have asked the question before and had got the answer. However I can't understand it. Hope someone can give me a more detail suggestion. Thanks!
 
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You need to show some work first. You say you've seen a proof already but are having trouble understanding it, well what is the proof and which part is giving you trouble. There is more than one way to do it.

It's a good idea to attempt the proof yourself first without looking at the answer.

Edit: I just noticed the other thread you're talking about which has the answer you don't understand, and which ironically was written by me along ways back :smile: I responded to your question in that thread.
 
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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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