Linear Algebra Proof: Prove r(CA) < n

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Homework Statement



Prove (without using determinants):

If A is an n x n matrix and r(A) < n, then for any n x n matrix C, r(CA) < n.
Hint: Can CA be invertible?

2. The attempt at a solution

Well, I'm running on almost very little sleep, so I can't really think today. What's the proof?

And...i spelled algebra wrong...
 
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I'm assuming that r(A) stands for rank(A), if so, then we have that the dimension of the row and column spaces are smaller than n. I'm not going to do the work for you, but take a look at the Fundamental Theorem of Invertible matrices, there's something in there that will greatly help you (One of the properties).

Try to show some work though.
 

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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