What is the Relationship Between Injectivity and Surjectivity in Linear Algebra?

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I'm having a lot of trouble figuring out how to do this proof:

Given real numbers a, b, c, d, let f: real^2 ----> real^2 be defined by f(x, y) = (ax + by, cx + dy). Prove that f is injective if and only if f is surjective.

If anybody could help, that would be great.
 
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Do you know what the rank-nullity theorem is?
 
Office_Shredder said:
Do you know what the rank-nullity theorem is?

A quick google search reveals that it appears to be dealing with some linear algebra I don't quite understand...

My book gives me the following hint for this proof: Consider two cases, depending on whether ad - bc = 0.

Not sure if that helps, I'm pretty confused here.
 
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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