April 20, 2014

Homework Help: math

Posted by mathstudent on Thursday, January 4, 2007 at 11:53am.

I'm having trouble understanding one step in a proof of the Cauchy-Schwarz inequality:

u = a non-zero vector
v = another vector
a = <u,u> (so a > 0 by positivity axiom)
b = 2<u,v>
c = <v,v> (so c >= 0 by positivity axiom)
t = any real number

0 <= <tu + v, tu + v> (by positivity axiom)
0 <= t^2<u,u> + 2t<u,v> + <v,v>
0 <= at^2 + bt + c

I understand thus far. But then, the book says, "This inequality implies that the quadratic polynomial at^2 + bt + c has either no real roots or a repeated real root. Therefore its discriminant must satisfy the inequality b^2 - 4ac <= 0"

Why does the inequality imply that the quadratic has no real roots or a repeated real root? Why can't the quadratic have two different roots if b^2 - 4ac > 0?

If there are two different roots, say y1 and y2 such that y1 < y2, then the function would factorize as:

at^2 + bt + c = A(t-y1)(t-y2)

This means that the function will change sign if you let t increase from a value smaller than y1 to a value larger than y1 but still smaller than y2. But a change of sign of this function is precluded because it has to be larger than zero for all t.

makes perfect sense. thanks!

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