- #1
RJLiberator
Gold Member
- 1,095
- 63
Homework Statement
Show that |<v|w>|^2 ≤ <v|v><w|w>
for any |v>,|w> ∈ ℂ^2
Homework Equations
The Attempt at a Solution
The Cauchy-Schwartz inequality is extremely relevant for the math/physics that I am interested in.
I feel like I have a very good proof here, but I am interested in a few things here.
1) Is my proof correct in the right context (Complex numbers here)
2) What is your favorite proof of the Cauchy Schwartz Inequality? (I want to master this one)
Proof:
Case 1: We have to show that if |v> or |w> = 0, then the sides are equal to 0. Which is somewhat trivial to prove by simply plugging in the values and reaching the conclusion using inner space axioms.
Case 2: Where |v> and |w> are not equal to 0.
Here, I let:
a = <v|v>
b = 2<v|w>
c = <w|w>
By Inner product axiom 3 stating that the inner product of any vector with itself is nonnegative we see
0 ≤ <tv+w | tv+w >
0 ≤ <v|v>t^2+2<v|w>t+<w|w>
0 ≤ at^2+bt+c
The conclusion here is that the inequality implies that the quadratic polynomial has either no real roots or a repeated real root. Therefore the discriminant must satisfy: b^2-4ac ≤ 0.
By plugging in the necessary values we see:
4<v|w>^2-4<v|v><w|w> ≤ 0
<v|w>^2 ≤ <v|v><w|w>
Which is what we set out to prove.
So, my one concern is that this may not be complete for complex inner products, but rather only for real numbers? Is there any cause for that concern?