How to prove Schwartz Inequality?

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SUMMARY

The discussion focuses on proving the Cauchy-Schwarz Inequality for the Hermitian product, specifically the statement that the inner product is less than or equal to the product of the norms ||u|| ||v||. The proof begins by assuming non-zero vectors u and v, utilizing complex numbers r and s to demonstrate that the expression ||rv + su||^2 is non-negative. The discussion emphasizes the importance of the Hermitian properties and concludes that the inequality holds by applying the Pythagorean theorem to the orthogonal decomposition of vector v.

PREREQUISITES
  • Understanding of Hermitian products in complex vector spaces
  • Familiarity with the Cauchy-Schwarz Inequality
  • Knowledge of vector decomposition and orthogonality
  • Basic principles of complex numbers
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  • Study the properties of Hermitian products in detail
  • Review the proof of the Cauchy-Schwarz Inequality in real vector spaces
  • Explore vector decomposition techniques in complex analysis
  • Learn about the implications of orthogonality in vector spaces
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Mathematics students, particularly those studying linear algebra and complex analysis, as well as educators seeking to deepen their understanding of the Cauchy-Schwarz Inequality in the context of Hermitian products.

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Homework Statement



Prove the Schwartz Inequality

/<u, v>/ is < or = //u//*//v//

for the Hermitian product.

Homework Equations


The Attempt at a Solution



Assume u, v are non-zero. For any r, s exist in set of complex numbers, //rv + su// is > or = 0. Then, //rv + su//^2 = <rv+su, rv+su> = <rv, rv> + <rv, su> + <su, rv> + <su, su> = r^2<v, v> + rs <v, u> + sr <u, v> + s^2 <u, u> is > or = 0

Not sure where to go from there...set r = <u u> and s = ...?? I don't know.

Note: I was working off of the proof of the inequality for the dot product. I am just not sure how the Hermitian properties affect it. is <v, u> = <u, v>, etc.
 
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We say (u.v)2 ≤ |u|2 |v|2, then if u or v = 0 then both sides of both sides are zero. But if they are not zero, the vector v can be rewritten as the sum of some scalar multiple of u, ku, and a vector w that is orthogonal to u. The k is obtained by w = v-ku & the orthogonality condition u.w=0

0 = u.w = u.(v-ku) = (u.v)-k(u.u), it follows that k = u.v/u.u, now apply the pythagoras theorem:

|v|2 = |ku|2 + |w|2 = k|u|2 + |w|2

|u|2|v|2 = (u.v) + |u|2|w|2

since |u|2|w|2 ≥ 0, you have (u.v)2 ≤ |u|2 |v|2
This establishes what we said first and hence the cauchy shwartz inequality.
 

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