Prove Theorem 7.1 (Cauchy-Schwarz): For u and v in a real...

Prove Theorem 7.1 (Cauchy-Schwarz): For $u$ and $v$ in a real inner product space $V$ $\langle u, u\rangle^{2} \leq\langle u, u\rangle\langle v, v\rangle \quad$ or $\quad|\langle u, v\rangle| \leq\|u\|\|v\|$ For any real number $t$ \[\langle t u+v, \quad t u+v\rangle=t^{2}\langle u, u\rangle+2 t(u, v)+\langle v, v\rangle=t^{2}\|u\|^{2}+2 t(u, v\rangle+\|v\|^{2}\] Let $a=\|u\|^{2}, b=2\langle u, v), c=\|v\|^{2} .$ Because $\|t u+v\|^{2} \geq 0,$ we have \[a t^{2}+b t+c \geq 0\] for every value of $t .$ This means that the quadratic polynomial cannot have two real roots, which implies that $b^{2}-4 a c \leq 0$ or $b^{2} \leq 4 a c .$ Thus, \[4\langle u, v\rangle^{2} \leq 4\|u\|^{2}\|v\|^{2}\] Dividing by 4 gives our result.

Prove Theorem 7.1 (Cauchy-Schwarz): For $u$ and $v$ in a real inner product space $V$ $\langle u, u\rangle^{2} \leq\langle u, u\rangle\langle v, v\rangle \quad$ or $\quad|\langle u, v\rangle| \leq\|u\|\|v\|$
For any real number $t$
\[\langle t u+v, \quad t u+v\rangle=t^{2}\langle u, u\rangle+2 t(u, v)+\langle v, v\rangle=t^{2}\|u\|^{2}+2 t(u, v\rangle+\|v\|^{2}\]
Let $a=\|u\|^{2}, b=2\langle u, v), c=\|v\|^{2} .$ Because $\|t u+v\|^{2} \geq 0,$ we have
\[a t^{2}+b t+c \geq 0\]
for every value of $t .$ This means that the quadratic polynomial cannot have two real roots, which implies that $b^{2}-4 a c \leq 0$ or $b^{2} \leq 4 a c .$ Thus,
\[4\langle u, v\rangle^{2} \leq 4\|u\|^{2}\|v\|^{2}\]
Dividing by 4 gives our result.

Key Concepts

Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality is a fundamental result in inner product spaces that bounds the absolute value of the inner product of two vectors by the product of their norms. It encapsulates the idea that the more 'aligned' two vectors are, the larger their inner product; conversely, if they are orthogonal, their inner product is zero. This inequality is instrumental in proving many other results in mathematics, including triangle inequalities and results in statistics and functional analysis.

Quadratic Forms and the Discriminant Condition

When considering expressions involving a real parameter and vectors, one often encounters quadratic forms. A quadratic form in a real variable, derived from the inner product and norm, takes on a non-negative value for all inputs if and only if its discriminant is non-positive. This condition, that the discriminant must be less than or equal to zero, is a key algebraic criterion that links the geometry of inner product spaces with algebraic properties of quadratic equations.

Norm (Induced by the Inner Product)

In a real inner product space, the norm of a vector is defined as the square root of the inner product of the vector with itself. This norm measures the length or magnitude of the vector. It is foundational in measuring distances and establishing convergence in analysis, and it forms the basis for many inequalities and estimations within the space.

Real Inner Product Space

A real inner product space is a vector space over the real numbers equipped with a function called an inner product that assigns a real number to every pair of vectors. This inner product satisfies properties such as linearity in its first argument, symmetry, and positive-definiteness. These spaces provide the geometric framework needed to define angles, lengths, and orthogonality, which are essential in many areas of mathematics and physics.

Transcript

00:03 We're asked to prove the kaushu schwartz inequality or theorem 7 .1.

00:10 We call this theorem says that if you and v are in a real inner product space, v, then the norm of you, the inner product of view with itself squared is less than or equal to the inner product of view with itself squared is less than or equal to the inner product of view with itself times the inner product of v with itself.

00:51 Or in other words, the magnitude or absolute value of the inner product of view with itself is less than or equal to the norm of you times the norm of v.

01:11 Well, as a hint, let's start with the left -hand inequality.

01:22 Consider any real number of t.

01:27 Well, consider the inner product t u plus v with itself.

01:37 Using linearity and symmetry, this is equal to t squared times the interproduct of u with itself, plus two times t times the inner product of u with v plus the inner product of v with itself.

02:00 And of course, by definition, this is the same as t squared times the normal, of u squared plus 2t times the inner product of u with v plus the norm of v squared now make things simpler let's call u squared a the inner product of u with v times 2 b and the norm of v squared call that c.

02:41 Well, because the norm of t u plus b squared is always greater than equal to zero, well, this is the same as the left -hand side of this equation up here, the norm of t -u plus b squared.

03:00 Therefore, it follows that a -t squared plus b -t plus c is always greater than it equal to zero...

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