Cauchy-Schwarz Inequality Proof Question

In summary, the Cauchy-Schwarz Inequality states that the dot product of two vectors is less than or equal to the product of their magnitudes. This can be proven by considering the dot product of a vector by itself, which is always greater than or equal to 0. By factoring and completing the square, it can be shown that the inequality holds for any possible value of t. Therefore, it holds for all possible values of t and is valid for all cases.
  • #1
Moridin
692
3

Homework Statement



Prove the Cauchy-Schwarz Inequality.

Homework Equations



[tex]|\mathbf{x \cdot y}| \leq |\mathbf{x}||\mathbf{y}|, \forall \mathbf{x,y} \in \mathbb{R}^{n}[/tex] (1)

The Attempt at a Solution



If x is equal to 0, then both sides are equal to 0. If x not equal to 0 the following applies. The dot product of a vector by itself is always greater or equal to 0. This means that the following holds:

[tex]\mathbf{0} \leq (t \mathbf{x} + \mathbf{y}) \cdot (t \mathbf{x} + \mathbf{y}) = t^{2} |\mathbf{x}|^{2} + 2t\mathbf{x} \cdot \mathbf{y} + |\mathbf{y}|^{2}[/tex] (2)

Factoring |x|2 (which is not equal to zero) gives:

[tex]|\mathbf{x}|^{2} (t^{2} + 2 \frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{x}|^{2}}t + \frac{|\mathbf{y}|^{2}}{|\mathbf{x}|^{2}})[/tex] (3)

Completing the square gives:

[tex]|\mathbf{x}|^{2} ( (t + \frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{x}|^{2}})^{2} - \frac{(\mathbf{x} \cdot \mathbf{y})^{2}}{|\mathbf{x}|^{4}} + \frac{|\mathbf{y}|^{2}}{|\mathbf{x}|^{2}})[/tex] (4)

Expansion gives:

[tex]|\mathbf{x}|^{2}(t + \frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{x}|^{2}})^2 - \frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{x}|^{2}} + |\mathbf{y}|^{2}[/tex] (5)

The parenthesis is equal to zero for an appropriate value of t, that is:

[tex]t = - \frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{x}|^2}[/tex] (6)

Then the following is true:

[tex]0 \leq |\mathbf{y}|^{2} - \frac{\mathbf{x} \cdot \mathbf{y}^{2}}{|\mathbf{x}|^{2}}[/tex] (7)

Which, rearranged and the root taken of both sides, becomes:

[tex]|\mathbf{x \cdot y}| \leq |\mathbf{x}||\mathbf{y}|[/tex] (8)

Analogous treatment for y. Q.E.D

My question regarding this attempt is with regards to the justification of moving from step 5 to 6 and 7. Is it valid just to pick a specific t, even though the first step does not fixate it? Or is that t that t which gives the lowest value of the RHS in the inequality, so all other t makes the RHS increase, so that if it is valid for the smallest possible RHS (which is the result of that t), then it is possible for all RHS larger (any other t)? The latter seems more intuitive, since the square cannot possible be negative, so the minimum it can get is 0?

Thank you for your time, have a nice day.
 
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  • #2
It's true for all t, so what could be wrong with picking a specific t? You could have picked any other t and still gotten a valid inequality, it just wouldn't be Cauchy-Schwarz.
 
  • #3
Right, since the inequality holds for all possible t, it must hold for any tn that is a possible t.

Thanks.
 

1. What is the Cauchy-Schwarz Inequality?

The Cauchy-Schwarz Inequality is a mathematical concept that states the relationship between the inner product of two vectors and their magnitudes. It states that the absolute value of the inner product of two vectors is less than or equal to the product of their magnitudes.

2. How is the Cauchy-Schwarz Inequality proven?

The Cauchy-Schwarz Inequality is proven using the concept of the dot product and the properties of vectors. It can be proven algebraically or geometrically.

3. What is the importance of the Cauchy-Schwarz Inequality in mathematics?

The Cauchy-Schwarz Inequality is an important concept in mathematics as it has a wide range of applications in various fields such as linear algebra, geometry, and statistics. It helps in proving other mathematical theorems and inequalities and also has practical applications in real-world problems.

4. Can the Cauchy-Schwarz Inequality be extended to higher dimensions?

Yes, the Cauchy-Schwarz Inequality can be extended to higher dimensions. In fact, it is known as the generalized Cauchy-Schwarz Inequality and it applies to any number of vectors in an n-dimensional space.

5. What are some common applications of the Cauchy-Schwarz Inequality?

The Cauchy-Schwarz Inequality has various applications in different fields such as optimization problems, signal processing, and statistics. It is also used in proving other important theorems in mathematics, such as the Triangle Inequality and the Arithmetic Mean-Geometric Mean Inequality.

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