Cauchy-Schwarz Inequality

Given two vectors $ v_1 $ and $ v_2 $ in the real two-dimensional space $ \mathbb{R}^2 $, the absolute value of their dot product $ |( v_1, v_2 )| $ is less than or equal to the product of their magnitudes $ |v_1||v_2| $. $$ ( v_1, v_2 ) \le |v_1||v_2| $$

This is one of the fundamental inequalities in linear algebra and analysis.

A Practical Example
Proof

A Practical Example

Let's consider the following vectors in $ \mathbb{R}^2 $:

$$ v_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix} $$

$$ v_2 = \begin{pmatrix} 3 \\ 4 \end{pmatrix} $$

The dot product $ (v_1, v_2) $ is calculated as:

$$ (v_1, v_2) = 1 \times 3 + 2 \times 4 = 3 + 8 = 11 $$

Next, let's find the magnitudes $ |v_1| $ and $ |v_2| $ of the two vectors.

The magnitude of $ v_1 $ is:

$$ |v_1| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} $$

And the magnitude of $ v_2 $ is:

$$ |v_2| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$

The Cauchy-Schwarz inequality tells us that for any two vectors $ v_1 $ and $ v_2 $ in the real two-dimensional space $ \mathbb{R}^2 $, the absolute value of their dot product is less than or equal to the product of their magnitudes.

$$ |(v_1, v_2)| \leq |v_1||v_2| $$

In this case, the dot product is $ (v_1, v_2) = 11 $ while the magnitudes are $ |v_1| = \sqrt{5} $ and $ |v_2| = 5 $.

$$ |11| \leq \sqrt{5} \times 5 $$

$$ 11 \leq 5 \times \sqrt{5} \approx 11.18 $$

Therefore:

$$ 11 \leq 11.18 $$

This confirms the Cauchy-Schwarz inequality, showing that this example satisfies the inequality:

Proof

Given $ v_1 = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} $ and $ v_2 = \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} $, the dot product is defined as:

$$ (v_1, v_2) = x_1 x_2 + y_1 y_2 $$

The magnitudes of the vectors $ v_1 $ and $ v_2 $ are:

$$ |v_1| = \sqrt{x_1^2 + y_1^2} $$

$$ |v_2| = \sqrt{x_2^2 + y_2^2} $$

The Cauchy-Schwarz inequality asserts that:

$$ |x_1 x_2 + y_1 y_2| \leq \sqrt{x_1^2 + y_1^2} \sqrt{x_2^2 + y_2^2} $$

Geometrically, this inequality expresses that the absolute value of the dot product between two vectors is at most equal to the product of their magnitudes.

This maximum is reached when the vectors are parallel (or antiparallel).

In other words, if $ \theta $ is the angle between the vectors $ v_1 $ and $ v_2 $, then the dot product is given by:

$$ (v_1, v_2) = |v_1||v_2| \cos \theta $$

Since $ |\cos \theta| \leq 1 $, it follows that the product of the magnitudes is always greater than or equal to the absolute value of the dot product:

$$ |(v_1, v_2)| \leq |v_1||v_2| $$

Thus, the Cauchy-Schwarz inequality can be expressed as:

$$ |(v_1, v_2)| \leq |v_1||v_2| $$

Note that it is incorrect to state that the dot product is less than or equal to the product of the magnitudes $ (v_1, v_2) \leq |v_1||v_2| $ because the dot product $ (v_1, v_2) $ could be negative. Therefore, the correct form involves the absolute value, meaning the absolute value of the dot product is less than or equal to the product of the magnitudes: $$ |(v_1, v_2)| \leq |v_1||v_2|. $$

Alternative Proof

Consider two vectors $ v_1 $ and $ v_2 $ in $ \mathbb{R}^2 $ with components:

$$v_1 = (x_1, y_1) $$

$$ v_2 = (x_2, y_2) $$

For any real value $ t $, the following expression is non-negative:

$$ 0 \leq (x_1 + t x_2)^2 + (y_1 + t y_2)^2 $$

This is obvious, as the square of any number is always greater than or equal to zero, i.e., non-negative.

Thus, the sum of two squares is always non-negative.

Let's expand the expression by performing the algebraic calculations:

$$ 0 \leq (x_1 + t x_2)^2 + (y_1 + t y_2)^2 $$

$$ 0 \leq x_1^2 + 2t x_1x_2 + t^2x_2^2 + y_1^2 + 2t y_1y_2 + t^2 y_2^2 $$

$$ 0 \leq x_1^2 + y_1^2 + 2t(x_1x_2 + y_1y_2) + t^2(x_2^2 + y_2^2) $$

The sum $ x_1^2 + y_1^2 $ is the square of the magnitude $ | v_1 | = \sqrt{x_1^2+y_1^2} $, i.e., $ | v_1 |^2 = x_1^2+y_1^2 $

$$ 0 \leq |v_1|^2+ 2t(x_1x_2 + y_1y_2) + t^2(x_2^2 + y_2^2) $$

Similarly, the sum $ x_2^2 + y_2^2 $ is the square of the magnitude $ | v_2 | = \sqrt{x_2^2+y_2^2} $, i.e., $ | v_2 |^2 = x_2^2+y_2^2 $

$$ 0 \leq |v_1|^2+ 2t(x_1x_2 + y_1y_2) + t^2 | v_2 |^2 $$

The expression $ x_1x_2 + y_1y_2 $ is the dot product $ (v_1, v_2) $

$$ 0 \leq |v_1|^2+ 2t(v_1, v_2) + t^2 | v_2 |^2 $$

Now, if we substitute $ \alpha = |v_2|^2 $, $ \beta = (v_1, v_2) $, and $ \gamma = |v_1|^2 $, we obtain a quadratic equation:

$$ 0 \leq \alpha t^ 2 + 2 \beta t + \gamma $$

Since this quadratic expression is non-negative for all $ t $, the discriminant of the associated quadratic equation must be less than or equal to zero, i.e.,

$$ \Delta = 4\beta^2 - 4\alpha \gamma \leq 0 $$

Explanation: If the discriminant $ \Delta > 0 $ were positive, then the polynomial $ P(t) = \alpha t^2 + 2 \beta t + \gamma $ would have two distinct real roots, implying that the polynomial $ P(t) $ is positive in some regions and negative in others. However, this is impossible because we have already established that the polynomial is non-negative as $ 0 \leq (x_1 + t x_2)^2 + (y_1 + t y_2)^2 $. The only case where $ P(t) \ge 0 $ for all $ t $ occurs when the polynomial has no distinct real roots, which happens when the discriminant of the equation $ \alpha t^2 + 2 \beta t + \gamma = 0 $ is less than or equal to zero, $ \Delta \le 0 $.

Let's now proceed with the algebraic calculations:

$$ \Delta = 4\beta^2 - 4\alpha \gamma \leq 0 $$

$$ 4\beta^2 \leq 4\alpha \gamma $$

Dividing both sides by 4, we get:

$$ \beta^2 \leq \alpha \gamma $$

Recalling that $ \beta = (v_1, v_2) $, $ \alpha = |v_2|^2 $, and $ \gamma = |v_1|^2 $, the last expression can be rewritten as:

$$ (v_1, v_2)^2 \leq |v_1|^2 |v_2|^2 $$

$$ \sqrt{ (v_1, v_2)^2 } \leq \sqrt{ |v_1|^2 |v_2|^2 } $$

$$ \sqrt{ (v_1, v_2)^2 } \leq |v_1| |v_2| $$

Knowing that the square root of a squared number is always non-negative $ \sqrt{n^2} = |n| $, we obtain the Cauchy-Schwarz inequality.

$$ |(v_1, v_2)| \leq |v_1| |v_2| $$

This proof relies on the fact that the discriminant of a quadratic polynomial must be non-positive to ensure that the polynomial remains non-negative at all points.

This condition is sufficient to conclude that the Cauchy-Schwarz inequality holds in $ \mathbb{R}^2 $.

And so on.