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Suppose \(v\) and \(w\) are elements of an inner product space. Show that $$ 2|\langle\mathbf{v}, \mathbf{w}\rangle| \leq\|\mathbf{v}\|^{2}+\|\mathbf{w}\|^{2} . $$

Short Answer

Expert verified
The proof that \( 2|\langle\mathbf{v}, \mathbf{w}\rangle| \leq \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2 \) is obtained through the application of the Cauchy-Schwarz inequality, subsequent simplification, and careful mathematical manipulation.

Step by step solution

01

Recalling the Cauchy-Schwarz Inequality

For any vectors \( \mathbf{u} \) and \( \mathbf{v} \) in an inner product space, the Cauchy-Schwarz Inequality states that \( |\langle\mathbf{u}, \mathbf{v}\rangle|^{2} \leq \|\mathbf{u}\|^{2} \|\mathbf{v}\|^{2} \) .
02

Applying the Cauchy-Schwarz Inequality to our vectors

Applying the Cauchy-Schwarz Inequality to the vectors \( \mathbf{v} \) and \( \mathbf{w} \), we find that \( |\langle\mathbf{v}, \mathbf{w}\rangle|^{2} \leq \|\mathbf{v}\|^{2} \|\mathbf{w}\|^{2} \) .
03

Simplifying the inequality

Taking the square root of both sides of the inequality, \( |\langle\mathbf{v}, \mathbf{w}\rangle| \leq \sqrt{\|\mathbf{v}\|^{2} \|\mathbf{w}\|^{2}} \) . This can be further simplified to \( |\langle\mathbf{v}, \mathbf{w}\rangle| \leq \|\mathbf{v}\| \|\mathbf{w}\| \).
04

Further Manipulation

Now, square both sides of this inequality to obtain \( |\langle\mathbf{v}, \mathbf{w}\rangle|^2 \leq \|\mathbf{v}\|^2 \|\mathbf{w}\|^2 \). Let \( x = \|\mathbf{v}\| \) and \( y = \|\mathbf{w}\| \) for easier manipulation. Then our inequality becomes \( |\langle\mathbf{v}, \mathbf{w}\rangle|^2 \leq x^2y^2 \).
05

The Final Proof

Divide both sides by \(2xy\), we get \( \frac{|\langle\mathbf{v}, \mathbf{w}\rangle|^2}{2xy} \leq \frac{(x+y)^2}{2} \). Square rooting both sides simplifies to \( |\langle\mathbf{v}, \mathbf{w}\rangle| \leq \frac{1}{2}(x+y) \), which is equivalent to the required \( 2|\langle\mathbf{v}, \mathbf{w}\rangle| \leq \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cauchy-Schwarz Inequality
The Cauchy-Schwarz Inequality is a fundamental concept in linear algebra and inner product spaces. It provides a vital bound on the absolute value of the inner product of two vectors. Given two vectors, \( \mathbf{u} \) and \( \mathbf{v} \), in an inner product space, the inequality is expressed as: \[ |\langle \mathbf{u}, \mathbf{v} \rangle|^2 \leq \|\mathbf{u}\|^2 \|\mathbf{v}\|^2. \] Here, \( |\langle \mathbf{u}, \mathbf{v} \rangle| \) represents the absolute value of their inner product, while \( \|\mathbf{u}\| \) and \( \|\mathbf{v}\| \) denote the norms of the vectors.
  • **Inner Product**: Think of it like multiplying both vectors which can give an idea of the vectors' alignment.
  • **Norm**: It's similar to length or magnitude of a vector.
This inequality essentially identifies the spread or angle between vectors, acting as a measure to confirm that their inner product is limited by the lengths of the vectors themselves.
Vector Norm
A vector norm is a crucial aspect of understanding vectors in any mathematical space. It essentially measures the size or length of a vector. Think of it as a way to quantify how long or large a vector is. An important property of the vector norm is that it is always non-negative. For a vector \( \mathbf{v} \), its norm is denoted as \( \|\mathbf{v}\| \). The standard or Euclidean norm, for instance, is computed as: \[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + \cdots + v_n^2}, \] where \( v_1, v_2, \ldots, v_n \) are the vector's components.
  • **Measures Distance**: It gives a straight-line distance from the origin in a geometric space.
  • **Used in Inequalities**: Norms are extensively employed in inequalities to bound vector quantities.
Understanding vector norms is pivotal to grasping how vectors relate to each other in an inner product space. It provides a concrete basis for discussing vector magnitude and later helps in analyzing more complex concepts like the Cauchy-Schwarz inequality.
Inequalities in Linear Algebra
Inequalities in linear algebra are fundamental tools that help understand relationships between vectors and their resultant quantities. They provide constraints and bounds on vector operations. The exercise demonstrated a specific type of inequality, but there are numerous others that you will encounter in linear algebra. These inequalities often involve:
  • **Bounding**: Restricting a vector's value or magnitude.
  • **Comparisons**: Assessing relationships between different vectors.
  • **Properties**: Often they lead to insights about the structure and behavior of vectors.
By leveraging inequalities, you can better understand how vectors behave and interact with one another in a space. They are instrumental in optimizing problems and can often reveal inherent symmetries or patterns in complex data. In this context, the exercise utilized the Cauchy-Schwarz inequality to derive a useful conclusion about the vectors \( \mathbf{v} \) and \( \mathbf{w} \), showing how inequalities can form the backbone of proofs and solutions in linear algebra.

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Most popular questions from this chapter

a. Use the norm defined in terms of the standard inner product on \(\mathrm{C}(1-\pi, \pi])\) to compute \(\|\sin \|\), \(\|\cos \|\), and \(\|\mathbf{1}\|\), where \(\mathbf{1}\) denotes the constant function with value 1 . b. Use the norm defined in terms of the standard inner product on \(\mathrm{C}([0, \pi])\) to compute \(\|\sin \|,\|\cos \|\), and \(\|\mathbf{1}\|\).

Verify that \(\left\langle\left[\begin{array}{ll}a & b \\ c & d\end{array}\right],\left[\begin{array}{ll}a^{\prime} & b^{\prime} \\\ c^{\prime} & d^{\prime}\end{array}\right]\right\rangle=a a^{\prime}+b b^{\prime}+c c^{\prime}+d d^{\prime}\) defines an inner product on \(\mathbb{M}(2,2)\).

a. Hundreds of proofs of the Pythagorean Theorem have been recorded. Look one up or try to discover one yourself. For instance, the altitude from the right angle to the hypotenuse gives two triangles that are similar to the original right triangle. The equality of the ratios of the lengths of corresponding sides of these triangles leads to the desired conclusion in a few simple algebraic steps. b. Use the Pythagorean Theorem to derive the formula for the distance between two points \(\left(x_{1}, x_{2}\right)\) and \(\left(y_{1}, y_{2}\right)\) in \(\mathbb{R}^{2}\). c. Use the Pythagorean Theorem to derive the formula for the distance between two points \(\left(x_{1}, x_{2}, x_{3}\right)\) and \(\left(y_{1}, y_{2}, y_{3}\right)\) in \(\mathbb{R}^{3}\). (Suggestion: First find the distance from \(\left(x_{1}, x_{2}, x_{3}\right)\) to \(\left(y_{1}, y_{2}, x_{3}\right)\).) d. Generalize your proof to derive the formula for the distance between two points in \(\mathbb{R}^{n}\).

Suppose \(S\) is a subspace of a finite-dimensional inner product space \(V\). With \(S^{\perp}\) defined as in the previous exercise, show that \(\operatorname{dim} S+\operatorname{dim} S^{\perp}=\operatorname{dim} V\).

Suppose the Gram-Schmidt process is applied to a basis \(\left\\{\mathbf{u}_{1}, \ldots, \mathbf{u}_{n}\right\\}\), where the first \(k\) vectors \(\mathbf{u}_{1}, \ldots, \mathbf{u}_{k}\) are orthonormal. Show that \(\mathbf{e}_{i}=\mathbf{u}_{i}\) for \(i=\) \(1, \ldots, k\).

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