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Solution by Steps

step 1

We need to find the distance from the point

\mathbf{a} = (2, 1)

to the line

L

defined by

\mathbf{x} = t \mathbf{e}

, where

\mathbf{e} = (-1, 3)

and

t \in \mathbb{R}

step 2

The line

L

can be parameterized as

\mathbf{x} = t(-1, 3)

step 3

The distance from a point to a line in

\mathbb{R}^2

can be found using the formula:

d = \frac{\|\mathbf{a} \times \mathbf{e}\|}{\|\mathbf{e}\|}

, where

\mathbf{a} \times \mathbf{e}

is the cross product of

\mathbf{a}

and

\mathbf{e}

step 4

Calculate the cross product

\mathbf{a} \times \mathbf{e}

:

\mathbf{a} \times \mathbf{e} = \begin{vmatrix} \mathbf{i} &amp; \mathbf{j} \\ 2 &amp; 1 \\ -1 &amp; 3 \\ \end{vmatrix} = (2 \cdot 3 - 1 \cdot (-1)) = 6 + 1 = 7

step 5

Calculate the norm of

\mathbf{e}

:

\|\mathbf{e}\| = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}

step 6

Substitute the values into the distance formula:

d = \frac{7}{\sqrt{10}} \approx 2.21359

Answer

The distance from the point

\mathbf{a} = (2, 1)

to the line

L

is approximately

2.21359

.

Key Concept

Distance from a point to a line in

\mathbb{R}^2

Explanation

The distance from a point to a line in

\mathbb{R}^2

can be calculated using the cross product of the point and the direction vector of the line, divided by the norm of the direction vector.

Solution by Steps

step 1

We need to determine if the operator

A: L^{2}(0,1) \mapsto L^{2}(0,1)

defined by

A x(t)=\int_{0}^{1} \sqrt{t+2 s} x(s) d s

is linear and continuous

step 2

To check linearity, we need to verify if

A(c_1 x_1 + c_2 x_2) = c_1 A x_1 + c_2 A x_2

for any

x_1, x_2 \in L^2(0,1)

and scalars

c_1, c_2

step 3

Compute

A(c_1 x_1 + c_2 x_2)

:

A(c_1 x_1 + c_2 x_2)(t) = \int_{0}^{1} \sqrt{t+2 s} (c_1 x_1(s) + c_2 x_2(s)) d s

step 4

By linearity of the integral, we have:

A(c_1 x_1 + c_2 x_2)(t) = c_1 \int_{0}^{1} \sqrt{t+2 s} x_1(s) d s + c_2 \int_{0}^{1} \sqrt{t+2 s} x_2(s) d s

step 5

This simplifies to:

A(c_1 x_1 + c_2 x_2)(t) = c_1 A x_1(t) + c_2 A x_2(t)

step 6

Therefore,

A

is linear

step 7

To check continuity, we need to show that there exists a constant

C

such that

\|A x\|_{L^2(0,1)} \leq C \|x\|_{L^2(0,1)}

for all

x \in L^2(0,1)

step 8

Compute

\|A x\|_{L^2(0,1)}

:

\|A x\|_{L^2(0,1)}^2 = \int_{0}^{1} |A x(t)|^2 d t = \int_{0}^{1} \left| \int_{0}^{1} \sqrt{t+2 s} x(s) d s \right|^2 d t

step 9

By the Cauchy-Schwarz inequality, we have:

\left| \int_{0}^{1} \sqrt{t+2 s} x(s) d s \right|^2 \leq \left( \int_{0}^{1} (\sqrt{t+2 s})^2 d s \right) \left( \int_{0}^{1} |x(s)|^2 d s \right)

step 10

Simplifying, we get:

\left| \int_{0}^{1} \sqrt{t+2 s} x(s) d s \right|^2 \leq \left( \int_{0}^{1} (t+2 s) d s \right) \|x\|_{L^2(0,1)}^2

step 11

Evaluate the integral:

\int_{0}^{1} (t+2 s) d s = t + \int_{0}^{1} 2 s d s = t + s^2 \bigg|_{0}^{1} = t + 1

step 12

Thus, we have:

\left| \int_{0}^{1} \sqrt{t+2 s} x(s) d s \right|^2 \leq (t + 1) \|x\|_{L^2(0,1)}^2

step 13

Integrating with respect to

t

, we get:

\|A x\|_{L^2(0,1)}^2 \leq \int_{0}^{1} (t + 1) \|x\|_{L^2(0,1)}^2 d t = \|x\|_{L^2(0,1)}^2 \int_{0}^{1} (t + 1) d t

step 14

Evaluate the integral:

\int_{0}^{1} (t + 1) d t = \left( \frac{t^2}{2} + t \right) \bigg|_{0}^{1} = \frac{1}{2} + 1 = \frac{3}{2}

step 15

Therefore, we have:

\|A x\|_{L^2(0,1)}^2 \leq \frac{3}{2} \|x\|_{L^2(0,1)}^2

step 16

Taking the square root, we get:

\|A x\|_{L^2(0,1)} \leq \sqrt{\frac{3}{2}} \|x\|_{L^2(0,1)}

step 17

Hence,

A

is continuous with norm

\|A\| \leq \sqrt{\frac{3}{2}}

Answer

The operator

A

is linear and continuous with norm

\|A\| \leq \sqrt{\frac{3}{2}}

.

Key Concept

Linearity and Continuity of Operators

Explanation

To determine if an operator is linear, we check if it satisfies the properties of additivity and homogeneity. To check continuity, we need to show that the operator is bounded, meaning there exists a constant such that the norm of the operator applied to any function is less than or equal to the constant times the norm of the function.

Solution by Steps

step 1

We need to show that the operator

A x = \frac{x(-1) + x(1)}{3}

belongs to the space

\mathcal{L}(C(-1,1], \mathbb{R})

step 2

The space

\mathcal{L}(C(-1,1], \mathbb{R})

consists of all bounded linear operators from

C(-1,1]

to

\mathbb{R}

step 3

To show that

A

is linear, we need to verify that

A(cx_1 + dx_2) = cA(x_1) + dA(x_2)

for any

x_1, x_2 \in C(-1,1]

and scalars

c, d \in \mathbb{R}

step 4

Let

x_1, x_2 \in C(-1,1]

and

c, d \in \mathbb{R}

. Then,

A(cx_1 + dx_2) = \frac{(cx_1 + dx_2)(-1) + (cx_1 + dx_2)(1)}{3}

step 5

Simplifying, we get

A(cx_1 + dx_2) = \frac{c(x_1(-1) + x_1(1)) + d(x_2(-1) + x_2(1))}{3} = cA(x_1) + dA(x_2)

step 6

Therefore,

A

is linear

step 7

Next, we need to show that

A

is bounded. The norm of the operator

A

is given by

\|A\| = \sup_{\|x\| \leq 1} \|A(x)\|

step 8

For any

x \in C(-1,1]

with

\|x\| \leq 1

, we have

|A(x)| = \left|\frac{x(-1) + x(1)}{3}\right| \leq \frac{|x(-1)| + |x(1)|}{3} \leq \frac{1 + 1}{3} = \frac{2}{3}

step 9

Therefore,

\|A\| \leq \frac{2}{3}

step 10

Hence,

A

is a bounded linear operator, and

A \in \mathcal{L}(C(-1,1], \mathbb{R})

step 11

The norm of the operator

A

is

\|A\| = \frac{2}{3}

Answer

A \in \mathcal{L}(C(-1,1], \mathbb{R})

and

\|A\| = \frac{2}{3}

Key Concept

Bounded Linear Operator

Explanation

A bounded linear operator is a linear transformation between normed vector spaces that maps bounded sets to bounded sets. In this problem, we showed that the operator

A

is linear and bounded, thus belonging to the space

\mathcal{L}(C(-1,1], \mathbb{R})

.

Solution by Steps

step 1

Given that

H

is a real Hilbert space and

\{x_n\}, \{y_n\} \subset H

with

\|x_n\| = 1

and

\|y_n\| = 1

for all

n \in \mathbb{N}

, we need to prove that if

x_n \cdot y_n \rightarrow 0

, then

\|x_n + y_n\| \rightarrow \sqrt{2}

as

n \rightarrow \infty

step 2

Start by using the norm property in a Hilbert space:

\|x_n + y_n\|^2 = \|x_n\|^2 + \|y_n\|^2 + 2(x_n \cdot y_n)

step 3

Substitute the given norms

\|x_n\| = 1

and

\|y_n\| = 1

into the equation:

\|x_n + y_n\|^2 = 1 + 1 + 2(x_n \cdot y_n)

step 4

Simplify the equation:

\|x_n + y_n\|^2 = 2 + 2(x_n \cdot y_n)

step 5

Given that

x_n \cdot y_n \rightarrow 0

as

n \rightarrow \infty

, we have

2(x_n \cdot y_n) \rightarrow 0

step 6

Therefore,

\|x_n + y_n\|^2 \rightarrow 2

as

n \rightarrow \infty

step 7

Taking the square root of both sides, we get

\|x_n + y_n\| \rightarrow \sqrt{2}

as

n \rightarrow \infty

Answer

\|x_n + y_n\| \rightarrow \sqrt{2}

as

n \rightarrow \infty

Key Concept

Norm in Hilbert Space

Explanation

The key concept is that the norm of the sum of two vectors in a Hilbert space can be expressed in terms of the norms of the individual vectors and their dot product. Given that the dot product tends to zero, the norm of the sum tends to the square root of the sum of the squares of the individual norms.

Solution by Steps

step 1

We start by identifying the vectors

x

and

y

in the orthonormal system. Given:

x = \frac{5}{2} e_{3} - \frac{e_{6} + e_{7} + e_{8}}{3}

y = \frac{e_{6} - e_{7}}{3}

step 2

To find the angle between the vectors

x

and

y

, we use the dot product formula:

\cos \theta = \frac{x \cdot y}{\|x\| \|y\|}

First, compute the dot product

x \cdot y

:

x \cdot y = \left( \frac{5}{2} e_{3} - \frac{e_{6} + e_{7} + e_{8}}{3} \right) \cdot \left( \frac{e_{6} - e_{7}}{3} \right)

x \cdot y = \frac{5}{2} e_{3} \cdot \frac{e_{6} - e_{7}}{3} - \frac{e_{6} + e_{7} + e_{8}}{3} \cdot \frac{e_{6} - e_{7}}{3}

Since

e_i \cdot e_j = 0

for

i \neq j

and

e_i \cdot e_i = 1

:

x \cdot y = 0 - \frac{1}{3} \left( e_{6} \cdot e_{6} - e_{6} \cdot e_{7} + e_{7} \cdot e_{6} - e_{7} \cdot e_{7} + e_{8} \cdot e_{6} - e_{8} \cdot e_{7} \right)

x \cdot y = 0 - \frac{1}{3} \left( 1 - 0 + 0 - 1 + 0 - 0 \right)

x \cdot y = 0 - \frac{1}{3} \left( 1 - 1 \right) = 0

step 3

Next, compute the magnitudes

\|x\|

and

\|y\|

:

\|x\| = \sqrt{\left( \frac{5}{2} \right)^2 + \left( -\frac{1}{3} \right)^2 + \left( -\frac{1}{3} \right)^2 + \left( -\frac{1}{3} \right)^2}

\|x\| = \sqrt{\frac{25}{4} + \frac{1}{9} + \frac{1}{9} + \frac{1}{9}}

\|x\| = \sqrt{\frac{25}{4} + \frac{3}{9}} = \sqrt{\frac{25}{4} + \frac{1}{3}} = \sqrt{\frac{75}{12} + \frac{4}{12}} = \sqrt{\frac{79}{12}}

\|x\| = \sqrt{\frac{79}{12}} = \frac{\sqrt{79}}{\sqrt{12}} = \frac{\sqrt{79}}{2\sqrt{3}} = \frac{\sqrt{79}}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{237}}{6}

\|y\| = \sqrt{\left( \frac{1}{3} \right)^2 + \left( -\frac{1}{3} \right)^2}

\|y\| = \sqrt{\frac{1}{9} + \frac{1}{9}} = \sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{3}

step 4

Since

x \cdot y = 0

, the angle

\theta

between

x

and

y

is:

\cos \theta = \frac{0}{\left( \frac{\sqrt{237}}{6} \right) \left( \frac{\sqrt{2}}{3} \right)} = 0

\theta = \cos^{-1}(0) = \frac{\pi}{2}

step 5

To find the distance between

x

and

y

, we use the formula:

\|x - y\| = \sqrt{(x - y) \cdot (x - y)}

x - y = \left( \frac{5}{2} e_{3} - \frac{e_{6} + e_{7} + e_{8}}{3} \right) - \left( \frac{e_{6} - e_{7}}{3} \right)

x - y = \frac{5}{2} e_{3} - \frac{e_{6} + e_{7} + e_{8}}{3} - \frac{e_{6} - e_{7}}{3}

x - y = \frac{5}{2} e_{3} - \frac{e_{6} + e_{7} + e_{8} + e_{6} - e_{7}}{3}

x - y = \frac{5}{2} e_{3} - \frac{2e_{6} + e_{8}}{3}

\|x - y\| = \sqrt{\left( \frac{5}{2} \right)^2 + \left( -\frac{2}{3} \right)^2 + \left( -\frac{1}{3} \right)^2}

\|x - y\| = \sqrt{\frac{25}{4} + \frac{4}{9} + \frac{1}{9}}

\|x - y\| = \sqrt{\frac{25}{4} + \frac{5}{9}} = \sqrt{\frac{225}{36} + \frac{20}{36}} = \sqrt{\frac{245}{36}} = \frac{\sqrt{245}}{6}

Answer

The angle between the vectors

x

and

y

is

\frac{\pi}{2}

radians, and the distance between them is

\frac{\sqrt{245}}{6}

.

Key Concept

Angle and distance between vectors in an orthonormal system

Explanation

The angle between two vectors can be found using the dot product, and the distance can be found using the Euclidean norm.

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