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2.1.1E: Exercises for Vectors in the Plane - Mathematics


For exercises 1 - 10, consider points (P(−1,3), Q(1,5),) and (R(−3,7)). Determine the requested vectors and express each of them

(a.) in component form and

(b.) by using standard unit vectors.

1) ( vecd{PQ})

Answer:
a. (vecd{PQ}=⟨2,2⟩)
b. (vecd{PQ}=2hat{mathbf i}+2hat{mathbf j})

2) (vecd{PR})

3) (vecd{QP})

Answer:
a. (vecd{QP}=⟨−2,−2⟩)
b. (vecd{QP}=−2hat{mathbf i}−2hat{mathbf j})

4) (vecd{RP})

5) (vecd{PQ}+vecd{PR})

Answer:
a. (vecd{PQ}+vecd{PR}=⟨0,6⟩)
b. (vecd{PQ}+vecd{PR}=6hat{mathbf j})

6) (vecd{PQ}−vecd{PR})

7) (2vecd{PQ}−2vecd{PR})

Answer:
a. (2vecd{PQ}→−2vecd{PR}=⟨8,−4⟩)
b. (2vecd{PQ}−2vecd{PR}=8hat{mathbf i}−4hat{mathbf j})

8) (2vecd{PQ}+frac{1}{2}vecd{PR})

9) The unit vector in the direction of (vecd{PQ})

Answer:
a. (leftlanglefrac{sqrt{2}}{2},frac{sqrt{2}}{2} ight angle)
b. (frac{sqrt{2}}{2}hat{mathbf i}+frac{sqrt{2}}{2}hat{mathbf j})

10) The unit vector in the direction of (vecd{PR})

11) A vector ({overset{scriptstyle ightharpoonup}{mathbf v}}) has initial point ((−1,−3)) and terminal point ((2,1)). Find the unit vector in the direction of (vecs v). Express the answer in component form.

Answer:
(⟨frac{3}{5},frac{4}{5}⟩)

12) A vector (vecs v) has initial point ((−2,5)) and terminal point ((3,−1)). Find the unit vector in the direction of (vecs v). Express the answer in component form.

13) The vector (vecs v) has initial point (P(1,0)) and terminal point (Q) that is on the (y)-axis and above the initial point. Find the coordinates of terminal point (Q) such that the magnitude of the vector (vecs v) is (sqrt{5}).

Answer:
(Q(0,2))

14) The vector (vecs v) has initial point (P(1,1)) and terminal point (Q) that is on the (x)-axis and left of the initial point. Find the coordinates of terminal point (Q) such that the magnitude of the vector (vecs v) is (sqrt{10}).

For exercises 15 and 16, use the given vectors (vecs a) and (vecs b).

a. Determine the vector sum (vecs a+vecs b) and express it in both the component form and by using the standard unit vectors.

b. Find the vector difference (vecs a −vecs b) and express it in both the component form and by using the standard unit vectors.

c. Verify that the vectors (vecs a, , vecs b,) and (vecs a+vecs b), and, respectively, (vecs a, , vecs b), and (vecs a−vecs b) satisfy the triangle inequality.

d. Determine the vectors (2vecs a, −vecs b,) and (2vecs a−vecs b.) Express the vectors in both the component form and by using standard unit vectors.

15) (vecs a=2hat{mathbf i}+hat{mathbf j}, vecs b=hat{mathbf i}+3hat{mathbf j})

Answer:
(a., vecs a+vecs b=⟨3,4⟩, quad vecs a+vecs b=3hat{mathbf i}+4hat{mathbf j})
(b., vecs a−vecs b=⟨1,−2⟩, quad vecs a−vecs b=hat{mathbf i}−2hat{mathbf j})
(c.) Answers will vary
(d., 2vecs a=⟨4,2⟩, quad 2vecs a=4hat{mathbf i}+2hat{mathbf j}, quad −vecs b=⟨−1,−3⟩, quad −vecs b=−hat{mathbf i}−3hat{mathbf j}, quad 2vecs a−vecs b=⟨3,−1⟩, quad 2vecs a−vecs b=3hat{mathbf i}−hat{mathbf j})

16) (vecs a=2hat{mathbf i}, vecs b=−2hat{mathbf i}+2hat{mathbf j})

17) Let (vecs a) be a standard-position vector with terminal point ((−2,−4)). Let (vecs b) be a vector with initial point ((1,2)) and terminal point ((−1,4)). Find the magnitude of vector (−3vecs a+vecs b−4hat{mathbf i}+hat{mathbf j}.)

Answer:
(15)

18) Let (vecs a) be a standard-position vector with terminal point at ((2,5)). Let (vecs b) be a vector with initial point ((−1,3)) and terminal point ((1,0)). Find the magnitude of vector (vecs a−3vecs b+14hat{mathbf i}−14hat{mathbf j}.)

19) Let (vecs u) and (vecs v) be two nonzero vectors that are nonequivalent. Consider the vectors (vecs a=4vecs u+5vecs v) and (vecs b=vecs u+2vecs v) defined in terms of (vecs u) and (vecs v). Find the scalar (λ) such that vectors (vecs a+λvecs b) and (vecs u−vecs v) are equivalent.

Answer:
(λ=−3)

20) Let (vecs u) and (vecs v) be two nonzero vectors that are nonequivalent. Consider the vectors (vecs a=2vecs u−4vecs v) and (vecs b=3vecs u−7vecs v) defined in terms of (vecs u) and (vecs v). Find the scalars (α) and (β) such that vectors (αvecs a+βvecs b) and (vecs u−vecs v) are equivalent.

21) Consider the vector (vecs a(t)=⟨cos t, sin t⟩) with components that depend on a real number (t). As the number (t) varies, the components of (vecs a(t)) change as well, depending on the functions that define them.

a. Write the vectors (vecs a(0)) and (vecs a(π)) in component form.

b. Show that the magnitude (∥vecs a(t)∥) of vector (vecs a(t)) remains constant for any real number (t).

c. As (t) varies, show that the terminal point of vector (vecs a(t)) describes a circle centered at the origin of radius (1).

Answer:
(a., vecs a(0)=⟨1,0⟩, quad vecs a(π)=⟨−1,0⟩)
(b.) Answers may vary
(c.) Answers may vary

22) Consider vector (vecs a(x)=⟨x,sqrt{1−x^2}⟩) with components that depend on a real number (x∈[−1,1]). As the number (x) varies, the components of (vecs a(x)) change as well, depending on the functions that define them.

a. Write the vectors (vecs a(0)) and (vecs a(1)) in component form.

b. Show that the magnitude (∥vecs a(x)∥) of vector (vecs a(x)) remains constant for any real number (x).

c. As (x) varies, show that the terminal point of vector (vecs a(x)) describes a circle centered at the origin of radius (1).

23) Show that vectors (vecs a(t)=⟨cos t, sin t⟩) and (vecs a(x)=⟨x,sqrt{1−x^2}⟩) are equivalent for (x=1) and (t=2kπ), where (k) is an integer.

Answer: Answers may vary

24) Show that vectors (vecs a(t)=⟨cos t, sin t⟩) and (vecs a(x)=⟨x,sqrt{1−x^2}⟩) are opposite for (x=1) and (t=π+2kπ), where (k) is an integer.

For exercises 25-28, find a vector (vecs v) with the given magnitude and in the same direction as the vector (vecs u).

25) (|vecs v|=7, quad vecs u=⟨3,4⟩)

Answer:
(vecs v=⟨frac{21}{5},frac{28}{5}⟩)

26) (‖vecs v‖=3,quad vecs u=⟨−2,5⟩)

27) (‖vecs v‖=7,quad vecs u=⟨3,−5⟩)

Answer:
(vecs v=⟨frac{21sqrt{34}}{34},−frac{35sqrt{34}}{34}⟩)

28) (‖vecs v‖=10,quad vecs u=⟨2,−1⟩)

For exercises 29-34, find the component form of vector (vecs u), given its magnitude and the angle the vector makes with the positive (x)-axis. Give exact answers when possible.

29) (‖vecs u‖=2, θ=30°)

Answer:
(vecs u=⟨sqrt{3},1⟩)

30) (‖vecs u‖=6, θ=60°)

31) (‖vecs u‖=5, θ=frac{π}{2})

Answer:
(vecs u=⟨0,5⟩)

32) (‖vecs u‖=8, θ=π)

33) (‖vecs u‖=10, θ=frac{5π}{6})

Answer:
(vecs u=⟨−5sqrt{3},5⟩)

34) (‖vecs u‖=50, θ=frac{3π}{4})

For exercises 35 and 36, vector (vecs u) is given. Find the angle (θ∈[0,2π)) that vector (vecs u) makes with the positive direction of the (x)-axis, in a counter-clockwise direction.

35) (vecs u=5sqrt{2}hat{mathbf i}−5sqrt{2}hat{mathbf j})

Answer:
(θ=frac{7π}{4})

36) (vecs u=−sqrt{3}hat{mathbf i}−hat{mathbf j})

37) Let (vecs a=⟨a_1,a_2⟩, vecs b=⟨b_1,b_2⟩), and (vecs c =⟨c_1,c_2⟩) be three nonzero vectors. If (a_1b_2−a_2b_1≠0), then show there are two scalars, (α) and (β), such that (vecs c=αvecs a+βvecs b.)

Answer: Answers may vary

38) Consider vectors (vecs a=⟨2,−4⟩, vecs b=⟨−1,2⟩,) and (vecs c =vecs 0) Determine the scalars (α) and (β) such that (vecs c=αvecs a+βvecs b).

39) Let (P(x_0,f(x_0))) be a fixed point on the graph of the differentiable function (f) with a domain that is the set of real numbers.

a. Determine the real number (z_0) such that point (Q(x_0+1,z_0)) is situated on the line tangent to the graph of (f) at point (P).

b. Determine the unit vector (vecs u) with initial point (P) and terminal point (Q).

Answer:
(a. quad z_0=f(x_0)+f′(x_0); quad b. quad vecs u=frac{1}{sqrt{1+[f′(x_0)]^2}}⟨1,f′(x_0)⟩)

40) Consider the function (f(x)=x^4,) where (x∈R).

a. Determine the real number (z_0) such that point (Q(2,z_0)) s situated on the line tangent to the graph of (f) at point (P(1,1)).

b. Determine the unit vector (vecs u) with initial point (P) and terminal point (Q).

41) Consider (f) and (g) two functions defined on the same set of real numbers (D). Let (vecs a=⟨x,f(x)⟩) and (vecs b=⟨x,g(x)⟩) be two vectors that describe the graphs of the functions, where (x∈D). Show that if the graphs of the functions (f) and (g) do not intersect, then the vectors (vecs a) and (vecs b) are not equivalent.

42) Find (x∈R) such that vectors (vecs a=⟨x, sin x⟩) and (vecs b=⟨x, cos x⟩) are equivalent.

43) Calculate the coordinates of point (D) such that (ABCD) is a parallelogram, with (A(1,1), B(2,4)), and (C(7,4)).

Answer:
(D(6,1))

44) Consider the points (A(2,1), B(10,6), C(13,4)), and (D(16,−2)). Determine the component form of vector (vecd{AD}).

45) The speed of an object is the magnitude of its related velocity vector. A football thrown by a quarterback has an initial speed of (70) mph and an angle of elevation of (30°). Determine the velocity vector in mph and express it in component form. (Round to two decimal places.)

Answer:
(⟨60.62,35⟩)

46) A baseball player throws a baseball at an angle of (30°) with the horizontal. If the initial speed of the ball is (100) mph, find the horizontal and vertical components of the initial velocity vector of the baseball. (Round to two decimal places.)

47) A bullet is fired with an initial velocity of (1500) ft/sec at an angle of (60°) with the horizontal. Find the horizontal and vertical components of the velocity vector of the bullet. (Round to two decimal places.)

Answer:
The horizontal and vertical components are (750) ft/sec and (1299.04) ft/sec, respectively.

48) [T] A 65-kg sprinter exerts a force of (798) N at a (19°) angle with respect to the ground on the starting block at the instant a race begins. Find the horizontal component of the force. (Round to two decimal places.)

49) [T] Two forces, a horizontal force of (45) lb and another of (52) lb, act on the same object. The angle between these forces is (25°). Find the magnitude and direction angle from the positive (x)-axis of the resultant force that acts on the object. (Round to two decimal places.)

Answer:
The magnitude of resultant force is (94.71) lb; the direction angle is (13.42°).

50) [T] Two forces, a vertical force of (26) lb and another of (45) lb, act on the same object. The angle between these forces is (55°). (Round to two decimal places.)

51) [T] Three forces act on object. Two of the forces have the magnitudes (58) N and (27) N, and make angles (53°) and (152°), respectively, with the positive (x)-axis. Find the magnitude and the direction angle from the positive (x)-axis of the third force such that the resultant force acting on the object is zero. (Round to two decimal places.)

Answer:
The magnitude of the third vector is (60.03) N; the direction angle is (259.38°).

52) Three forces with magnitudes 80 lb, 120 lb, and 60 lb act on an object at angles of (45°, 60°) and (30°), respectively, with the positive (x)-axis. Find the magnitude and direction angle from the positive (x)-axis of the resultant force. (Round to two decimal places.)

53) [T] An airplane is flying in the direction of (43°) east of north (also abbreviated as (N43E) at a speed of (550) mph. A wind with speed (25) mph comes from the southwest at a bearing of (N15E). What are the ground speed and new direction of the airplane?

Answer:
The new ground speed of the airplane is (572.19) mph; the new direction is (N41.82E.)

54) [T] A boat is traveling in the water at (30) mph in a direction of (N20E) (that is, (20°) east of north). A strong current is moving at (15) mph in a direction of (N45E). What are the new speed and direction of the boat?

55) [T] A 50-lb weight is hung by a cable so that the two portions of the cable make angles of (40°) and (53°), respectively, with the horizontal. Find the magnitudes of the forces of tension (vecs T_1) and (vecs T_2) in the cables if the resultant force acting on the object is zero. (Round to two decimal places.)

Answer:
(|vecs T_1|=30.13 , lb, quad |vecs T_2|=38.35 , lb)

56) [T] A 62-lb weight hangs from a rope that makes the angles of (29°) and (61°), respectively, with the horizontal. (Round to two decimal places.)

57) [T] A 1500-lb boat is parked on a ramp that makes an angle of (30°) with the horizontal. The boat’s weight vector points downward and is a sum of two vectors: a horizontal vector (vecs v_1) that is parallel to the ramp and a vertical vector (vecs v_2) that is perpendicular to the inclined surface. The magnitudes of vectors (vecs v_1) and (vecs v_2) are the horizontal and vertical component, respectively, of the boat’s weight vector. Find the magnitudes of (vecs v_1) and (vecs v_2). (Round to the nearest integer.)

Answer:
(|vecs v_1|=750 , lb, quad |vecs v_2|=1299 , lb)

58) [T] An 85-lb box is at rest on a (26°) incline. Determine the magnitude of the force parallel to the incline necessary to keep the box from sliding. (Round to the nearest integer.)

59) A guy-wire supports a pole that is (75) ft high. One end of the wire is attached to the top of the pole and the other end is anchored to the ground (50) ft from the base of the pole. Determine the horizontal and vertical components of the force of tension in the wire if its magnitude is (50) lb. (Round to the nearest integer.)

Answer:
The two horizontal and vertical components of the force of tension are (28) lb and (42) lb, respectively.

60) A telephone pole guy-wire has an angle of elevation of (35°) with respect to the ground. The force of tension in the guy-wire is (120) lb. Find the horizontal and vertical components of the force of tension. (Round to the nearest integer.)

Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.


2.1.1E: Exercises for Vectors in the Plane - Mathematics

This is a fairly short chapter. We will be taking a brief look at vectors and some of their properties. We will need some of this material in the next chapter and those of you heading on towards Calculus III will use a fair amount of this there as well.

Here is a list of topics in this chapter.

Basic Concepts – In this section we will introduce some common notation for vectors as well as some of the basic concepts about vectors such as the magnitude of a vector and unit vectors. We also illustrate how to find a vector from its starting and end points.

Vector Arithmetic – In this section we will discuss the mathematical and geometric interpretation of the sum and difference of two vectors. We also define and give a geometric interpretation for scalar multiplication. We also give some of the basic properties of vector arithmetic and introduce the common (i), (j), (k) notation for vectors.

Dot Product – In this section we will define the dot product of two vectors. We give some of the basic properties of dot products and define orthogonal vectors and show how to use the dot product to determine if two vectors are orthogonal. We also discuss finding vector projections and direction cosines in this section.

Cross Product – In this section we define the cross product of two vectors and give some of the basic facts and properties of cross products.


Multivariate Statistics

It will be useful to talk about vector spaces. These are sets of vectors that can be added together, or multiplied by a scalar. You should be familiar with these from your undergraduate degree. We don’t provide a formal definition here, but you can think of a real vector space (V) as a set of vectors such that for any (mathbf v_1, mathbf v_2 in V) and (alpha_1, alpha_2 in mathbb) , we have [alpha_1 mathbf v_1 + alpha_2 mathbf v_2 in V] i.e., vector spaces are closed under addition and scalar multiplication.


A subset (U subset V) of a vector space (V) is called a vector subspace if (U) is also a vector space.

2.2.1 Linear independence

2.2.2 Row and column spaces

We can think about the matrix-vector multiplication (mathbf Amathbf x) in two ways. The usual way is as the inner product between the rows of (A) and (x) .

But a better way to think of (mathbf Amathbf x) is as a linear combination of the columns of (A) .

For [A=left( egin 1 & 2 3&45&6end ight) ] we can see that the column space is a 2-dimensional plane in (mathbb^3) . The matrix (mathbf B) has the same column space as (mathbf A) [mathbf B=left( egin 1 & 2&3 &4 3&4 &7&105&6&11&16end ight) ]

The number of linearly independent columns of (mathbf A) is called the column rank of (mathbf A) , and is equal to the dimension of the column space of (mathcal(mathbf A)) . The column rank of (mathbf A) and (mathbf B) is 2.

The row space of (mathbf A) is defined to be the column space of (mathbf A^ op) , and the row rank is the number of linearly independent rows of (mathbf A) .

Thus we can simply refer to the rank of the matrix.

2.2.3 Linear transformations

We can view an (n imes p) matrix (mathbf A) as a linear map between two vector spaces: [egin mathbf A: mathbb^p & ightarrow mathbb^n mathbf x&mapsto mathbf Amathbf x end]

The image of (mathbf A) is precisely the column space of (mathbf A) : [operatorname(mathbf A) = ^p>=mathcal(mathbf A) subset mathbb^n]

The kernel of (A) is the set of vectors mapped to zero: [operatorname(mathbf A)=subset mathbb^p] and is sometimes called the null-space of (mathbf A) and denoted (mathcal(mathbf A)) .

If we’re thinking about matrices, then (dim mathcal(mathbf A)+dim mathcal(mathbf A)=p) , or equivalently that
(operatorname(mathbf A)+dim mathcal(mathbf A)=p) .

We’ve already said that the row space of (mathbf A) is (mathcal(mathbf A^ op)) . The left-null space is (^n: mathbf x^ op mathbf A=0>) or equivalently (^n: mathbf A^ op mathbf x=0>=mathcal(mathbf A^ op)) . And so by the rank-nullity theorem we must have [n=dim mathcal(mathbf A^ op) + dim mathcal(mathbf A^ op)= operatorname(mathbf A)+dim operatorname(mathbf A^ op).]

Example 2.10 Consider again the matrix (D: mathbb^3 ightarrow mathbb^2) [ D=left( egin 1 & 2&3 2&4&6 end ight)= left( egin 1 2 end ight)left(egin1&2&3end ight) ] We have already seen that [mathcal(D)=operatornameleft12end ight) ight>] and so (dim mathcal(D)=operatorname(D)=1) . The kernel, or null-space, of (mathbf D) is the set of vectors for which (mathbf Dmathbf x=oldsymbol 0) , i.e., [x_1+2x_2+3x_3=0] This is a single equation with three unknowns, and so there must be a plane of solutions. We need two linearly independent vectors in this plane to describe it. Convince yourself that [mathcal(D) = operatornameleft03-2end ight), left(egin2-1end ight) ight>] So we have [dim mathcal(D)+dim mathcal(D)=1+2=3] as required by the rank-nullity theorem.

If we consider (D^ op) , we already know (dim mathcal(D^ op)=1) (as row-rank=column rank), and the rank-nullity theorem tells us that the dimension of the null space of (D^ op) must be (2-1=1) . This is easy to confirm as (D^ op x=0) implies [x_1+2x_2=0] which is a line in (mathbb^2) [mathcal(D^ op) = operatornameleft< left(egin-21end ight) ight>]

Question: When does a square matrix (mathbf A) have an inverse?

  • Precisely when the kernel of (mathbf A) contains only the zero vector, i.e., has dimension 0. In this case the column space of (mathbf A) is the original space, and (mathbf A) is surjective and so must have an inverse. A simpler way to determine if (mathbf A) has an inverse is to consider its determinant.

Question: Suppose we are given a (n imes p) matrix (mathbf A) , and a n-vector (mathbf y) . When does [mathbf Amathbf x= mathbf y] have a solution?


2.1.1E: Exercises for Vectors in the Plane - Mathematics

Matrices and determinants appear in two other important contexts one is in solving simultaneous linear equations in several variables. The other is in representing linear transformations of vectors. The first of these is discussed in detail in Chapter 32 .

In the latter context a matrix represents the transformation that takes the column basis vectors into the vectors that are the corresponding columns of the matrix.

Sums of original basis vectors are transformed into the same sums of the corresponding columns. This fact defines the transformation on all vectors.

When the determinant of a matrix is zero, the volume of the region with sides given by its columns or rows is zero, which means the matrix considered as a transformation takes the basis vectors into vectors that are linearly dependent and define 0 volume.

This happens, the determinant is zero, when the columns (and rows) of the matrix are linearly dependent.

4.2 What is the matrix of the transformation which takes a unit vector in the direction of the x axis into one in the direction of the y axis, and similarly one along the y axis into one along the z axis, and one along the z axis into one along the x axis?

4.3 What matrix describes the transformation which doubles the component of vectors in the x direction, halves components in the y direction, and leaves components in the z direction alone.

4.4 What matrix describes the transformation in three dimensions which projects vectors into the ( x , y ) plane? onto the x axis? onto the diagonal in the ( x , y ) plane?


From the coplanar section above,
c=λa+μb

When position vectors are used,

r=(1-λ-u)a+ λb+μc is the vector equation of the plane.

Since λ and b are variable, there will be many possible equations for the plane.

Find a vector equation of the plane through the points
A (-1,-2,-3) , B(-2,0,1) and C (-4,-1,-1)

When A is a known point on the plane,
R is any old point on the plane and b and c are vectors
parallel to the plane,

the vector equation of the plane is
r=a+λb+μc


A vector has a magnitude and direction.

Vectors can be used to represent many physical quantities that have a magnitude and direction, like forces.

Vectors may be represented as arrows where the length of the arrow indicates the magnitude and the arrowhead indicates the direction of the vector.

Vectors in two dimensions can be drawn on the Cartesian plane.

Vectors can be added graphically using the head-to-tail method or the tail-to-tail method.

A closed vector diagram is a set of vectors drawn on the Cartesian using the tail-to-head method and that has a resultant with a magnitude of zero.

Vectors can be added algebraically using Pythagoras' theorem or using components.

The direction of a vector can be found using simple trigonometric calculations.

The components of a vector are a series of vectors that, when combined, give the original vector as their resultant.

Components are usually created that align with the Cartesian coordinate axes. For a vector (vec) that makes an angle of ( heta) with the positive (x)-axis the (x)-component is (vec_x=Rcos( heta)) and the (y)-component is (vec_y=Rsin( heta)).


Vectors – IGCSE Year 11 revision questions


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Linear Algebra and Its Applications, Exercise 3.4.24

Exercise 3.4.24. As discussed on page 178, the first three Legendre polynomials are 1, , and . Find the next Legendre polynomial it will be a cubic polynomial defined for and will be orthogonal to the first three Legendre polynomials.

Answer: The process of finding the fourth Legendre poloynomial is essentially an application of Gram-Schmidt orthogonalization. The first three polynomials are

We can find the fourth Legendre polynomial by starting with and subtracting off the projections of on the first three polynomials:

For the first term we have

so that the first term does not appear in the expression for .

The third term drops out for the same reason: its numerator is

That leaves the second term with numerator of

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fifth Edition and the accompanying free online course, and Dr Strang’s other books.


Exercise 3.4.24. As discussed on page 178, the first three Legendre polynomials are 1, , and . Find the next Legendre polynomial it will be a cubic polynomial defined for and will be orthogonal to the first three Legendre polynomials. &hellip Continue reading &rarr

Exercise 3.4.23. Given the step function with for and for , find the following Fourier coefficients: Answer: For the numerator is and the denominator is so that . For the numerator is so that . For the numerator is and &hellip Continue reading &rarr


Subsection 2.2.2 Spans

It will be important to know what are all linear combinations of a set of vectors

In other words, we would like to understand the set of all vectors

such that the vector equation (in the unknowns

has a solution (i.e. is consistent).

Definition

is the collection of all linear combinations of

is the subset spanned by or generated by the vectors

The above definition is the first of several essential definitions that we will see in this textbook. They are essential in that they form the essence of the subject of linear algebra: learning linear algebra means (in part) learning these definitions. All of the definitions are important, but it is essential that you learn and understand the definitions marked as such.

Set Builder Notation

reads as: “the set of all things of the form

” The vertical line is “such that” everything to the left of it is “the set of all things of this form”, and everything to the right is the condition that those things must satisfy to be in the set. Specifying a set in this way is called set builder notation.

All mathematical notation is only shorthand: any sequence of symbols must translate into a usual sentence.