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Asignatura: Matemàtiques, Profesor: r r, Carrera: Enginyeria de Tecnologies i Serveis de Telecomunicació, Universidad: UPV
Tipo: Ejercicios
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Flatland: http://www.eldritchpress.org/eaa/FL.HTM
Definition I.3. An n × n matrix (that is, with the same number of rows and columns, n) will be called a square matrix of order n.
Example I.4.
is a 2 × 3 matrix.
is a square matrix of order 2.
is row vector.
(^) is a column vector or vector).
(^) is the 3 × 2 zero matrix.
The set of all vectors of n components will be denoted by Rn.
Definition I.5. The vector sum of ~u and ~v is the vector of the sums.
~u + ~v =
u 1 .. . un
v 1 .. . vn
u 1 + v 1 .. . un + vn
Note that the vectors must have the same number of entries for the addition to be defined. This entry-by-entry addition works for any pair of matrices, not just vectors, provided that they have the same number of rows and columns:
Example I.6. [ 1 − 1 0 0 2 − 1
Definition I.7. The scalar multiplication of the real number r and the vector ~v is the vector of the multiples.
r · ~v = r ·
v 1 .. . vn
rv 1 .. . rvn
We write scalar multiplication in either order, as r · ~v or ~v · r, or without the ‘·’ symbol: r~v. (Do not refer to scalar multiplication as ‘scalar product’ because we use that name for a different operation.)
Example I.8.
Notice that the definitions of vector addition and scalar multiplication agree where they overlap, for instance, ~v + ~v = 2~v. As with the addition operation, the entry-by-entry scalar multiplication operation extends beyond just vectors to any matrix:
Example I.9.
Definition I.10. A vector obtained from other vectors combining the two operations (sum of vectors and scalar multiplication) is a linear combination.
Example I.11. For example: 2
is a linear combination of the vectors [ 1 3
and
with coefficients 2, 1 and −.
Example I.12.
~u
~v
2 ~u + 3~v − 2 ~u + 3~v
−~u − ~v
4 ~u
−~v
2 ~u − 12 ~v
Definition I.13. Given some vectors ~v 1 ,... , ~vm ∈ Rn, we will call the span of ~v 1 ,... , ~vm to the set of all linear combinations of those vectors. We will denote it by span(~v 1 ,... , ~vm).
“Higher-dimensional geometry” sounds exotic. It is exotic — interesting and eye-opening. But it isn’t distant or unreachable. We begin by defining one-dimensional space to be R^1. To see that the definition is reasonable, we picture a one-dimensional space
and make a correspondence with R by picking a point to label 0 and another to label 1.
0 1 Now, with a scale and a direction, finding the point corresponding to, say, +2.17, is easy — start at 0 and head in the direction of 1, but don’t stop there, go 2.17 times as far. The basic idea here, combining magnitude with direction, is the key to extending to higher dimensions. An object comprised of a magnitude and a direction is a vector (we use the same word as in the prior section because we shall show below how to describe such an object with a column vector). We can draw a vector as having some length, and pointing somewhere.
There is a subtlety here — these vectors
are equal, even though they start in different places, because they have equal lengths and equal directions. Again: those vectors are not just alike, they are equal. How can things that are in different places be equal? Think of a vector as representing a displacement (the word vector is Latin for “carrier” or “traveler”). These two squares undergo equal displacements, despite that those displacements start in different places.
Sometimes, to emphasize this property vectors have of not being anchored, we can refer to them as free vectors. Thus, these free vectors are equal as each is a displacement of one over and two up.
More generally, vectors in the plane are the same if and only if they have the same change in first components and the same change in second components: the vector extending from the point (a 1 , a 2 ) to the point (b 1 , b 2 ) equals the vector from (c 1 , c 2 ) to (d 1 , d 2 ) if and only if b 1 − a 1 = d 1 − c 1 and b 2 − a 2 = d 2 − c 2. Saying ‘the vector that, were it to start at (a 1 , a 2 ), would extend to (b 1 , b 2 )’ would be unwieldy. We instead describe that vector as [ b 1 − a 1 b 2 − a 2
so that the ‘one over and two up’ arrows shown above picture this vector. [ 1 2
We often draw the arrow as starting at the origin, and we then say it is in the canonical position (or natural position or standard position). When the vector [ v 1 v 2
is in canonical position then it extends to the endpoint (v 1 , v 2 ). We typically just refer to “the point [ 1 2
rather than “the endpoint of the canonical position of ” that vector. Thus, we will call each of these R^2. R^2 = {
x 1 x 2
∣ (^) x 1 , x 2 ∈ R}
In the prior section we defined vectors and vector operations with an algebraic motivation;
r ·
v 1 v 2
rv 1 rv 2
v 1 v 2
w 1 w 2
v 1 + w 1 v 2 + w 2
we can now understand those operations geometrically. For instance, if ~v represents a displacement then 3~v represents a displacement in the same direction but three times as far, and − 1 ~v represents a displacement of the same distance as ~v but in the opposite direction. ~v
−~v
3 ~v
And, where ~v and w~ represent displacements, ~v + w~ represents those displacements combined.
~v
w~
~v + w~
The long arrow is the combined displacement in this sense: if, in one minute, a ship’s motion gives it the displacement relative to the earth of ~v and a passenger’s motion gives a displacement relative to the ship’s deck of w~, then ~v + w~ is the displacement of the passenger relative to the earth. Another way to understand the vector sum is with the parallelogram rule. Draw the parallelogram formed by the vectors ~v and w~. Then the sum ~v + w~ extends along the diagonal to the far corner. ~v + w~
~v
w~
In R^3 , a line uses one parameter so that a particle on that line is free to move back and forth in one dimension, and a plane involves two parameters. For example, the plane through the points (1, 0 , 5), (2, 1 , −3), and (− 2 , 4 , 0 .5) consists of (endpoints of) the vectors in
(^) + t
(^) + s
∣ (^) t, s ∈ R} =
(^) + span(
(the column vectors associated with the parameters
are two vectors whose whole bodies lie in the plane). As with the line, note that we describe some points in this plane with negative t’s or negative s’s or both. Generalizing,
Definition I.18. A set of the form ~p + span(~v 1 ,... , ~vk), where ~p, ~v 1 ,... , ~vk ∈ Rn, is an affine subspace.
For example, in R^4
π 3 − 0. 5
+^ t
t ∈ R} =
π 3 − 0. 5
+ span(
is a line,
+^ t
+^ s
t, s ∈ R} =
+ span(
is a plane, and
+^ r
+^ s
+^ t
∣ (^) r, s, t ∈ R} =
+ span(
is a three-dimensional affine subspace (in Lesson 4 we will see the concrete notion of dimension). Again, the intuition is that a line permits motion in one direction, a plane permits motion in combinations of two directions, etc.
Definition I.19. Any affine subspace of Rn^ that contains ~ 0 is called a vector subspace of Rn.
For example:
Notice that any vector subspace of Rn^ has the form span(~v 1 ,... , ~vk), for ~v 1 ,... , ~vk ∈ Rn. That is, the vector subspaces of Rn^ are the spans of vectors of Rn. In Lesson 4 we will generalize the notion of vector subspace and we will go deeper into this concept.
UTILITARIAN SUMMARY I.20.
The notions of length (or norm) of a vector, angle between vectors and orthogonality have a clear geometrical meaning in the plane and in the three-dimensional space. The objective of this section is to extend these notions to Rn.
Definition I.21. The length (or norm) of a vector ~v ∈ Rn^ is the square root of the sum of the squares of its components. ‖~v ‖ =
v^21 + · · · + v^2 n
Remark I.22. This is a natural generalization of the Pythagorean Theorem.
Note that for any nonzero ~v, the vector ~v/‖~v‖ has length one. We say that the second vector normalizes ~v to length one. We can use that to get a formula for the angle between two vectors. Consider two vectors in R^3 where neither is a multiple of the other
~v ~u
Still reasoning with letters but guided by the pictures, we use the next theorem to argue that the triangle formed by ~u, ~v, and ~u − ~v in Rn^ lies in the planar subset of Rn^ generated by ~u and ~v.
Theorem I.26 (Triangle Inequality). For any ~u, ~v ∈ Rn,
‖~u + ~v ‖ ≤ ‖~u ‖ + ‖~v ‖
with equality if and only if one of the vectors is a nonnegative scalar multiple of the other one.
This is the source of the familiar saying, “The shortest distance between two points is in a straight line.”
~u
~u + ~v ~v
start
finish
Since all the numbers are positive, the inequality holds if and only if its square holds.
‖~u + ~v ‖^2 ≤ ( ‖~u ‖ + ‖~v ‖ )^2 ( ~u + ~v ) •^ ( ~u + ~v ) ≤ ‖~u ‖^2 + 2 ‖~u ‖ ‖~v ‖ + ‖~v ‖^2 ~u •^ ~u + ~u •^ ~v + ~v •^ ~u + ~v •^ ~v ≤ ~u •^ ~u + 2 ‖~u ‖ ‖~v ‖ + ~v •^ ~v 2 ~u •^ ~v ≤ 2 ‖~u ‖ ‖~v ‖
This holds if and only if the relationship obtained by multiplying both sides by the nonnegative numbers ‖~u ‖ and ‖~v ‖ 2 ( ‖~v ‖ ~u ) •^ ( ‖~u ‖ ~v ) ≤ 2 ‖~u ‖^2 ‖~v ‖^2 holds. And this inequality can be rewritten as
0 ≤ ‖~u ‖^2 ‖~v ‖^2 − 2 ( ‖~v ‖ ~u ) •^ ( ‖~u ‖ ~v ) + ‖~u ‖^2 ‖~v ‖^2.
But factoring shows that it is true
0 ≤ ( ‖~u ‖ ~v − ‖~v ‖ ~u ) •^ ( ‖~u ‖ ~v − ‖~v ‖ ~u )
since it only says that the square of the length of the vector ‖~u ‖ ~v − ‖~v ‖ ~u is not negative. Finally, equality holds when, and only when, ‖~u ‖ ~v − ‖~v ‖ ~u is ~ 0. The check that ‖~u ‖ ~v = ‖~v ‖ ~u if and only if one vector is a nonnegative real scalar multiple of the other is easy. QED
This result supports the intuition that even in higher-dimensional spaces, lines are straight and planes are flat. We can easily check from the definition that linear surfaces have the property that for any two points in that surface, the line segment between them is contained in that surface. But if the linear surface were not flat then that would allow for a shortcut.
P Q
Because the Triangle Inequality says that in any Rn^ the shortest cut between two endpoints is simply the line segment connecting them, linear surfaces have no bends. Back to the definition of angle measure. The heart of the Triangle Inequality’s proof is the ~u •^ ~v ≤ ‖~u ‖ ‖~v ‖ line. We might wonder if some pairs of vectors satisfy the inequality in this way: while ~u •^ ~v is a large number, with absolute value bigger than the right-hand side, it is a negative large number. The next result says that does not happen.
Corollary I.27 (Cauchy-Schwartz Inequality). For any ~u, ~v ∈ Rn,
| ~u •^ ~v | ≤ ‖ ~u ‖ ‖~v ‖
with equality if and only if one vector is a scalar multiple of the other.
PROOF:
The Triangle Inequality’s proof shows that ~u •^ ~v ≤ ‖~u ‖ ‖~v ‖. So, if ~u •^ ~v is positive or zero then we are done. If ~u •^ ~v is negative then this holds:
| ~u •^ ~v | = −( ~u •^ ~v ) = (−~u ) •^ ~v ≤ ‖ − ~u ‖ ‖~v ‖ = ‖~u ‖ ‖~v ‖.
We leave the equality condition as an exercise. QED
The Cauchy-Schwartz inequality assures us that the next definition makes sense because the fraction has absolute value less than or equal to one.
Definition I.28. The angle between two nonzero vectors ~u, ~v ∈ Rn^ is
θ = arccos( ~u •^ ~v ‖~u ‖ ‖~v ‖
(by definition, the angle between the zero vector and any other vector is right).
Definition I.29. Two vectors from Rn^ are orthogonal, if their dot product is zero.
Example I.30. These vectors are orthogonal. [ 1 − 1
We’ve drawn the arrows away from canonical position but nevertheless the vectors are orthogonal.
Example I.31. The angle between these angles of R^3
0 3 2
^11 0
column of B. That is:
ci,j = ai, 1 b 1 ,j + ai, 2 b 2 ,j + · · · + ai,nbn,j =
∑^ n
s=
ai,sbs,j
Example I.35. [ 1 2 3 0 1 − 1
The purpose of the following figure is to emphasize that two matrices can be multiplied only if their dimensions satisfy a certain relation:
m × n n × p =^ m × p
Observe now the following example:
We have multiplied a matrix by a column vector and the result is a linear combination of the columns of the matrix. It is evident that this is true in general, that is,
Proposition I.36. The product of a matrix by a vector ~u is a linear combination of the columns of the matrix (whose coefficients are the components of ~u):
Col. 1 Col. 2 · · · Col.n
u 1 u 2 .. . un
= u 1 (Col.1) + u 2 (Col.2) + · · · + un(Col.n).
The next result (that we assume without proof) provides some properties of the multiplication of matrices:
Proposition I.37. Let A, B and C be matrices. Whenever the products have sense, the following properties hold:
Notice that the multiplication of matrices is not commutative:
Example I.38. If A =
and B =
then
Notice also that the product of two non-zero matrices can be a zero matrix:
Example I.39. (^) [ 1 2 1 2
Definition I.41. The main diagonal (or principle diagonal or diagonal) of a square matrix goes from the upper left to the lower right.
Definition I.42. An identity matrix is square and every entry is 0 except for 1’s in the main diagonal.
In×n =
If the size n of the identity matrix is understood, we write I instead of In×n, for short.
Example I.43. Here is the 2×2 identity matrix leaving its multiplicand unchanged when it acts from the right. (^)
Definition I.50. A lower triangular matrix is square and all its entries above the main diagonal are 0’s, that is, it has the form
a 1 , 1 0... 0 a 2 , 1 a 2 , 2... 0
... an, 1 an, 2... an,n
Example I.51. This is a lower triangular matrix
Definition I.52. Given a m × n matrix A, its transpose, denoted by At^ is the n × m matrix whose rows are the columns of A.
Example I.53. Consider the matrix
. Its transpose is
Dt^ =
The next result is an evident property of the transpose:
Proposition I.54. If A is any matrix then (At)t^ = A.
The following proposition, whose proof is trivial, shows that the operation “taking transpose” behaves well with respect to the sum of matrices and the scalar multiplication:
Proposition I.55. If A and B are m × n matrices and α ∈ R then (A + B)t^ = At^ + Bt^ and (αA)t^ = αAt.
Now we will show that the transpose of the product of two matrices is the product of the trasposes, but the order of the factors reverses.
Proposition I.56. If A is an m × k matrix and B is a k × n matrix then (AB)t^ = BtAt.
PROOF:
Notice that (AB)t^ is an n × m matrix. The element located at the position (j, i) of this matrix (that is, at row j and column i) is the same as the element (i, j) of AB, by the definition
of transpose matrix. But this is the product of the ith row of A by the jth column of B, that coincides with the product of the jth row of Bt^ by the ith column of At; and this is the element of BtAt^ located at the position (j, i). QED
Using induction, this result generalizes to any number of factors:
Proposition I.57. If A 1 ,... , As are m × k matrices then (A 1 · · · As)t^ = Ats · · · At 1.
Definition I.58. A square matrix A is symmetric if At^ = A.
Example I.59. The following matrix is symmetric:
Definition I.60. A square matrix A is antisymmetric if At^ = −A.
Example I.61. The following matrix is antisymmetric:
Notice that the diagonal elements of an antisymmetric matrix are equal to 0 (why?).
UTILITARIAN SUMMARY I.62.