0% found this document useful (0 votes)
37 views12 pagesLec 01
Mathematics Camp - Yale University - lec 01
Uploaded by
Alexy FlemmingsCopyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
0% found this document useful (0 votes)
37 views12 pagesLec 01
Mathematics Camp - Yale University - lec 01
Uploaded by
Alexy FlemmingsCopyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
/ 12
Lecture #1
Introduction
For those of you who have previously studied the material presented in these lectures, |
hope it will revive and organize old memories. For those who have not seen much of the
material, there is a risk that the course will be overwhelming, because a great deal will be
presented in a short time. | will ry to address the needs of both types of students by providing a
Structure that | hope will link the various topics tightly enough as to make it possible to retain
them. You should also realize that to learn mathematics you need to do many practice problems.
Mathematics is like a foreign language. tis learned not only through insight but by frequent
repetition. | will assign daily problem sets. You should try to do these and then perhaps make
up some more of your own to do.
‘An important thing to remember about mathematics is that the main purpose of its
abstractions is to simplify problems and put you in a position to calculate in situations where
you might otherwise be confused by complex impressions. The realization that an abstract
concept can simplify life should make mathematics less daunting and easier to grasp and control
Linear Algebra
4) Simultaneous Linear Equations
We start with a simple example of how one solves two simultaneous linear equations,
Example’
x + 7x =-57
12x | + 3x, = 485.
‘Subtract one third of the second equation from four times the first.
4x + 28x = -228
ax + 15
27x = -243
43
27
It follows that
Substituting this value for x, into the first equation, we obtain
-Ix, - 87 = 63-57 = 6 Matrix Representation of Simultaneous Linear Equations
We need convenient notation to describe systematically the solution procedure just used.
Write M simultaneous linear equations in N unknowns as follows:
ax+ax+..tax=y
axta
ny Fe AK Me
where the coefficients a andy are numbers, and x,, x,, .. , x, are unknowns. These
equations may be written as the single equation Ax = y, where x is the Nevector (xj. 1X). ¥
is the Meveetor (y , ...,¥.), and Ais the matrix
faa a)
aa... a!
A=
a a
Ais called an MxN matrix, which means that it has M rows and N columns. In writing the
equation Ax = y, think of x as an Nx1 matrix and y as an Mx1 matrix, so that the equation may
be visualized as
aa) fs
As ee Bu [Xe Ye
I: |-|- (1.4)
Fn Baw Mu
oras [axe tax) iy
ax ton a
= (1.2)
|
(uk oe a) Oe
In going from system 1.1 to system 1.2, take the Nx1 matrix x, rotate it counterclockwise by
90 degrees, place it on each of the successive rows of the MxN matrix A, multiply superimposed
entries, and add the products.
Elementary Row Operations and Row Equivalence of Matrices
Definition: The following operations on the MxN matrix A, are known as elementary row
operations:
a) multiply a row of A by a non-zero number,
b) interchange two rows, and
c) replace a row by that row plus c times another row, where ¢ is a non-zero number.
Definition: If the MxN matrix B is obtained from A by a finite sequence of such
operations, then B and A are said to be row equivalent.
Lemma 1.1: If A and B are row equivalent, then the equations Ax = 0 and Bx = 0 have the
same solutions,
Proof: It is sufficient to show that the equations Ax = 0 and Bx = 0 have the same
solutions if B is obtained from A by one elementary row operation. This is clearly true for
operations of types (a) and (b) above. Consider an operation of type (c). Let a_be the mth row
of A. Suppose that row m is replaced by row m plus c times row k, where c #0 and k +m.
IfAx=0, then Ya x =O and Ya x =0, sothat
+ca x = LaxscYax
and hence Bx = 0. If Bx=0, then Ya x = 0 and
axsofax- faxso, sothat Ya x = 0 andhence Ax =0. .
We will soon see that elementary row operations may be used to find a matrix B row
equivalent to any matrix A such that the solutions of Bx = 0 are obvious. A similar technique
can be used to find solutions of the equation Ax = y.
Ity is an M vector, let (Ay) be the Mx(N+1) matrix
Fa Ya
By an argument similar to the proof of lemma 1.1, if the Mx(N+1) matrix (B +2) is obtained
from the Mx(N+1) matrix (A:y) by elementary row operations, then the equations Bx = z and
(x
=Oand so(A:y)
Ax = y have the same solutions. For instance, it Ax = y, then Ax —
*|
where 1) written as an (N#1)x1 matrix. Hence
|is the vector (x,
1} "
)
x
(B:z)|~ |=0, and so Bx =z. Similarly Ax = y, if Bx
M4
Definition: A matrix B is said to be row reduced if
a) the first non-zero entry in any row is 1, and
b) each column that contains the first non-zero entry of some row has all its other entries equal
to 0.
Definition: The first non-zero entry of a row is said to be the leading non-zero entry of
that row.
Lemma 1.2: Elementary row operations may be used to reduce any MxN matrix A to @
row reduced matrix B.
Proof: The proof is almost self-evident . Example: The following matrix is row reduced.
jo140
0000
1030
looo4
Definition: A matrix is a row reduced echelon matrix if
a) it is row reduced,
) any row of zeros lies below all non-zero rows, and
©) if the non-zero rows are rows 1 through r and the leading non-zero entry of row m is in
column n , form=1,...,fthenn M, then Wy way W, are
linearly dependent. Since V., «., ¥, span V,w. = }¢a_v., foralln and for some numbers
Asse Ws ey, aF@ numbers, then
Exw-SxSav=$Saxv=-S(Saxv
Sn EE Bane Ee Boeke” eels ceonn Je
not all zero, such that
Since M < N, theorem 1.4 implies that there exist numbers x, ...,
Ya x =0, form
1M, Hence x w+... + XW, = 0 andsow,
ware linearly
dependent. .
Definition: A vector space is finite dimensional, if it has a finite basis.
Corollary 1.7: If V is a finite dimensional vector space, then any two bases have the same
number of elements,
Drool: Hy yoy VandW,, ..., W, are bases of V, then because , ..., v, span V and
Wn.) W_ are independent, it follows that N- 0,v, #0, for some n. Without loss of generality, we may assume
that v #0. The vector v_ by itself is a linearly independent set of vectors, so that there is a
subset of v.,...., v, that is linearly independent. Therefore there is a largest subset of
¥, that is linearly independent. Without loss of generality, we may let this subset be
Voce Wy where KN, IFK=N, then v0 ¥, is abasis for V. If K < Nand n> K, then v
‘must be a linear combination of v., ..., v,, for otherwise by lemma 1.10, the sev... we v
Vv os ¥, Spans V, the set, ..., V Spans V and so forms a basis for V. .
A similar argument proves the following.
Theorem 1.14: If V is a finite dimensional, non-zero vector space, any largest or
maximal set of linearly independent vectors in V is a basis for V.
This theorem suggests a way to construct a basis for a non-zero vector space V. Choose a
non-zero vector vin V. If v_ spans Y, it is a basis for V. We continue by induction on the
number n of vectors chosen to belong to the basis. Suppose that the linearly independent vectors
V., s+ ¥_ Rave been chosen from V. If they span V, they form a basis. If not, choose a non-zero
vector v from V that does not belong to the span ofv , ...., v. By lemma 1.10,
Vy o0+V,V__ are linearly independent. If V is finite dimensional, theorem 1.6 implies that
this process must eventually end. That is, for large enough N, the linearly independent vectors
Vo, on) ¥_ must span V and so form a basis for it
Suppose that the dimension of the vector space V is N and we have K linearly independent
vectors v,....,¥_ in V, where K < N. The inductive process described in the previous
paragraph may be used to construct a basis v,....V.V_
he Year owen, fOF'V. The new basis is
called an extension of v , ...., v, to a basis for V.
Theorem 1.15 Let W be a non-zero subspace of a finite dimensional vector space V such
that W#V. Then W has a finite basis and dim W < dim V.
Proof: Since W is not zero, there is a non-zero vector w in W. Because w is non-zero,
itis independent. From the inductive process described above, we know there is a sequence
10 W., W,, .. Of independent vectors in W. Since these vectors are in V, which is finite
dimensional, we know from theorem 1.6 that this sequence has no more than dim V members.
Let, w,, ...,W, be @ maximal such sequence, where M< dim V. By the argument used in the
previous two paragraphs, w., w,, .... W, span W and so form a basis for W, and so M = dim W,
Since W + V, there is a non-zero vector v in V such that v does not belong to W. By lemma 1.10,
WW, W,. V are independent and so by theorem 1.14, dim V 2 M1 > M-= dim W. i
The Row and Column ranks of a Matrix
IFA is an MxN matrix, its rows are N-vectors and so belong to RY and its columns are
M-vectors and so belong to R. The linear span of the rows of A is a subspace of R called the
row space of A, and the linear span of the columns of A is a subspace of R called the column
space of A.
Definition: f A is an MxN matrix, the column rank of A is the dimension of the column
space of A and the row rank of A is the dimension of the row space of A
The objective of this section is to show that the column and row ranks of any matrix are
equal. | prove this assertion by reducing the matrix to row reduced echelon form.
Lemma 1.16: If the MxN matrices A and B are row equivalent, then their row spaces are
the equal.
Proof: It is sufficient to show that each of the elementary row operations does not change
the row space. Clearly neither interchanging two rows nor multiplying a row by a non-zero
number changes the row space. Consider replacing a row by the sum of that row and a multiple
of another row. Without loss of generality, suppose that B is the matrix A with the first row of
A replaced by that row plus d times the second row, where d 0. Iv belongs to the span of the
rows of A, then v = E ca, where ais the mit row of Aandc,, .....¢ are numbers. Then
v=o(a +da) +(e-eda+ Soa = opt (e-ogb,+ ep, where b_ is the mth
row of B. Therefore v belongs to the span of the rows of B. Si
rly if v belongs to the span of
the rows of B, v = i cb, for some numbers ¢., ...¢,, andsov= ca +(od+c)a,
+ 3.c@ and hence lies in the span of the rows of A. This proves that the row spaces of A and B
are the same, .
The preceding lemma implies that if 6 is obtained from A by elementary operations, then
the row ranks of A and 8 are the same.
"1 Lemma 1.17: If the MxN matrices A and B are row equivalent, then their column ranks
are the equal.
Proof: Let K be the column rank of A. By theorem 1.18, we may assume that K columns
of A form a basis for its column space. Without loss of generality, we may assume that this
basis consists of the first K columns of A. If K Kandizn. By
lemma 1.1, Bx = 0 and therefore b” = )) ¢b', where b* denotes the kth column of 8. Therefore
the first K columns of B span the column space of B. | complete the proof by showing that the
first K columns of B are independent. If they are not, then there exist numbers ¢, .. ,¢ , not
all of which are zero, such that ¢b* = 0. Let x be the N-vector defined by x = c, if k
1,..,Kandx =0, ifm=K+4,....,N. Then Bx =0, so that by lemma 1.1, Ax = 0 and
hence J ¢.a' = 0 and so the first K columns of A are not independent, contrary to hypothesis.
Therefore the first K columns of B are linearly independent and so form a basis for the column
space of B. Hence the column rank of B is K. .
‘Theorem 1.18: The row and column ranks of any matrix are equal.
Proof: Let A be an MXN matrix and let B be @ row reduced echelon matrix that is row
equivalent to A. By lemmas 1.16 and 1.17, the row rank of A equals that of B and the column
rank of A equals that of B, so that it is sufficient to show that the row and column ranks of B are
equal. Let K be the number of non-zero rows of B, these being the first K rows, since B is a row
reduced echelon matrix. | show that both the row and column ranks of B equal K.
For k= 1, K, let the leading non-zero entry of the kt" row be in column ny, where
11