Class Notes (1,100,000)
CA (620,000)
U of A (10,000)
MATH (1,000)
MATH125 (100)
Lecture 20

# MATH125 Lecture Notes - Lecture 20: List Of Forgotten Realms Nations, Linear Combination

Department
Mathematics
Course Code
MATH125
Professor
George Peschke
Lecture
20

Page:
of 3 November 6 2014
CCIS 1-160
Subvector spaces (subspace)
The orthogonal compliment of a subset S of = {x ^n :|: x
S }
The orthogonal compliment of the 0-vector is ^n.
The orthogonal compliment of ^n is the 0-vector
For any subset S of is a subversion space of i.e. 0, is
v+w in S^ s = 0 for every s in S?
(s w) = 0
Because (s w) = 0, yes!.
(3) if v in S^ v in S^ v) s) = 0 Yes!
Null space of a matrix
Given an (m,n)-matrix A, write it in terms of row vectors Ri
A = [a11 ... a1n] [R1]
[ . . ] = [..]
[am1 ... amn] [Rm]
The null space of A is the orthogonal compliment of S =
{R1, ..., Rm}
Null(A) = S^
(a11, ..., a1n) (x1, ..., xn) = 0
x] 
[ . . ] [..] = [ .. ] = [.]
[am1 ... amn] [xn] [Rm ^n is linearly
independant if, for any choice of
pairwise distinct vectors a1, ..., am, from S, the vector
equation t1am = 0 has exactly one solution, namely, t1 = ... =
tm = 0
Recall: a || b if there are numbers s,t not both 0,
such that sb = 0
When not parallel, only solution of sb = 0 is s = t = 0
Two vectors are linearly dependant exactly when they are
parallel. - This is
the definition of being parallel.
Suppose a1, ..., am are linear independant vectors. If
s1a1 + ... + smam = x = t1a1 + ... + tmam.
then, s1 = t1, ..., sm = tm.
That is, if a vector x is in the span of a set of
linearly independant
vectors a1, ..., am, then there is exactly one way of
expressing x as a
linear combination of a1, ..., am.
Because, (t1-s1)a1 + ... + (tm-sm)am = 0
t1-s1 = ... = tm-sm = 0
n, the rows of A are linearly independant if and only
if there are m columns in A whose determinant is
not 0
ii) If n 0, then a, b, c are linearly independant
b) For each choice of 3 columns from A, the
resulting (3,3)-matrix has det = 0, then a, b, c are linearly
dependant.
det(Asub(3,3))[2 0 3] = det[2 3] = 23 = 4 -
9 = -5 a, b, c are linearly independant.
a, b, c are linearly independant.
If the (n,n)-matrix is invertable, then its columns form
If the (n,n)-matrix is invertable, then its columns form
a linearly independant set.
a linearly independant set.
Orthogonal/orthonormal set of vectors
Orthogonal/orthonormal set of vectors
Let S be a set of non-zero vectors in for x y = 0
Let S be a set of non-zero vectors in for x y = 0
S is called orthonormal x x = 1
S is called orthonormal x x = 1
Every set S of orthogonal vectors in linearly independant
Every set S of orthogonal vectors in linearly independant
Consider the orthogonal set {a1, a2, ..., an}
Consider the orthogonal set {a1, a2, ..., an}
t1ak + ... + tna1 + ... + tkan)ak = 0
t1ak + ... + tna1 + ... + tkan)ak = 0
t1(a1ak) + ... + tn(anak) = 0
t1(a1ak) + ... + tn(anak) = 0
(ak tk = 0
(ak tk = 0
So, {a1, a2, ..., an} are linearly independant.
So, {a1, a2, ..., an} are linearly independant.