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MATH125 (100)

George Peschke (20)

Lecture 20

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] [0]

[ . . ] [..] = [ .. ] = [.]

[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.

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###### Document Summary

The orthogonal compliment of a subset s of = {x ^n :|: x. 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! Given an (m,n)-matrix a, write it in terms of row vectors ri. The null space of a is the orthogonal compliment of s = Null(a) = s^ (a11, , a1n) (x1, , xn) = 0 x] [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.

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