This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Range and nullspaces Range space y = All, A E RHIXII Example: _ 8 5 —2 _ 8371+5372—2m3
” _ 0 0 0 "'2 _ 0 :53 3/6722 but span{yy=AJ;, $6723} = span{[(1)]} #722. (span{[(l)]} C R2) Range of A
Range(A) =: R(A) = {y  y = Ax, I ERW’} Q R777 Range of a function: f : R” —> R’” g = m, Rm ={f(rI:)rI‘ER"} g Rm Roy Smith, ECE 210a, 3: 1 Range and nullspaces Range(A) is a subspace The key property is closure (Via addition and scalar multiplication) Additive closure: Given 3113312 6 R (A) implies that there exists $1,111) 6 R” such that7
1/1 = A 1‘1 and 1/2 = ATz Now y1+ 1/2 = A2751 + A252 2 14(171 + 172) and therefore yl + 1/2 E R Multiplicative closure: Given 1/1 E R(A) and a E .7", We have 171 E R" such that7
yr = Ali Now ayl : (r1411 : Am;er and therefore 04/1 6 R Roy Smith, ECE 210a, 3: 2 Range and nullspaces Range space
Every subspace of R’” is the range of some linear function (or matrix).
Proof:
Choose a basis for the subspace, V Q Rm,
B]; = {1,117 . . . ., '07.}, r S m. V = span{By} and every 2) E V can be written as (11
7‘
, , , a2
“:2 :aivi = 01 02 U7» .
{:1
a,
Expand with zero columns to get,
051
u = ’01 Ur 0 ' and all :1; e R” give a v E V.
a,
*
ﬁ—z Roy Smith, ECE 210a, 3: 3 Range and nullspaces Range space Of AT (Or AT if A E Rmxn)
Rm {I  g m
Summary R (A) = Span ' ' [Aw]    } (column space) R (AI) = span  [Amy    } (row space) y E 'R, (A) <=> 5:1; 6 R" such that y 2 Art at E R (AT) <=> 3y 6 R’“ such that acT = yTA ( or ac : ATy) Roy Smith, ECE 210a, 3: 4 Range and nullspaces Nullspace of A 6 RM” (also known as the kernel of A)
N(A) = {xAx=0} Q R”. Nullspace of a function: f : R" —> ’R’“ = I fob) = g Rn. The nullspace of A is a subspace For all .1‘1,.’L‘2 €N(A),
Axl=0 and Axg=0 => Ax1+Ax2=A(:m+:r2)=0 => 171+:cheN(A) For all 171 EN(A) and all a E 7",
Am = 0 => aAm'l = A(a:z;1) = 0 => (171:1 E N(A) Nullspace of AT:
MAT) = {LUIATFO} E Roy Smith, ECE 210a, 3: 5 Range and nullspaces Relationship between R (A), R (AT), N(A) and N (AT):
Basis for N(A):
As N (A) is a subspace7 we can ﬁnd a spanning set of basis vectors: BN(A) = {flanyjk} and extend this to all R” with arbitrary linearly independent vectors, B XA. BX={ 5:1,...7Ek, $1,...,a:,» } k+r=n XA := span{BXA} isasubspace.
*1 EH
BN(A) Em A
span{ABXA};R(A> ABXA A BXA dim(XA>=r
0 x 0 W
A dim<N<A>)
=n—r Roy Smith, ECE 210a, 3: 6 Range and nullspaces
Relationship between R (A), 7?, (AT), N (A) and N (AT):
Claim: span {A BXA} = R(A) and so dim (R = 1“.
First we show that dim(’R, 2 1‘.
To see this, apply A to each of the basis vectors, BXA, Aml = y1, etc.,
to get a set of linearly independent vectors: {3/1, . . . , yr} 6 R If they were not linearly independent then there would exist, 017,, i = 1, . . . , r (not all zero)
7‘ 7" 7"
such that, 0 = Zaiyi = ZoeA xi = A aim)
71:1 71:1 71:1 but this would imply that either 07%;) E N (A) or 20:73:77 2 0.
7:1 7:1 This is a contradiction as neither can be the case: 1. A linear combination of vectors in XA is also in XA (not N 2. The m are a linearly independent set. Roy Smith, ECE 210a, 3: 7 Range and nullspaces Relationship between R(A), R (AT), AMA) and N (AT):
Claim: Span {A BXA} = R (A) and so dim (R = 7‘,
And now we show that R (A) cannot contain more than 7" linearly independent vectors. Suppose that there was another linearly independent vector, yrﬂ E ’R (i.e. yr+1 = ACCT+1 )
This would mean that 017 : 0 is the only solution to
r+1 r+1 7’+1 r+1
0 = Zaiyi = ZealAccl = A dim) and so Zeal$1 = 0 => 012 = 0.
71:1 71:1 71:1 71:1 This is a contradiction because it would mean that dim(XA) 2 7’ + 1. A
dim (72 (A)) : 7 A BM dim(XA) : 7
<\ 0 R"
A dim(N (14))
=n—r Roy Smith, ECE 210a, 3: 8 Range and nullspaces Relationship between R (A), R (AT), N‘ (A) and N (AT):
Fundamental spaces for AT: N (AT) and R (AT). All of our above arguments apply to the matrix: AT : R771 —> R". Find a basis for N (AT) and pick any other linearly independent vectors to span the rest of R”. ByAT = {y1,...,ys} and By = BRm = BN(AT) U ByAT. dim (yAT) = s T dim (R (AT)) = s
\A/ .
R771 —>
dim (N (AT)) AT/ =m—s Roy Smith, ECE 210a, 3: 9 Range and nullspaces Relationship between R (A), R (AT), N(A) and N (AT): A
. A .
A ; R" _) Rm dim (R = 1" d1m(XA) : 7"
<\ R71
A dimUV (14))
= n — 7" dim (32“) = s dim (R (AT)) = s
W
AT Z Rm _> R71, Rm, dim (N (AT)) AT =m—s Roy Smith, ECE 210a, 3: 10 Range and nullspaces Relationship between ’R, (A), R (AT), N (A) and N (AT) Claim: For all y E ’R, (A), 3; ¢ N (AT).
This means that dim(R S m—dim(N (AT)). Equivalently, r S 3.
To see this suppose there was a vector, :1) = An”: (i.e. 17 E ’R,(A))
That also satisﬁed, AT?) = 0. (i.e. g] E N (AT))
But the sum of the squares of the components of g is given by,
My = yTAcfc = (ATgf a = 0 => y = o. (contradiction)
Claim: For all :1: E ’R (AT), 2: ¢ N (Which means that s S r) The same argument, using N (A) and ’R (AT), veriﬁes this, so: 7' = 5. Roy Smith, ECE 210a, 3: 11 Range and nullspaces
Relationship between R (A), R (AT), N(A) and N (AT)
For Lg, choose any basis of ’R, (Our previous arguments are good for any basis) For XA, choose any basis of R (AT). Complete picture: A
A H
dim (R (A)) = 7“ BR(A) dim (R (A1 = T
dim (N (“D AT A dim (N (A)) Another consequence: A E 72"an : dim(R + dim(N = n. Aside: It is possible to choose BR<AT) = and BRM) = {yi} such that y) 2 Am and $1 = ATyi. Roy Smith, ECE 210a, 3: 12 Matrix rank Matrix rank
The rank of a matrix, A E 73””X", is r = dim(’R, = dim(’R, If A 6 RM” (square) then, det(A) # 0 <=> A’1 exists <=> rank(A) = n. This is because if rank(A) = n, then dim(N = 0. So there is no vector :1: e R", such that Am = 0. Roy Smith, ECE 210a, 3: 13 Matrix rank Matrix products: A E R’"”, B E RWY), AB 6 73m”
rank(AB) : rank(B) — dim(N (A) F] R We can (roughly) picture this as7 dim=r dim=r dimzs
A
A B
Bum) /— BR(BT)
0 . 0
A . B
dimzm—r dimzn—r dimzp—s Technical details: See Meyer, section 4.5.1. Roy Smith, ECE 210a, 3: 14 Matrix rank Matrix products: A E Rm”, B E 73"”, AB 6 Rmxp. Upper bound: rank(AB) S min{rank(A),rank(B)} Proof:
rank(AB) = rank(B)—dim(N(A)ﬂR(B)) g rank(B).
rank(AB) = rank((AB)r) = rank(BTAT) S rank(AT) = rank(A). Lower bound: rank(AB) 2 rank(B) + rank(A) — 71.
Proof:
dim(.N(A) ﬂ g dim(N = n — rank(A). So,
rank(AB) = rank(B) i dim(N(A)ﬂR(B)) Z rank(B) + rank(A) i n. Roy Smith, ECE 210a, 3: 15 ...
View
Full
Document
 Spring '08
 Chandrasekara
 Linear Algebra, linearly independent vectors, column space, Roy Smith

Click to edit the document details