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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 ...
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 Spring '08
 Chandrasekara
 Linear Algebra, linearly independent vectors, column space, Roy Smith

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