## Monday, May 14, 2007

### Linear Combinations

Today's blog is taken straight from Matrices and Linear Algebra by Hans Schneider and George Philip Barker.

If you need to review vectors (see Definition 1 here), vector spaces (see Definition 2, here) or family of vectors (see Definition 4, here), start here.

Definition 1: linear combination

Let (x1, ..., xt) be a family of vectors in a vector space V. We call u in V a linear combination of (x1, ..., xt) if and only if there exists scalars α1, ..., αt such that:

u = α1x1 + ... + αtxt = ∑ (i=1,t) αixi.

Definition 2: [[ X ]]

Let X = (x1, ..., xn) is a family of vectors.

Then [[X]] is the set of all vectors that are a linear combination from X.

Note: This is not the standard notation but I am using it because of limitations with the blogging software that I use.

Definition 3: spanning a vector space

The set of linear combinations of a family of vectors [[X]] is said to span a vector space V if all vectors v that are elements of V are also elements of [[X]].

Lemma 1:

Let V be a vector space and let X = (x1, ..., xt) be a family of vectors in V. Let W be a subspace of V that contains each xi.

Then, [[X]] ⊆ W.

Proof:

(1) Since W is a subspace, it closed under addition and scalar multiplication. [See Definition 2, here]

(2) So α1x1, ..., αtxt ∈ W by the closure of scalar multiplication.

(3) α1x1 + ... + αtxt ∈ W by the closure of addition.

(4) Thus, it is clear that every linear combination of (x1, ..., xt) is contained in W.

QED

Corollary 1.1:

Let V be a vector space and let X = (x1, ..., xt) and Y = (y1, ..., yu) be families of vectors in V.

If each xi is a linear combination of (y1, ..., yu), then [[X]] ⊆ [[Y]]

Proof:

(1) Assume that for each i, xi ∈ [[y1, ..., yu]].

(2) Then [[y1, ..., yu]] is a subspace of V. [See Theorem 2, here]

(3) Using Lemma 1 above, we can conclude that:

[[x1, ..., xt]] ⊆ [[y1, ..., yu]]

QED

Corollary 1.2:

Let X and Y be two families of vectors in V.

Then [[X]] ⊆ [[X,Y]]

Proof:

(1) Let X = (x1, ..., xt).

(2) Each xi is a member of (X,Y)

(3) Hence, each xi is a linear combination of (X,Y).

(4) Using Corollary 1.1 above, we can conclude that [[X]] ⊆ [[X,Y]]

QED

Definition 3: ai,*

Let this designate a row such that: ai,* = [ ai,1 ai,2 ... ai,p ]

Now, I will use this notation in the following lemma.

Lemma 2:

If C = BA, then the ith row of C is a linear combination of the rows A with coefficients from the ith row of B.

Proof:

(1) Let B be an m x n matrix

(2) Let A be an n x p matrix

(3) Let C = BA so that:

C is an m x p matrix [see Definition 1, here for review of matrix multiplication]

such that:

ci,j = ∑ (k=1,n) bi,kak,j

(4) So,

ci,* = [ ci,1 ci,2 ... ci,p ] =

= [ ∑(k=1,n) bi,kak,1 ∑(k=1,n) bi,kak,2 ... ∑(k=1,n) bi,kak,p ] =

= [ bi,1a1,1 bi,1a1,2 ... bi,1a1,p ] +
[ bi,2a2,1 bi,2a2,2 ... bi,2a2,p ] + ... +
[ bi,nan,1 bi,nan,2 ... bi,nan,p ] =

= bi,1[ a1,1 a1,2 ... a1,p] +
bi,2[ a2,1 a2,2 ... a2,p] + ... +
bi,n[ an,1 an,2 ... an,p] =

= bi,1(a1,*) +
bi,2( a2,* ) + ... +
bi,n( an,* )

= bi,*A

QED

Lemma 3:

If C = AB, then the ith row of C is a linear combination of the columns A with coefficients from the jth row of B.

Proof:

(1) Let B be an m x n matrix

(2) Let A be an p x m matrix

(3) Let C = AB so that:

C is an p x n matrix [see Definition 1, here for review of matrix multiplication]

such that:

ci,j = ∑ (k=1,m) ai,kbk,j

(4) So,

c*,j = [ c1,j c2,j ... cp,j ] =

= [ ∑(k=1,m) a1,kbk,j ∑(k=1,m) a2,kbk,j ... ∑(k=1,m) ap,kbk,j ] =

= [ a1,1b1,j a2,1b1,j ... ap,1b1,j ] +
[ a1,2b2,j a2,2b2,j ... ap,2b2,j ] + ... +
[ a1,nbn,j a2,nbn,j ... ap,nbn,j ] =

= b1,j[ a1,1 a2,1 ... ap,1] +
b2,j[ a1,2 a2,2 ... ap,2] + ... +
bn,j[ a1,n a2,n ... ap,n] =

= b1,j(a*,1) +
b2,j( a*,2 ) + ... +
bn,j( a*,n )

= b*,jA

QED

References
• Hans Schneider, George Philip Barker, , 1989.