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

u = α₁x¹ + ... + α_tx^t = ∑ (i=1,t) α_ixⁱ.

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, ..., x^t) 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, ..., α_tx^t ∈ W by the closure of scalar multiplication.

(3) α₁x¹ + ... + α_tx^t ∈ W by the closure of addition.

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

QED

Corollary 1.1:

Let V be a vector space and let X = (x¹, ..., x^t) and Y = (y¹, ..., y^u) be families of vectors in V.

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

Proof:

(1) Assume that for each i, xⁱ ∈ [[y¹, ..., y^u]].

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

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

[[x¹, ..., x^t]] ⊆ [[y¹, ..., y^u]]

QED

Corollary 1.2:

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

Then [[X]] ⊆ [[X,Y]]

Proof:

(1) Let X = (x¹, ..., x^t).

(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: a_i,*

Let this designate a row such that: a_i,* = [ a_i,1 a_i,2 ... a_i,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:

c_i,j = ∑ (k=1,n) b_i,ka_k,j

(4) So,

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

= [ ∑(k=1,n) b_i,ka_k,1 ∑(k=1,n) b_i,ka_k,2 ... ∑(k=1,n) b_i,ka_k,p ] =

= [ b_i,1a_1,1 b_i,1a_1,2 ... b_i,1a_1,p ] +
[ b_i,2a_2,1 b_i,2a_2,2 ... b_i,2a_2,p ] + ... +
[ b_i,na_n,1 b_i,na_n,2 ... b_i,na_n,p ] =

= b_i,1[ a_1,1 a_1,2 ... a_1,p] +
b_i,2[ a_2,1 a_2,2 ... a_2,p] + ... +
b_i,n[ a_n,1 a_n,2 ... a_n,p] =

= b_i,1(a_1,*) +
b_i,2( a_2,* ) + ... +
b_i,n( a_n,* )

= b_i,*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:

c_i,j = ∑ (k=1,m) a_i,kb_k,j

(4) So,

c_*,j = [ c_1,j c_2,j ... c_p,j ] =

= [ ∑(k=1,m) a_1,kb_k,j ∑(k=1,m) a_2,kb_k,j ... ∑(k=1,m) a_p,kb_k,j ] =

= [ a_1,1b_1,j a_2,1b_1,j ... a_p,1b_1,j ] +
[ a_1,2b_2,j a_2,2b_2,j ... a_p,2b_2,j ] + ... +
[ a_1,nb_n,j a_2,nb_n,j ... a_p,nb_n,j ] =

= b_1,j[ a_1,1 a_2,1 ... a_p,1] +
b_2,j[ a_1,2 a_2,2 ... a_p,2] + ... +
b_n,j[ a_1,n a_2,n ... a_p,n] =

= b_1,j(a_*,1) +
b_2,j( a_*,2 ) + ... +
b_n,j( a_*,n )

= b_*,jA

QED


References

• Hans Schneider, George Philip Barker, Matrices and Linear Algebra, 1989.