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 (x

^{1}, ..., x

^{t}) be a family of vectors in a vector space V. We call u in V a linear combination of (x

^{1}, ..., x

^{t}) if and only if there exists scalars α

_{1}, ..., α

_{t}such that:

u = α

_{1}x

^{1}+ ... + α

_{t}x

^{t}= ∑ (i=1,t) α

_{i}x

^{i}.

Definition 2: [[ X ]]

Let X = (x

^{1}, ..., x

^{n}) 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 = (x

^{1}, ..., x

^{t}) be a family of vectors in V. Let W be a subspace of V that contains each x

^{i}.

Then, [[X]] ⊆ W.

Proof:

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

(2) So α

_{1}x

^{1}, ..., α

_{t}x

^{t}∈ W by the closure of scalar multiplication.

(3) α

_{1}x

^{1}+ ... + α

_{t}x

^{t}∈ W by the closure of addition.

(4) Thus, it is clear that every linear combination of (x

^{1}, ..., x

^{t}) is contained in W.

QED

Corollary 1.1:

Let V be a vector space and let X = (x

^{1}, ..., x

^{t}) and Y = (y

^{1}, ..., y

^{u}) be families of vectors in V.

If each x

^{i}is a linear combination of (y

^{1}, ..., y

^{u}), then [[X]] ⊆ [[Y]]

Proof:

(1) Assume that for each i, x

^{i}∈ [[y

^{1}, ..., y

^{u}]].

(2) Then [[y

^{1}, ..., y

^{u}]] is a subspace of V. [See Theorem 2, here]

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

[[x

^{1}, ..., x

^{t}]] ⊆ [[y

^{1}, ..., 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 = (x

^{1}, ..., x

^{t}).

(2) Each x

^{i}is a member of (X,Y)

(3) Hence, each x

^{i}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,k}a

_{k,j}

(4) So,

c

_{i,*}= [ c

_{i,1}c

_{i,2}... c

_{i,p}] =

= [ ∑(k=1,n) b

_{i,k}a

_{k,1}∑(k=1,n) b

_{i,k}a

_{k,2}... ∑(k=1,n) b

_{i,k}a

_{k,p}] =

= [ b

_{i,1}a

_{1,1}b

_{i,1}a

_{1,2}... b

_{i,1}a

_{1,p}] +

[ b

_{i,2}a

_{2,1}b

_{i,2}a

_{2,2}... b

_{i,2}a

_{2,p}] + ... +

[ b

_{i,n}a

_{n,1}b

_{i,n}a

_{n,2}... b

_{i,n}a

_{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,k}b

_{k,j}

(4) So,

c

_{*,j}= [ c

_{1,j}c

_{2,j}... c

_{p,j}] =

= [ ∑(k=1,m) a

_{1,k}b

_{k,j}∑(k=1,m) a

_{2,k}b

_{k,j}... ∑(k=1,m) a

_{p,k}b

_{k,j}] =

= [ a

_{1,1}b

_{1,j}a

_{2,1}b

_{1,j}... a

_{p,1}b

_{1,j}] +

[ a

_{1,2}b

_{2,j}a

_{2,2}b

_{2,j}... a

_{p,2}b

_{2,j}] + ... +

[ a

_{1,n}b

_{n,j}a

_{2,n}b

_{n,j}... a

_{p,n}b

_{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

_{*,j}A

QED

References

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