LECTURE 2

Defintion. A subset W of a vector space V is a \underline{subspace} if (1) W is non-empty

(2) For every v̄, w̄ ∈ W and a, b ∈ 𝔽, av̄ + bw̄ ∈ W.

Expressions like av̄ + bw̄, or more generally

∑_{i=1}^{k} a_{i}v̄ + i

are called \underline{linear combinations}. So a non-empty subset of V is a subspace if it is closed under linear combinations. Much of today’s class will focus on properties of subsets and subspaces detected by various conditions on linear combinations.

Theorem. If W is a subspace of V, then W is a vector space over 𝔽 with operations coming from those of V.

In particular, since all of those axioms are satisfied for V, then they are for W. We only have to check closure!

Examples:

Defintion. Let 𝔽^{n} = {(a_{1}, …, a_{n})|a_{i} ∈ 𝔽} with coordinate-wise addition and scalar multiplication.

This gives us a few examples. Let W ⊂ 𝔽^{n} be those points which are zero except in the first coordinate:

W = {(a, 0, …, 0)} ⊂ 𝔽^{n}.

Then W is a subspace, since

a · (α, 0, …, 0) + b · (β, 0, …, 0) = (aα + bβ, 0, …, 0) ∈ W.

If 𝔽 = ℝ, then W′ = {(a_{1}, …, a_{n})|a_{i} ≥ 0} is not a subspace. It’s closed under addition, but not scalar multiplication.

We have a number of ways to build new subspaces from old.

Proposition. If W_{i} for i ∈ I is a collection of subspaces of V, then

W = ⋂_{i∈I} W_{i} = {w̄ ∈ V |w̄ ∈ W_{i}∀i ∈ I}

is a subspace.

Proof. Let v̄, w̄ ∈ W. Then for all i ∈ I, v̄, w̄ ∈ W_{i}, by definition. Since each W_{i} is a subspace, we then learn that for all a, b ∈ 𝔽,

av̄ + bw̄ ∈ W_{i},

and hence av̄ + bw̄ ∈ W. □

Thought question: Why is this never empty?

The union is a little trickier.

Proposition. W_{1} ∪ W_{2} is a subspace iff W_{1} ⊂ W_{2} or W_{2} ⊂ W_{1}.

Proof. ⇐ is obvious. We need to show the other, so assume that we can find w̄_{1} ∈ W_{1}−W_{2} and w̄_{2} ∈ W_{2}−W_{1}. Then if W_{1}∪W_{2} is a subspace, then w̄_{1}+w̄_{2} ∈ W_{1}∪W_{2}. This means it’s in one of them, and without loss of generality, we may assume it’s in W_{1}. But this means

w_{2} = (w̄_{1} + w̄_{2}) − w̄_{1} ∈ W_{1},

a contradiction. □

Example to keep in mind: ℝ^{2} with W_{1} = x-axis and W_{2} = y-axis.

Instead of the union, we consider the smallest subspace containing the union.

Defintion. If W_{1} and W_{2} are subspaces of V, then the \underline{sum} is

W_{1} + W_{2} = {aw̄_{1} + bw̄_{2}|w̄_{i} ∈ W_{i}}.

In other words, we consider all linear combinations of elements of W_{1} and W_{2}. Clearly if W is any subspace that contains W_{1} and W_{2}, then W contains W_{1} + W_{2}. On the other hand, it’s also closed under linear combinations and non-empty, so this is a subspace.

Special attention is given to the case where W_{1} ∩ W_{2} = {¯0}.

Defintion. If W_{1} ∩ W_{2} = {¯0}, then we say the sum of W_{1} and W_{2} is the (internal) \underline{direct sum}, and we write it W_{1} ⊕ W_{2}.

So what’s so special about direct sums?

Proposition. Every element in W_{1} ⊕ W_{2} can be uniquely written as w̄_{1} + w̄_{2}.

Proof. Assume w̄_{1} + w̄_{2} = v̄_{1} + v̄_{2}, where the subscript indicates the subspace from which the element is drawn. Then rearranging, we see

w̄_{1} − v̄_{1} = v̄_{2} − w̄_{2}.

The left-hand side is in W_{1}, while the right-hand side is in W_{2}, so both are in the intersection. This means both are ¯0, and hence w̄_{i} = v̄_{i}. □

Our notion of sums therefore includes two distinct notions:

(1) being able to write elements as linear combinations

(2) being able to do so uniquely.

The former is related to span, the latter to linear independence.

Defintion. Let X ⊂ V. Then the \underline{span} of X is the collection of all linear combinations of elements of X:

Span(X) = ⟨X⟩ = {a_{1}x̄_{1} + ⋯ + a_{n}x̄_{n}|x̄_{i} ∈ X, n ≥ 0}.

If X is empty, then ⟨X⟩ = {¯0}.

The following is immediate.

Proposition. The set ⟨X⟩ is a subspace of V.

Examples: For any V, ⟨V ⟩ = V. If X = W ∪ U, then ⟨X⟩ = W + U.

Just as before, if W is a subspace of V and W contains X, then ⟨X⟩ ⊂ W. Thus ⟨X⟩ is the smallest subspace containing X, and the elements of X provide convenient names for every element of their span.

Proposition. If w̄ ∈ ⟨X⟩, then

⟨{w̄} ∪ X⟩ = ⟨X⟩.

Proof. One inclusion is obvious. For the other, since w̄ ∈ ⟨X⟩, we know we can write it as a linear combination of elements of X. Thus if we have a linear combination of elements of X and w̄, then we can substitute for w̄ and get a linear combination with just elements of X. □

This helps us greatly in cutting out redundant elements: anything we can name using a smaller set of elements is extraneous.

Defintion. A \underline{spanning set} is a set X such that ⟨X⟩ = V.

In general, these can be very big. One of the goals of much of linear algebra is to give a very compact spanning set for an arbitrary vector space.

The corresponding small notion is linear independence.

Defintion. A set X is \underline{linearly independent} if

a_{1}v̄_{1} + ⋯ + a_{n}v̄_{n} = ¯0

implies a_{1} = ⋯ = a_{n} = 0 for any v̄_{i} ∈ X.

If X is not linearly independent, then it is linearly dependent.

We again see only ¯0 showing up. We can restate this definition as “¯0 has a unique presentation as a linear combination of elements of X.

Theorem. If X is linearly independent and

a_{1}v̄_{1} + … a_{n}v̄_{n} = b_{1}w̄_{1} + ⋯ + b_{m}w̄_{m},

where all vectors are from X and all coefficients are non-zero, then n = m and up to reordering, a_{i} = b_{i} and v̄_{i} = w̄_{i} for all i.

Proof. Order the v̄s and w̄s so that v̄_{1} = w̄_{1}, …, v̄_{t} = w̄_{t}. Moving all terms to one side then gives

(a_{1} − b_{1})v̄_{1} + ⋯ + (a_{t} − b_{t})v̄_{t} + a_{t+1}v̄_{t+1} + ⋯ − (b_{t+1}w̄_{t+1} + …) = ¯0.

Since X is assumed to be linearly independent, we learn that a_{t+1} = ⋯ = a_{n} = 0 and b_{t+1} = ⋯ = 0, and by our assumption on the non-zeroness of the coefficients, this forces t = n = m. Moreover, we learn that a_{i} = b_{i} for all i. □

Thus knowing that ¯0 has a unique presentation as a linear combination ensures that everything in the span does so. We can use span more directly to test for linear independence.

Proposition. A set X is linearly independent if and only if for all v̄ ∈ X, v̄ ∉ ⟨X − {v̄}⟩.

Proof. We show the negative of this. X is linearly dependent if there is a linear dependence relation

a_{1}v̄_{1} + ⋯ + a_{n}v̄_{n} = ¯0,

with some coefficient (say a_{1}) not equal to zero. Solving for v̄_{1} then expresses v̄_{1} as an element of the span of the remaining vectors. □

Thus if a set is linearly independent, then we can add any vector not in the span of it to it and still have a linearly independent set. On the other hand, if we add in a vector in the span, then the set becomes linearly dependent.