Linear Algebra and Analysis

3.2 Subspaces

The notion of a subspace is of great importance in applications. Here is what is meant by a subspace.

The fundamental result about subspaces is that they are themselves vector spaces.

Proof: Suppose W is a subspace. Why is it a vector space? To be a vector space, the operations of addition and scalar multiplication must satisfy the axioms for a vector space. However, all of these are obvious because it is a subset of V . The only thing which is not obvious is whether 0 is in W and whether −u ∈ W whenever u is. But these follow right away from Proposition 3.0.2 because if u ∈ W,

u = −u ∈ W by the fact that W is closed with respect to linear combinations, in particular multiplication by the scalar −1. Similarly, take u ∈ W. Then 0 = 0u ∈ W. As to + being an operation on W, this also follows because for u,v ∈ W,u + v ∈ W. Thus if it is a subspace, it is indeed a vector space.

Conversely, suppose it is a vector space. Then by definition, it is closed with respect to linear combinations and so it is a subspace. ■

This leads to the following simple result.

Proof: Suppose span

= V . Using the same construction of Proposition 3.1.8, the same process must stop after k ≤ n steps since otherwise one could obtain a linearly independent set of vectors with more vectors in it than a spanning set. Thus it has a basis with no more than n vectors. ■

You have from the equation that x = 2y + z and so any vector in this set is of the form

( ) 2y + z |( y |) : y,z ∈ ℝ z

Conversely, any vector which is of the above form satisfies the condition to be in W. Therefore, W is of the form

( ) ( ) 2 1 y |( 1 |) + z|( 0 |) 0 1

where y,z are scalars. Hence it equals the span of the two vectors in ℝ³ in the above. Are the two vectors linearly independent? If so, they will be a basis. Suppose then that

( ) ( ) ( ) | 2 | | 1 | | 0 | y( 1 ) + z( 0 ) = ( 0 ) 0 1 0

Then from the second position, y = 0. It follows then that z = 0 also and so the two vectors form a linearly independent set. Hence a basis for W is

( ( ) ( ) ) |{ | 2 | | 1 | |} | ( 1 ) ,( 0 ) | ( 0 1 )

The dimension of this subspace is also 2.

There are two things to show, that the set of vectors is independent and that it spans ℝ³. Thus we need to verify that there is exactly one solution to the system of equations

( ) ( ) ( ) ( ) | 1 | | 1 | | 0| | a | x( 1 ) + y( 3 ) + z( 1) = ( b ) 1 3 4 c

for any choice of the right side. Recall how to do this. You set up the augmented matrix and then row reduce it.

( ) 1 1 0 a |( 1 3 1 b |) 1 3 4 c

After some row operations, this yields

( 1 0 0 3a− 2b + 1c ) | 22 13 61 | ( 0 1 0 3b− 2a − 6c ) 0 0 1 13c− 13b

Thus there is a unique solution to the system of equations. This shows that the set of vectors is a basis because one solution when the right side of the system equals the zero vector is x = y = z = 0. Therefore, from what was just done, it is the only solution and so the vectors are linearly independent. As to the span of the vectors equalling ℝ³, this was just shown also.

You can do it the same way. It is really a question about whether there exists a unique solution to the system

( ) ( ) ( ) ( ) 1 1 1 a x |( 1 |) + y|( 3 |) + z|( 1 |) = |( b |) 1 3 − 4 c

for any choice of the right side. The augmented matrix is

( ) 1 1 1 a |( 1 1 1 b|) 1 3 − 4 c

After row reduction, this yields

( ) | 1 1 1 a | ( 0 2 − 5 c− a ) 0 0 0 b− a

Thus there is no solution to the equation unless b = a. It follows the span of the given vectors is not all of ℝ³ and so this cannot be a basis.

If the span of these vectors were all of ℝ³, this would contradict Theorem 3.1.5 because it would be a spanning set which is shorter than a linearly independent set

.

If it were a basis, then it would need to be linearly independent but this cannot happen because it would contradict Theorem 3.1.5 by being an independent set of vectors which is longer than a spanning set.

Proof: Say

is linearly independent. Is span = V ? If not, there would be wspan and then by Lemma 3.1.7 would be linearly independent which contradicts Theorem 3.1.5. As to the second claim, cannot be linearly independent because this would contradict Theorem 3.1.5 and so it is dependent.

Now say V = span

. By Theorem 3.1.5 again, you must have m ≥ n since spanning sets are at least as long as linearly independent sets, one of which is a basis having n vectors. If w₁ is in the span of the other vectors, delete it. Then consider w₂. If it is in the span of the other vectors, delete it. Continue this way till a shorter list is obtained with the property that no vector is a linear combination of the others, but its span is still V . By Proposition 3.1.3, the resulting list of vectors is linearly independent and is therefore, a basis since it spans V .

Now suppose for k < n,

is linearly independent. Follow the process of Proposition 3.1.8, adding in vectors not in the span and obtaining successively larger linearly independent sets till the process ends. The resulting list must be a basis. ■