Engineering Math

6.1 Subspaces

A subspace is a set of vectors with the property that linear combinations of these vectors remain in the set. Geometrically, subspaces are like lines and planes which contain the origin. More precisely, the following definition is the right way to think of this.

There is no substitute for the above definition or equivalent algebraic definition! However, it is sometimes helpful to look at pictures at least initially. The following are four subsets of ℝ². The first is the shaded area between two lines which intersect at the origin, the second is a line through the origin, the third is the union of two lines through the origin, and the last is the region between two rays from the origin. Note that in the last, multiplication of a vector in the set by a nonnegative scalar results in a vector in the set as does the sum of two vectors in the set. However, multiplication by a negative scalar does not take a vector in the set to another in the set.

Observe how the above definition indicates that the claims posted on the picture are valid. Now here are the two main examples of subspaces.

Proof: Suppose Ax_i is in Im

and a,b are scalars. Does it follow that aAx₁ + bAx₂ is in Im? The answer is yes because

aAx1 + bAx2 = A (ax1 + bx2) ∈ Im (A )

this because of the above properties of matrix multiplication. Note that A0 = 0 so 0 ∈ Im

and so Im≠∅.

Now suppose x,y are both in N

and a,b are scalars. Does it follow that ax + by ∈ N? The answer is yes because

A(ax + by) = aAx + bAy = a0+ b0 = 0.

Thus the condition is satisfied. Of course N

≠∅ because A0 = 0. ■

Subspaces are exactly those subsets of Fⁿ which are themselves vector spaces. Recall that a vector space is something which satisfies the vector space axioms on Page 56.

Proof: Suppose first that V is a subspace. It is obvious all the algebraic laws hold on V because it is a subset of Fⁿ and they hold on Fⁿ. Thus u + v = v + u along with the other axioms. Does V contain 0? Yes because it contains 0u = 0. Are the operations defined on V ? That is, when you add vectors of V do you get a vector in V ? When you multiply a vector in V by a scalar, do you get a vector in V ? Yes. This is contained in the definition. Does every vector in V have an additive inverse? Yes because −v =

v which is given to be in V provided v ∈ V .

Next suppose V is a vector space. Then by definition, it is closed with respect to linear combinations. Hence it is a subspace. ■

There is a fundamental result in the case where m < n. In this case, the matrix A of the linear transformation looks like the following.

Proof: First consider the case where A is a 1 × n matrix for n > 1. Say

( ) A = a1 ⋅⋅⋅ an

If a₁ = 0, consider the vector x = e₁. If a₁≠0, let

( ) b || 1 || x = || .. || ( . ) 1

where b is chosen to satisfy the equation

∑n a1b+ ak = 0 k=2

Suppose now that the theorem is true for any m × n matrix with n > m and consider an

× n matrix A where n > m + 1. If the first column of A is 0, then you could let x = e₁ as above. If the first column is not the zero vector, then by doing row operations, the equation Ax = 0 can be reduced to the equivalent system

A x = 0 1

where A₁ is of the form

( ) 1 aT A1 = 0 B

where B is an m×

matrix. Since n > m + 1, it follows that > m and so by induction, there exists a nonzero vector y ∈ Fⁿ⁻¹ such that By = 0. Then consider the vector

( b ) x = y

A₁x has for its top entry the expression b + a^Ty. Letting B =

, the i^th entry of A ₁x for i > 1 is of the form b_i^Ty = 0. Thus if b is chosen to satisfy the equation b + a^Ty = 0, then A₁x = 0.■

Now here is a very fundamental definition.

With this preparation, here is a major theorem.

Proof: Let u_i = ∑ _j=1^sa_jiv_j. This is merely giving names to the scalars in the linear combination which yields u_i. Now suppose that s < r. Then if A is the matrix which has a_ji in the j^th row and the i^th column, it follows from Theorem 6.1.4 that there exists a vector in F^r such that Ax = 0 but x≠0. However, then

Which contradicts the assertion that the set of vectors is linearly independent. Indeed, there is a nonzero vector x for which x = 0. Thus we cannot have s < r and so the only other possibility is that s ≥ r. ■

Proof: From Theorem 6.1.6, r ≤ s because each u_i is in the span of

and is independent. Then also r ≥ s by the same reasoning. ■

The next lemma says that if you have a vector not in the span of a linearly independent set, then you can add it in and the resulting longer list of vectors will still be linearly independent.

Proof: Suppose ∑ _i=1^kc_iu_i + dv = 0. It is required to verify that each c_i = 0 and that d = 0. But if d≠0, then you can solve for v as a linear combination of the vectors,

,

∑k ( ) v = − ci- ui i=1 d

contrary to assumption. Therefore, d = 0. But then ∑ _i=1^kc_iu_i = 0 and the linear independence of

implies each c_i = 0 also.  ■

It turns out that every subspace equals the span of some vectors. This is the content of the next theorem.

Proof: Pick a nonzero vector of V,u₁. If V = span

, then stop. You have found your basis. If V ≠span, then there exists u₂ a vector of V which is not a vector in span. Consider span. By Lemma 6.1.11, is linearly independent. If V = span, stop. You have found a basis. Otherwise, pick u₃span. Continue this way until you obtain a basis. The process must stop after fewer than n + 1 iterations because if it didn’t, then there would be a linearly independent set of more than n vectors which is impossible because there is a spanning set of n vectors from the above observation. ■

The following is a fundamental result.

Proof: If the dimension of V is m, then it has a basis of m vectors. It follows m ≤ n because if not, you would have an independent set of vectors which is longer than a spanning set of vectors

contrary to Theorem 6.1.6.

Next, if

is an independent set of vectors of V , then if it fails to span V, it must be there is a vector w which is not in this span. But then by Lemma 6.1.11, you could add w to the list of vectors and get an independent set of m + 1 vectors. However, the fact that V is of dimension m means there is a spanning set having only m vectors and so this contradicts Lemma 6.1.11. Thus must be a spanning set.

Finally, if k = m, the vectors

must span V since if not, you could add another vector which is not in this list to the list and get an independent set which is longer than a spanning set contrary to Theorem 6.1.6. Thus assume k < m. The set of vectors cannot span V because if it did, the dimension of V would be k not m. Thus there is a vector v_k+1 not in this span. Then by Lemma 6.1.11, is independent. If it spans V , stop. You have your basis. Otherwise, there is a v_k+2 not in the span and so you can add it in and get an independent set . Continue this process till it stops. It must stop since otherwise, you would be able to get an independent set of vectors larger than m which is the dimension of V, contrary to Theorem 6.1.6. ■

The row reduced echelon form for the above matrix is

( ) 1 0 − 5 − 7 − 3 |( 0 1 6 6 4 |) (6.2) 0 0 0 0 0

(6.2)

and so its rank is 2 because every column is in the span of the first two columns. You can think of the above as the row reduced version of several systems of equations, those which have the following augmented matrices.

( ) ( ) ( ) | 1 0 − 5 | | 1 0 − 7 | | 1 0 − 3 | ( 1 1 1 ) ,( 1 1 − 1 ) ,( 1 1 1 ) , 0 1 6 0 1 6 0 1 4

In each case, you can obtain the third column as a linear combination of the first two. Thus the last three columns in 6.1 are linear combinations of the first two columns in 6.1. Therefore, any linear combination of the columns of 6.1 can also be written as a linear combination of the first two columns of 6.1. In other words, the span of the columns of 6.1 equals the span of the first two columns of 6.1. Also, from 6.2, we can see that the first two columns of 6.1 are independent. Therefore, these columns are a basis for Im

.

Similar considerations apply to determining whether some vectors are independent. Remember the definition. To determine whether some vectors are independent, make them the columns of a matrix A and determine the solution set to Ax = 0. If there is only the zero solution, then the vectors are independent. If there are more solutions then these vectors are not independent.