Vector Spaces and Subspaces CAAM 335: Matrix Analysis (updated September 17, 2010)
Definition [Vector Space] Let V be a nonempty set with rules for addition and scalar multiplica- tion which assigns to any u,v ∈ V a sum u+v ∈ V and to any v ∈ V and α ∈ R a product αv ∈ V. This set V is called a (real) vector space (and the elements of V are called vectors) if the following properties hold:
1. For any vectors u,v,w ∈ V, (u + v) + w = u + (v + w).
2. There is a vector in V denoted by 0 and called the zero vector, for which u + 0 = u for any vector u ∈ V.
3. For each vector u ∈ V there exists a vector in V, denoted by −u, for which u + (−u) = 0.
4. For any vectors u,v ∈ V, u + v = v + u.
5. For any scalar α ∈ R and any vectors u,v ∈ V, α(u + v) = αu + αv.
6. For any scalars α,β ∈ R and any vector v ∈ V, (α + β)v = αv + βv.
7. For any scalars α,β ∈ R and any vector v ∈ V, (αβ)v = α(βv). 8. For the scalar 1 and any vector v ∈ V, 1v = v.
Examples of vector spaces:
n • The R is a vector space. m×n • The set of real m × n matrices, R , is a vector space.
Note that for each u ∈ V and scalar α ∈ R, • 0u = 0 Proof: 0u = (0 + 0)u = 0u + 0u. If we add (−0u) to both sides, we get 0 = 0u + (−0u) = 0u + 0u + (−0u) = 0u.
• (−1)u = −u Proof: u + (−1)u = (1 − 1)u = 0u = 0. Hence −u = (−1)u.
1 Definition [Subspace] Let V be a vector space. A subset W of V is called a subspace if
1. W is closed under vector addition, that is: For any vectors v,w ∈ W, v + w ∈ W.
2. W is closed under scalar multiplication, that is: For any scalar α ∈ R and any vector w ∈ W, αw ∈ W.
Note that a subspace is itself a vector space. Furthermore, if W is a subspace, then 0w = 0 ∈ W. That is, if 0 ∈/ W, then W cannot be a subspace.
Examples: x1 2 • Is W = : 2x1 − x2 = 0 a subspace of R ? x2 Yes. Proof:
– Let x,y be vectors in W. Then 2x1 − x2 = 0 and 2y1 − y2 = 0. Consequently, 2(x1 + y1) − (x2 + y2) = (2x1 − x2) + (2y1 − y2) = 0, which implies that x + y ∈ W.
– Let x be a vector in W and α ∈ R. Then 2x1 − x2 = 0. Consequently, 2(αx1) − (αx2) = α(2x1 − x2) = 0, which implies that αx ∈ W. x1 2 0 • Is W = : x1 − x2 = 1 a subspace of R ? No, since the zero vector satisfies x2 0 0 − 0 = 0 6= 1 and is not contained in W.
n • If V,W are subspaces of R , then their sum is defined as the set
V +W = {v + w : v ∈ V,w ∈ W}.
Is V +W a subspace? Yes.
– Let v1 + w1 and v2 + w2 be vectors in V +W. Then
(v1 + w1) + (v2 + w2) = v1 + v2 +w1 + w2 ∈ V +W. | {z } | {z } ∈V ∈W
– Let v + w be a vector in V +W and α ∈ R. Then
α(v + w) = αv + αw ∈ V +W. |{z} |{z} ∈V ∈W
n×n T • Recall that a matrix A ∈ R is called orthogonal if A A = I. Is the set of orthogonal n×n matrices a subspace (of the vector space R of n × n matrices)? n×n T No, because the zero matrix 0 ∈ R satisfies 0 0 = 0 6= I and therefore is not an orthogonal matrix.
2 Two important subspaces related to a matrix:
m×n • The column space (range space) of A ∈ R defined by
m n R (A) = {Ax ∈ R : x ∈ R }
m is a subspace of R . Proof:
n – Let y,z ∈ R (A). By definition of the column space, there exist vectors x,v ∈ R such that y = Ax and z = Av. Consequently, y + z = Ax + Av = A(x + v), which implies that y + z ∈ R (A). n – Let y ∈ R (A) and α ∈ R. By definition of the column space, there exists a vector x ∈ R such that y = Ax. Hence αy = αAx = A(αx), which implies that αy ∈ R (A). m×n • The null space of A ∈ R defined by
n N (A) = {x ∈ R : Ax = 0}
n is a subspace of R . Proof: – Let x,v ∈ N (A). Then using the definition of N (A), A(x + v) = Ax + Av = 0 + 0 = 0, which implies that x + v ∈ N (A). – Let x ∈ N (A) and α ∈ R. A(αx) = αAx = α0 = 0, which implies that αx ∈ N (A).
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