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 2 V a sum u+v 2 V and to any v 2 V and a 2 R a product av 2 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 2 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 2 V. 3. For each vector u 2 V there exists a vector in V, denoted by −u, for which u + (−u) = 0. 4. For any vectors u;v 2 V, u + v = v + u. 5. For any scalar a 2 R and any vectors u;v 2 V, a(u + v) = au + av. 6. For any scalars a;b 2 R and any vector v 2 V, (a + b)v = av + bv. 7. For any scalars a;b 2 R and any vector v 2 V, (ab)v = a(bv). 8. For the scalar 1 and any vector v 2 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 2 V and scalar a 2 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 2 W, v + w 2 W. 2. W is closed under scalar multiplication, that is: For any scalar a 2 R and any vector w 2 W, aw 2 W. Note that a subspace is itself a vector space. Furthermore, if W is a subspace, then 0w = 0 2 W. That is, if 0 2= 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 2 W. – Let x be a vector in W and a 2 R. Then 2x1 − x2 = 0. Consequently, 2(ax1) − (ax2) = a(2x1 − x2) = 0, which implies that ax 2 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 = fv + w : v 2 V;w 2 Wg: 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 2 V +W: | {z } | {z } 2V 2W – Let v + w be a vector in V +W and a 2 R. Then a(v + w) = av + aw 2 V +W: |{z} |{z} 2V 2W n×n T • Recall that a matrix A 2 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 2 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 2 R defined by m n R (A) = fAx 2 R : x 2 R g m is a subspace of R . Proof: n – Let y;z 2 R (A). By definition of the column space, there exist vectors x;v 2 R such that y = Ax and z = Av. Consequently, y + z = Ax + Av = A(x + v), which implies that y + z 2 R (A). n – Let y 2 R (A) and a 2 R. By definition of the column space, there exists a vector x 2 R such that y = Ax. Hence ay = aAx = A(ax), which implies that ay 2 R (A). m×n • The null space of A 2 R defined by n N (A) = fx 2 R : Ax = 0g n is a subspace of R . Proof: – Let x;v 2 N (A). Then using the definition of N (A), A(x + v) = Ax + Av = 0 + 0 = 0, which implies that x + v 2 N (A). – Let x 2 N (A) and a 2 R. A(ax) = aAx = a0 = 0, which implies that ax 2 N (A). 3.