Subspaces Definition and Examples Let V be a vector space and let S be a subset of V such that S is a vector space with the same + and * from V. Then S is called a subspace of V. Remark: Every vector space V contains at least two subspace, namely V and the set {0}.
Example Let V be the vector space R^{3} and let S be the set of points that lie on the plane z = x  y Then S is a subspace of V. This is true since S is closed under + and *. A point belonging to S has the form (x, y, x  y) If (x_{1}, y_{1}, x_{1}  y_{1}) and (x_{2}, y_{2}, x_{2}  y_{2}) are in S then (x_{1}, y_{1}, x_{1}  y_{1}) + (x_{2}, y_{2}, x_{2}  y_{2}) = (x_{1} + x_{2}, y_{1}+ y_{2}, x_{1}  y_{1} + x_{2}  y_{2}) = (x_{1} + x_{2}, y_{1}+ y_{2}, (x_{1} + x_{2})  (y_{1} + y_{2})) is in S. We also have c(x_{1}, y_{1}, x_{1}  y_{1}) = (cx_{1}, cy_{1}, cx_{1}  cy_{1}) is in S. The rest of the properties follow immediately since they are true in V. In fact, the two closure properties are all we need to show when we want to check that any subspace S is a subspace of any vector space V. Theorem Let V be a vector space and S be a subset of V. If S is closed under + and * then S is a subspace of V. The proof of this theorem only involves noticing that the properties are all true in S by virtue of being true in V.
Example Set V be the vector space of all differentiable functions and let S be the subset of V such that for any f in V f '(0) = 1 Then S is not a subspace of V since if f and g are in S, then (f + g)'(0) = f '(0) + g'(0) = 1 + 1 = 2 Hence S is not closed under +. Example Let S be the subset of M^{2x2} of trace 0, that is the sum of the diagonal entries is zero. Then S is a subspace of M^{2x2}. Elements of S have the form
so if
then
has zero trace. And
also has trace zero. Hence S is closed under + and *. We can conclude that S is a subspace of V. The Range and Null Space of a Matrix There are two subspaces that deserve special attention. Recall that an m x n matrix A is associated with a linear transformation from R^{n} > R^{m}. We defined subspaces of R^{n} and R^{m} as follows. Definition Let A be an m by n matrix. Define the null space of A to be the subspace consisting of vectors x in R^{n} with the property Ax = 0 and define the range of A to be the subspace consisting of vectors y in R^{m} such that there is a b in R^{n} with Ab = y
To prove that the null space and range are subspaces of R^{n} and R^{m} respectively, we need to show that closure is satisfied. We will show this for the null space and leave the range as an exercise. Let u and v be elements of the null space. Then A(u + v) = Au + Av = 0 + 0 = 0 and A(cu) = cAu = c* 0 = 0 hence the null space is a subspace of R^{n}. Linear Combinations and Span Example Suppose that V = P_{2} and let f(t) = t^{2}  t and g(t) = t + 1 Let S be the subset of P_{2} that consists of all polynomials of the form c_{1}f + c_{2}g where c_{1} and c_{2} are constants. Then S is called the span of f and g and is a subspace of P_{2}. We will show closure under *. Closure under + is also not difficult to show and is left to the reader to check. Let u = c_{1}f + c_{2}g and c be a constant. Then cu = cc_{1}f + cc_{2}g = af + bg so is in S. In general, if V is a vector space and S = {v_{1},v_{2}, ...,v_{n}} is a subset of V, then we call c_{1}v_{1} + c_{2}v_{2} + ... + c_{n}v_{n} a linear combination of S. The set of all linear combinations of S is called the span of S and is a subspace of V.
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