Math 2331 Linear Algebra

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1 4.5 The Dimension of a Vector Space Math 233 Linear Algebra 4.5 The Dimension of a Vector Space Shang-Huan Chiu Department of Mathematics, University of Houston schiu@math.uh.edu math.uh.edu/ schiu/ Shang-Huan Chiu, University of Houston Math 233, Linear Algebra Fall, 27 / 4

2 4.5 The Dimension of a Vector Space The Dimension of a Vector Space: Theorems The Dimension of a Vector Space: Definition The Dimension of a Vector Space: Example Dimensions of Subspaces of R 3 Dimensions of Subspaces: Theorem The Basis Theorem Dimensions of Col A and Nul A: Examples Shang-Huan Chiu, University of Houston Math 233, Linear Algebra Fall, 27 2 / 4

3 The Dimension of a Vector Space: Theorems Theorem (9) If a vector space V has a basis β = {b,..., b n }, then any set in V containing more than n vectors must be linearly dependent. Proof: Suppose {u,..., u p } is a set of vectors } in V where p > n. Then the coordinate vectors {[u ] β,, [u p ] β are in R n. } Since p > n, {[u ] β,, [u p ] β are linearly dependent and therefore {u,..., u p }are linearly dependent. Shang-Huan Chiu, University of Houston Math 233, Linear Algebra Fall, 27 3 / 4

4 The Dimension of a Vector Space: Theorems (cont.) Theorem () If a vector space V has a basis of n vectors, then every basis of V must consist of n vectors. Proof: Suppose β is a basis for V consisting of exactly n vectors. Now suppose β 2 is any other basis for V. By the definition of a basis, we know that β and β 2 are both linearly independent sets. By Theorem 9, if β has more vectors than β 2, then linearly dependent set (which cannot be the case). is a Again by Theorem 9, if β 2 has more vectors than β, then a linearly dependent set (which cannot be the case). is Therefore β 2 has exactly n vectors also. Shang-Huan Chiu, University of Houston Math 233, Linear Algebra Fall, 27 4 / 4

5 The Dimension of a Vector Space: Definition Dimension of a Vector Space If V is spanned by a finite set, then V is said to be finite-dimensional, and the dimension of V, written as dim V, is the number of vectors in a basis for V. The dimension of the zero vector space {} is defined to be. If V is not spanned by a finite set, then V is said to be infinite-dimensional. Example The standard basis for P 3 is { }. So dim P 3 =. Example In general, dim P n = n +. The standard basis for R n is {e,..., e n } where e,..., e n are the columns of I n. So, for example, dim R 3 = 3. Shang-Huan Chiu, University of Houston Math 233, Linear Algebra Fall, 27 5 / 4

6 The Dimension of a Vector Space: Example Example Find a basis and the dimension of the subspace a + b + 2c W = 2a + 2b + 4c + d b + c + d : a, b, c, d are real. 3a + 3c + d Solution: Since a + b + 2c 2a+2b+4c+d b + c + d 3a + 3c + d = a b 2 + c d Shang-Huan Chiu, University of Houston Math 233, Linear Algebra Fall, 27 6 / 4

7 The Dimension of a Vector Space: Example (cont.) where v = 2 3 W =span{v, v 2, v 3, v 4 }, v 2 = 2, v 3 = 2 4 3, v 4 =. Note that v 3 is a linear combination of v and v 2, so by the Spanning Set Theorem, we may discard v 3. v 4 is not a linear combination of v and v 2. a basis for W. Also, dim W =. So {v, v 2, v 4 } is Shang-Huan Chiu, University of Houston Math 233, Linear Algebra Fall, 27 7 / 4

8 Dimensions of Subspaces of R 3 Example (Dimensions of subspaces of R 3 ) -dimensional subspace contains only the zero vector = (,, ). 2 -dimensional subspaces. Span{v} where v is in R 3. 3 These subspaces are through the origin. 4 2-dimensional subspaces. Span{u, v} where u and v are in R 3 and are not multiples of each other. 5 These subspaces are through the origin. 6 3-dimensional subspaces. Span{u, v, w} where u, v, w are linearly independent vectors in R 3. This subspace is R 3 itself because the columns of A = [u v w] span R 3 according to the IMT. Shang-Huan Chiu, University of Houston Math 233, Linear Algebra Fall, 27 8 / 4

9 Dimensions of Subspaces: Theorem Theorem () Let H be a subspace of a finite-dimensional vector space V. Any linearly independent set in H can be expanded, if necessary, to a basis for H. Also, H is finite-dimensional and Example Let H =span, dim H dim V.. Then H is a subspace of R3 and dim H < dim R 3. We could expand the spanning set, to,, for a basis of R3. Shang-Huan Chiu, University of Houston Math 233, Linear Algebra Fall, 27 9 / 4

10 The Basis Theorem Theorem (2 The Basis Theorem) Let V be a p dimensional vector space, p. Any linearly independent set of exactly p vectors in V is automatically a basis for V. Any set of exactly p vectors that spans V is automatically a basis for V. Example Show that { t, t, + t t 2} is a basis for P 2. Solution: Let v = t, v 2 = t, v 3 = + t t 2 and β = {, t, t 2}. Corresponding coordinate vectors [v ] β =, [v 2 ] β =, [v 3 ] β = Shang-Huan Chiu, University of Houston Math 233, Linear Algebra Fall, 27 / 4

11 The Basis Theorem (cont.) [v 2 ] β is not a multiple of [v ] β [v 3 ] β is not a linear combination of [v ] β and [v 2 ] β } = {[v ] β, [v 2 ] β, [v 3 ] β is linearly independent and therefore {v, v 2, v 3 } is also linearly independent. Since dim P 2 = 3, {v, v 2, v 3 } is a basis for P 2 according to The Basis Theorem. Shang-Huan Chiu, University of Houston Math 233, Linear Algebra Fall, 27 / 4

12 Dimensions of Col A and Nul A: Example Recall our techniques to find basis sets for column spaces and null spaces. Example Suppose A = [ Solution [ ]. Find dim Col A and dim Nul A. ] [ So {[ ] [ ]}, is a basis for Col A and dim Col A = 2. ] Shang-Huan Chiu, University of Houston Math 233, Linear Algebra Fall, 27 2 / 4

13 Dimensions of Col A and Nul A: Example (cont.) Now solve Ax = by row-reducing the corresponding augmented matrix. Then we arrive at [ ] [ 2 4 ] x = 2x 2 4x 4 x 3 = x x 2 x 3 x 4 = x x 4 4 Shang-Huan Chiu, University of Houston Math 233, Linear Algebra Fall, 27 3 / 4

14 Dimensions of Col A and Nul A: Example (cont.) So 2, 4 is a basis for Nul A and dim Nul A = 2. Note dim Col A = number of pivot columns of A dim Nul A = number of free variables of A. Shang-Huan Chiu, University of Houston Math 233, Linear Algebra Fall, 27 4 / 4