The Dimension of a Vector Space

4.5 The Dimension of a Vector Space Math 2331 – Linear Algebra 4.5 The Dimension of a Vector Space Math 2331, Linear Algebra 1 / 14 4.5 The Dimension of a Vector Space Dimension Basis Theorem 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 Math 2331, Linear Algebra 2 / 14 4.5 The Dimension of a Vector Space Dimension Basis Theorem The Dimension of a Vector Space: Theorems Theorem (9) If a vector space V has a basis β = {b1 , . . . , bn }, then any set in V containing more than n vectors must be linearly dependent. Proof: Suppose {u1 , . . . , up }nis a set of vectors o in V where p > n. Then the coordinate vectors [u1 ]β , · · · , [up ]β are in Rn . n o Since p > n, [u1 ]β , · · · , [up ]β are linearly dependent and therefore {u1 , . . . , up }are linearly dependent.  Math 2331, Linear Algebra 3 / 14 4.5 The Dimension of a Vector Space Dimension Basis Theorem The Dimension of a Vector Space: Theorems (cont.) Theorem (10) If a vector space V has a basis of n vectors, then every basis of V must consist of n vectors. Proof: Suppose β1 is a basis for V consisting of exactly n vectors. Now suppose β2 is any other basis for V . By the definition of a basis, we know that β1 and β2 are both linearly independent sets. By Theorem 9, if β1 has more vectors than β2 , then linearly dependent set (which cannot be the case). Again by Theorem 9, if β2 has more vectors than β1 , then a linearly dependent set (which cannot be the case). Therefore β2 has exactly n vectors also. is a is  Math 2331, Linear Algebra 4 / 14 4.5 The Dimension of a Vector Space Dimension Basis Theorem 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 {0} is defined to be 0. If V is not spanned by a finite set, then V is said to be infinite-dimensional. Example The standard basis for P3 is { . P3 = }. So dim In general, dim Pn = n + 1. Example The standard basis for Rn is {e1 , . . . , en } where e1 , . . . , en are the columns of In . So, for example, dim R3 = 3. Math 2331, Linear Algebra 5 / 14 4.5 The Dimension of a Vector Space Dimension Basis Theorem The Dimension of a Vector Space: Example Example Find a basis and the dimension of the subspace    a + b + 2c        2a + 2b + 4c + d  : a, b, c, d are real . W =    b+c +d       3a + 3c + d Solution: Since  a + b + 2c  2a+2b+4c+d   b+c +d 3a + 3c + d     1 1   2   2  = a +b   0   1 3 0    2     +c 4 +d   1   3 Math 2331, Linear Algebra   0 1   1  1 6 / 14 4.5 The Dimension of a Vector Space Dimension Basis Theorem The Dimension of a Vector Space: Example (cont.) W =span{v1 , v2 , v3 , v4 }    1 1  2   2   where v1 =   0  , v2 =  1 3 0     2 0   4   1  , v3 =   , v4 =    1   1 3 1   .  Note that v3 is a linear combination of v1 and v2 , so by the Spanning Set Theorem, we may discard v3 . v4 is not a linear combination of v1 and v2 . So {v1 , v2 , v4 } is a basis for W . Also, dim W = . Math 2331, Linear Algebra 7 / 14 4.5 The Dimension of a Vector Space Dimension Basis Theorem Dimensions of Subspaces of R 3 Example (Dimensions of subspaces of R 3 ) 1 0-dimensional subspace contains only the zero vector 0 = (0, 0, 0). 2 1-dimensional subspaces. Span{v} where v 6= 0 is in R3 . 3 These subspaces are 4 2-dimensional subspaces. Span{u, v} where u and v are in R3 and are not multiples of each other. 5 These subspaces are 6 3-dimensional subspaces. Span{u, v, w} where u, v, w are linearly independent vectors in R3 . This subspace is R3 itself because the columns of A = [u v w] span R3 according to the IMT. through the origin. through the origin. Math 2331, Linear Algebra 8 / 14 4.5 The Dimension of a Vector Space Dimension Basis Theorem Dimensions of Subspaces: Theorem Theorem (11) 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 dim H ≤ dim V . Example     1   1 Let H =span  0  ,  1  . Then H is a subspace of R3 and   0 0 dim H < dim R3 . We could expand the spanning set           1  1 0   1  1  0  ,  1  to  0  ,  1  ,  0  for a basis of R3 .     0 0 0 0 1 Math 2331, Linear Algebra 9 / 14 4.5 The Dimension of a Vector Space Dimension Basis Theorem The Basis Theorem Theorem (12 The Basis Theorem) Let V be a p− dimensional vector space, p ≥ 1. 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, 1 − t, 1 + t − t 2 is a basis for P2 . Solution: Let v1 = t, v2 = 1 − t, v3 = 1 + t − t 2 Corresponding coordinate vectors     0 1 [v1 ]β =  1  , [v2 ]β =  −1  , 0 0 and  β = 1, t, t 2 .  1 [v3 ]β =  1  −1 Math 2331, Linear Algebra  10 / 14 4.5 The Dimension of a Vector Space Dimension Basis Theorem The Basis Theorem (cont.) [v2 ]β is not a multiple of [v1 ]β [v3 ]β is not a linear combination of [v1 ]β and [v2 ]β n o =⇒ [v1 ]β , [v2 ]β , [v3 ]β is linearly independent and therefore {v1 , v2 , v3 } is also linearly independent. Since dim P2 = 3, {v1 , v2 , v3 } is a basis for P2 according to The Basis Theorem. Math 2331, Linear Algebra 11 / 14 4.5 The Dimension of a Vector Space Dimension Basis Theorem 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 So  1 2 3 4 . Find dim Col A and dim Nul A. 2 4 7 8  1 2 3 4 2 4 7 8    ∼ 1 2 3 4 0 0 1 0   ,   is a basis for Col A and dim Col A = 2. Math 2331, Linear Algebra 12 / 14 4.5 The Dimension of a Vector Space Dimension Basis Theorem Dimensions of Col A and Nul A: Example (cont.) Now solve Ax = 0 by row-reducing the corresponding augmented matrix. Then we arrive at  1 2 3 4 0 2 4 7 8 0   ∼ ··· ∼ 1 2 0 4 0 0 0 1 0 0  x1 = −2x2 − 4x4 x3 = 0       x1 −2 −4  x2   1   0         x3  = x2  0  + x4  0  x4 0 1 Math 2331, Linear Algebra 13 / 14 4.5 The Dimension of a Vector Space Dimension Basis Theorem Dimensions of Col A and Nul A: Example (cont.) So     −2 −4        0  1  ,   0   0       0 1 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 . Math 2331, Linear Algebra 14 / 14