44.5© 2012 Pearson Education, Inc.Vector SpacesTHE DIMENSION OF A VECTOR SPACESlide 4.5- 2© 2012 Pearson Education, Inc.DIMENSION OF A VECTOR SPACE Theorem 9: If a vector space V has a basis  , then any set in V containing more than n vectors must be linearly dependent. Proof: Let {u1, …, up} be a set in V with more than n vectors. The coordinate vectors [u1], …, [up]form a linearly dependent set in , because there are more vectors (p) than entries (n) in each vector.1{b ,...,b}nnSlide 4.5- 3© 2012 Pearson Education, Inc.DIMENSION OF A VECTOR SPACE So there exist scalars c1, …, cp, not all zero, such that Since the coordinate mapping is a linear transformation,110u... u0ppcc BBThe zero vector inn110u... u0ppcc BSlide 4.5- 4© 2012 Pearson Education, Inc.DIMENSION OF A VECTOR SPACE The zero vector on the right displays the n weights needed to build the vector from the basis vectors in . That is, . Since the ciare not all zero, {u1, …, up} is linearly dependent. Theorem 9 implies that if a vector space V has a basis  , then each linearly independent set in V has no more than n vectors.11u ... uppcc11 1u... u 0b...0b 0pp ncc1{b ,...,b}nSlide 4.5- 5© 2012 Pearson Education, Inc.DIMENSION OF A VECTOR SPACE Theorem 10: If a vector space V has a basis of nvectors, then every basis of V must consist of exactly nvectors. Proof: Let 1be a basis of n vectors and 2be any other basis (of V).  Since 1is a basis and 2is linearly independent, 2has no more than n vectors, by Theorem 9. Also, since 2is a basis and 1is linearly independent, 2has at least n vectors. Thus 2consists of exactly n vectors.Slide 4.5- 6© 2012 Pearson Education, Inc.DIMENSION OF A VECTOR SPACE Definition: 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 zero. If V is not spanned by a finite set, then V is said to be infinite-dimensional. Example 1: Find the dimension of the subspace3654:,,, in 25abcadH abcdbcddSlide 4.5- 7© 2012 Pearson Education, Inc.DIMENSION OF A VECTOR SPACE H is the set of all linear combinations of the vectors, , , Clearly, , v2is not a multiple of v1, but v3is a multiple of v2.  By the Spanning Set Theorem, we may discard v3and still have a set that spans H.115v00230v10360v20404v151v0Slide 4.5- 8© 2012 Pearson Education, Inc.SUBSPACES OF A FINITE-DIMENSIONAL SPACE Finally, v4is not a linear combination of v1and v2. So {v1, v2, v4} is linearly independent and hence is a basis for H. Thus dim . 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 and3Hdim dimHVSlide 4.5- 9© 2012 Pearson Education, Inc.SUBSPACES OF A FINITE-DIMENSIONAL SPACE Proof: If , then certainly . Otherwise, let be any linearly independent set in H. If S spans H, then S is a basis for H. Otherwise, there is some in H that is not in Span S.{0}Hdim 0 dimHV1{u ,...,u}kS1ukSlide 4.5- 10© 2012 Pearson Education, Inc.SUBSPACES OF A FINITE-DIMENSIONAL SPACE But then will be linearly independent, because no vector in the set can be a linear combination of vectors that precede it (by Theorem 4). So long as the new set does not span H, we can continue this process of expanding S to a larger linearly independent set in H. But the number of vectors in a linearly independent expansion of S can never exceed the dimension of V, by Theorem 9.11{u ,...,u ,u}kkSlide 4.5- 11© 2012 Pearson Education, Inc.THE BASIS THEOREM So eventually the expansion of S will span H and hence will be a basis for H, and . Theorem 12: Let V be a p-dimensional vector space,. Any linearly independent set of exactly pelements in V is automatically a basis for V. Any set of exactly p elements that spans V is automatically a basis for V.  Proof: By Theorem 11, a linearly independent set Sof p elements can be extended to a basis for V.dim dimHV1p Slide 4.5- 12© 2012 Pearson Education, Inc.THE BASIS THEOREM But that basis must contain exactly p elements, since dim . So S must already be a basis for V. Now suppose that S has p elements and spans V. Since V is nonzero, the Spanning Set Theorem implies that a subset of S is a basis of V. Since dim , must contain p vectors. Hence . VpSVpSSSSlide 4.5- 13© 2012 Pearson Education, Inc.THE DIMENSIONS OF NUL A AND COL A Let A be an matrix, and suppose the equation has k free variables. A spanning set for Nul A will produce exactly klinearly independent vectors—say, —one for each free variable. So is a basis for Nul A, and the number of freevariables determines the size of the basis.mnx0A1u ,...,uk1{u ,...,u}kSlide 4.5- 14© 2012 Pearson Education, Inc.DIMENSIONS OF NUL A AND COL A Thus, the dimension of Nul A is the number of free variables in the equation , and the dimension of Col A is the number of pivot columns in A. Example 2: Find the dimensions of the null space and the column space of x0A361171223124584A Slide 4.5- 15© 2012 Pearson Education, Inc.DIMENSIONS OF NUL A AND COL A Solution: Row reduce the augmented matrix to echelon form: There are three free variable—x2, x4and x5. Hence the dimension of Nul A is 3. Also dim Col because A has two pivot