LECTURE 6: VECTOR SPACES II (CHAPTER 3 IN THE BOOK)

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1 LECTURE 6: VECTOR SPACES II (CHAPTER 3 IN THE BOOK) In this lecture, F is a fixed field. One can assume F = R or C. 1. More about the spanning set 1.1. Let S = { v 1, v n } be n vectors in V, we have defined last time: (1) Span(S) = Span(v 1,, v n ) = { n i=0 α i v i α i F }. Note that Span(S) is the smallest vector subspace in V contains S (can serve as an altenative definition of span. (2) S is called a spanning set of V if Span(S) = V. (3) Linearly independent: all vectors in Span(S) can be written as a linear combination of vectors in S in a unique way. Example 1.1. If v 1 is a linear combination of v 2, v 3,..., v n, that is, v 1 = c 2 v 2 + c 3 v c n v n. Then the vectors v 1, v 2,..., v n are always linearly dependent. To see this we rewrite the above equation as (c 1 = 1) 0 = ( 1)v 1 + c 2 v 2 + c 3 v c n v n. Lemma 1.2. The following statements are equivalent for a set of vectors { v 1, v 2,..., v n }: (i) v 1, v 2,..., v n are linearly dependent. (ii) One of the vectors, say v j, is a linear combination of the rest of the vectors. (iii) Span(v 1,..., v n ) = Span(v 1,..., /v j,..., v n ). Proof. (i) (ii) If v 1,..., v n are linearly dependent, that is, where some c j are nonzero. Then c 1 v 1 + c 2 v c n v n = 0 v j = c 1 j (c 1 v /v j + + c n v n ) = ( c 1 j c i )v i. Therefore v j is represented by a linear combination of the rest of vectors. (ii) (iii) Clearly Span(v 1,, v n ) Span(v 1,..., /v j,..., v n ). We may assume (1) v j = c i v i. For any vector v = i α i v i Span(v 1,, v n ). Substitute (1) into above equation, we get v =α 1 v α j ( c i v i ) + + α n v n = (α i + α j c i )v i Therefore, Span(v 1,, v n ) Span(v 1,..., /v j,..., v n ). Now the claim follows. (iii) (i) Since v j Span(v 1,, v n ). By the assumption, we have v j Span(v 1,..., /v j,..., v n ), i.e. v j is a linear combination of rest vectors. The claim follows from the obervation in Example

2 2 LECTURE 6: VECTOR SPACES II (CHAPTER 3 IN THE BOOK) Definition 1.3. A vector space V is said to be finite dimesional if there is a finite set of vectors that span V ; otherwise, we say that V is infinite dimesional. Example 1.4. (1) F n, Mat m n (F ), a null space N(A) of a matrix and the space P n of polynormials with degree less than n are finite dimension spaces. (2) the space of all polynomials F [x] and the space C n ([a, b]) of n-th differentiable functions on a interval [a, b] are not finite dimension spaces (they are infinite dimension spaces) Form now on, assume that V is a finite dimensinoal vector space. Let S = {v 1, v 2,..., v n } be a finite spanning set, that is, Span(S) = V. If v 1, v 2,..., v n are linearly dependent, then by Lemma 1.2 (iii) we can delete one vector from S such that V = Span(v 1,..., v n ) = Span(v 1,..., /v j,..., v n ). We get a smaller spanning set S {v j } = {v 1,..., /v j,..., v n }. In other words, v j is redundant as a spanning vector. If v 1,..., /v j,..., v n are linearly dependent, then we can again delete another vector from the spanning set. We can keep doing this and we get a smaller and smaller spanning set. Eventually this process will stop when all the vectors in the spanning set are linearly independent. We call this final spanning set S a minimal spanning set. We cannot get a smaller spanning set because otherwise by Lemma 1.2, the vectors in S are linearly dependent. Conclusion: Let V be a finite dimensional vector space, then there are vectors v 1,, v n such that V = Span(v 1,..., v n ) and v 1,..., v n are linearly independent. The set is called a minimal spanning set. S = {v 1,..., v n } 2. Basis and Dimension Use minimal spanning set to describe a vector space. Definition 2.1. The vectors v 1, v 2,, v n form a basis for a vector space V if (1) v 1,, v n are linearly independent. (2) v 1,, v n span V. In other word, a basis of V is a minimal spanning set of V. Example 2.2. (1) the standard basis e 1,, e n is a basis of R n But, there are much more other basis of R n (2) 1, x, x 2,, x n is a basis of P n. 1, x, x(x 1), x(x 1)(x 2),, x(x 1)(x 2) (x (n 1)) is also a basis of P n. (3) Let E ij be the matrix with (i, j)-th entry 1 and other entries are 0. Then { E ij 1 i m, 1 j n } form a basis of Mat m,n (F ). Using the method in Section 1.2, we get: Proposition 2.3. Let V be a finite dimensional vector space V with a finite spanning set S, that is V = Span(S). Then there exists a subset S of S such that S is a minimal spanning set of V, i.e. S is a basis of V.

3 LECTURE 6: VECTOR SPACES II (CHAPTER 3 IN THE BOOK) In this section, we show that the number of elements of a basis is an invariant of the vector space, i.e. independent of the choice of basises. Theorem 2.4. If { v 1,, v n } is a spanning set for a vector space V, then any collection of m vectors in V with m > n is linearly dependent. Proof. Let u 1,, u m be m vectors in V. Since { v 1,, v n } is a spanning set, there exits a ij F such that u j = a ij v i i Let A = (a ij ) be the corresponding n m matrix. Now j c j u j = j c j i a ij v i = i ( j a ij c j )v i. j a ij c j = 0 gives a system linear equation, A c = 0. It is a underdetermined system, so it has a non-trivial solution, say ĉ 1, ĉ m. Hence k ĉ j u j = 0 for ĉ 1,, ĉ m which are not all zeros, i.e. u 1,, u m is linearly dependent. Corollary 2.5. If S 1 = { v 1,, v n } and S 2 = { u 1,, u m } are two basis of V, then n = m. Proof. S 1 spanning and S 2 linearly independent implies n m by the above theorem. Silmilary, S 2 spanning + S 1 linearly independent implies n m. Hence n = m. Definition 2.6. If V has a basis consisting of n vectors, we say that V has dimension n. The zero vector space is said to have dimension zero. Example 2.7. (1) F n, P n 1 have dimension n. (2) Mat m n (F ) has dimension mn How to find basis? Lemma 2.8. Let S = { v 1,..., v m } be a set of m linearly independent vectors in a vector space V. Suppose V Span(S) and let v / Span(S), then the m + 1 vectors are linearly independent. v 1, v 2,..., v m, v Proof. Suppose v 1, v 2,..., v m, v are linearly dependent, then c 1 v c m v m + cv = 0 where not all c i, c are zero scalars. If c = 0, then c 1 v c m v m = 0 where not all c i s are zero. This means that v 1,..., v n are linear dependent. This contradicts our assumption. On the other hand if c 0 then v = ( c 1 c 1 )v 1 + ( c 1 c 2 )v ( c 1 c m )v m so v Span(S). This contradicts the fact that v / Span(S). We conclude that are linearly independent. v 1, v 2,..., v m, v Lemma 2.9. Let V be a vector space of dimension n, then (i) If m < n, no set of m-vectors can span V. (ii) If m > n, no set of m vectors is linearly independent. (iii) any n vectors that span V are linearly independent, i.e. form a basis. (iv) any set of n linearly independent vectors span V.

4 4 LECTURE 6: VECTOR SPACES II (CHAPTER 3 IN THE BOOK) Proof. (i) Otherwise, any set of n vectors will be linearly dependent by Theorem 2.4. This contradicts to that V has dimension n. (ii) Let S be a basis of V, then S is a set of n-vectors spanning V. So any m vectors with m > n must be linearly dependent by Theorem 2.4. (iii) Suppose a set S of n-vectors are linearly dependent and spanning V, by procedure in Section 1.2, one can find a proper subset S with m such that m < n, and S is a basis of V. This contradicts the assumption. (iv) Suppose S = { v 1,, v n } are linearly independent but not spanning V, then there is v V Span(S). Hence the set of n + 1 vectors S = { v 1,, v n, v } is linearly independent by Lemma 2.8. On the other hand, S must be linearly dependent by (ii). We get a contradiction. Therefore, S must span V. Corollary A set of vectors { v 1,, v n } form a basis in F n if and only if A is invertiable (or equivalently det(a) 0) where A = (v 1 v 2 v n ). Example The following set if a basis for R , By Lemma 2.9 (iv), we need only show that they are linearly independent. This follows from that the follow determinant is nonzero: = Theorem (Going Down and Going Up) Let V be a vector space of dimension n. (i) If S is a spanning set containing more than n vectors, then there is a subset S of S with n-vectors such that S is a basis of V. (ii) If S is a set of m linearly independent vectors with m < n, then we can add n m vectors to S so as to form a basis of V. Proof. (i) Use the procedure in Section 1.2 to find S. Note that S must consist exactly n-vectors. (ii) Use Lemma 2.8 to extand S: Suppose S = { v 1,, v m }. We note that Span(S) V because S is not a basis (c.f. Lemma 2.9 (i)). Therefore we find a nonzero v m+1 / Span(S) and by Lemma 2.8, v 1, v 2,..., v m, v m+1 are linearly independent. If these m + 1 vectors spans V, we stop. If not, repeat the process until we get a spanning set consisting of linearly independent vectors. This will form a basis. There is one danger. The process may never stop. This cannot happen because any n + 1 vectors must be linearly dependent. (So this process stops when we get n-vectors in fact.) 3. Change of basis Definition 3.1. Let V be a vector space and let B = { v 1,, v n } be an ordered basis for V. If v is any element of V, then v can be written in the form v = c 1 v 1 + c 2 v 2 + c n v n. where c 1, c 2,, c n are scalars. Thus, we can associate with each vector v a unique vector c = (c 1, c 2,, c n ) T in R n. The vector c defined in this way is called the coordinate vector of v with respect to the ordered basis B and is denoted [v] B. The c i s are called the coordinates of v relative to B.

5 LECTURE 6: VECTOR SPACES II (CHAPTER 3 IN THE BOOK) 5 Let B = { u 1,, u n } be another basis of V. We want to find a relation between [v] B and [v] B. Since that B is a basis, we have scalars s ij F such that v j = s ij u i = s 1j u 1 + s 2j u 2 + s nj u n. i Let S = (s ij ) be the corresponding n n-matrix. Then [v] B = S[v] B. Let x = (x 1,, x n ) T = [v] B. The following diagram clarifiy the situation: x 1 x 2 v = (v 1 v 2 v n ) x n x 1 x 2 = ( i u i s i1 i u i s i2 i u i s in ) x n s 11 s 12 s 1n x 1 s = (u 1 u 2 u n ) 21 s 22 s 2n x 2 s n1 s n2 s nn x n = (u 1 u 2 u n ) S[v] B Clearly S is an invertiable matrix. (Since one also can represent u j by v i, this will give a matrix S. It is clear that S S = I n ) Definition 3.2. The matrix S is called the transition matrix from basis B to B. We have [v] B = S[v] B and S 1 [v] B = [v] B, i.e. S 1 is the transiton matrix from basis B to basis B. Example 3.3. Let b 1 = (1, 1) T and b 2 = ( 2, 3) T. Find the transition matrix form { e 1, e 2 } to { b 1, b 2 }. Find the coordinates of x = (1, 2) T with respect to { b 1, b 2 }. The transition matrix from { b 1, b 2 } to { e 1, e 2 } is U = ( ). Hence The transition matrix from { e 1, e 2 } to { b 1, b 2 } is S = U 1 = ( ). Now [x] { b1,b 2 } = S[x] { e1,e 2 } = ( ) (1 2 ) = (7 3 ).

6 6 LECTURE 6: VECTOR SPACES II (CHAPTER 3 IN THE BOOK) 4. Row Space and Column Space Definition 4.1. Let A be an m n matrix (1) The row space is defined as the span of the m row vectors of A. It is a vector subspace in F n. (2) The column space is defined as the span of the n column vectors of A. It is a vector subspace in F m. Theorem 4.2. Two row equivalent matrices have the same row space. Proof. Let A and B are two row equivelant matrices with rows a i and b i respectively. Then there exists a matrix M = (m ij ) such that B = MA. We get b i = m ij a i. j Hence Span( b 1,, b n ) Span( a 1,, a n ). Reverse the role of A and B we will get Span( b 1,, b n ) Span( a 1,, a n ). Hence, the row space of A and B are the same. Definition 4.3. The rank of a matrix A, denote rank(a), is the dimension of the row space of A. The dimension of the null space of A is called the nullity of the matrix. Example 4.4. Find the row space of A = The row echelon form of A is U = Therefore, (1, 2, 3) and (0, 1, 5) form a basis of the row space of U, which is also the row space of A by Theorem 4.2. Now A has rank 2. We recollect a result already discussed in previous lectrues. Theorem 4.5 (Consistency Theorem for Linear Systems). A linear system Ax = b is consistent if and only if b is in the column space of A. Theorem 4.6. Let A be an m n matrix. (i) The linear system Ax = b is consistent for every b F m if and only if the column vectors of A span F m. (ii) The system Ax = b has at most one solution for every b F m if and only if the column vectors of A are linearly independent. Proof. (i) Clear. (ii) Ax = b has at most one solution is equivalent to Ax = 0 has only trivial solution (by disscusions in previous lectures). But this is equivalent to the column space is linearly independent. Corollary 4.7. An n n matrix A is nonsingular if and only if the column vectors of A form a basis for F n. Theorem 4.8 (The Rank-Nullity Theorem). If A is an m n matrix,then the rank of A plus the nullity of A equals n.

7 LECTURE 6: VECTOR SPACES II (CHAPTER 3 IN THE BOOK) 7 Proof. Let U be the reduced row echelon form of A. The system Ax = bfzero is equivalent to the system Ux = 0. If rank(a) = r, then U will have r nonzero rows, and consequently the system Ux = 0 will involve r lead variables and n r free variables. The dimension of N(A) will equal the number of free variables. Recall a result Lemma 4.9. Let V be any subspace in F n. Let S = { v 1,, v m } be a finite set of vectors in V. Suppose M is an invertiable matrix. Then S is linearly independent if and only if { Mv 1,, Mv n } is linearly independent. Proof. Let A = (v 1,, v m ) and U = MA. This is follows from that Ax = 0 has only trivial solution if and only if Ux = 0 has only trivial solution. Theorem If A is an m n matrix, the dimension of the row space of A equals the dimension of the column space of A. Proof. Let U be the reduced row echelon form of A. Suppose it have r leading 1 s. The columns of U corresponding to the lead 1s will be linearly independent. By Lemma 4.9, the r columns in A at same positions of the leading 1 s in U are also linearly independent. Therefore, we get dim(column space of A) r = dim(row space of A). Applying this result to the matrix A T, we see that dim(row space of A) = dim(column space of A T ) dim(row space of A T ) = dim(column space of A) Thus, the dimension of the row space of A must equal the dimension of the column space of A. Example Find a basis and the dimension of the subspace of R 4 spanned by x 1 =, x 1 2 =, x 3 3 =, x 2 4 = Solution The subspace Span(x 1, x 2, x 3, x 4 ) is the same as the column space of the matrix X = The row echelon form of X is Therefore, the first two columns x 1 and x 2 of X will form a basis for the column space of X and dim Span(x 1, x 2, x 3, x 4 ) = 2.

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