Subspaces of lRn
Vector Spaces
CS 130: Mathematical Methods in Computer Science Vector Spaces and Linear Systems Nestine Hope S. Hernandez Algorithms and Complexity Laboratory Department of Computer Science University of the Philippines, Diliman nshernandez@dcs.upd.edu.ph Day 5
Subspaces of lRn
Vector Spaces
Vector Spaces and Linear Systems
Vector Spaces Basis and Dimension
Subspaces of lRn Row(Column) Space of a Matrix Orthonormal Bases in lRn
Subspaces of lRn
Vector Spaces
Vector Spaces and Linear Systems
Vector Spaces Basis and Dimension
Subspaces of lRn Row(Column) Space of a Matrix Orthonormal Bases in lRn
Subspaces of lRn
Vector Spaces
Basis
The vectors v~1 , v~2 , · · · , v~k in a vector space V are said to form a basis for V if (a) v~1 , v~2 , · · · , v~k span V and (b) v~1 , v~2 , · · · , v~k are linearly independent.
Subspaces of lRn
Vector Spaces
Basis
The vectors v~1 , v~2 , · · · , v~k in a vector space V are said to form a basis for V if (a) v~1 , v~2 , · · · , v~k span V and (b) v~1 , v~2 , · · · , v~k are linearly independent.
Example Show that S = {t2 + 1, t − 1, 2t + 2} is a basis for the vector space P2 .
Subspaces of lRn
Vector Spaces
Basis
The vectors v~1 , v~2 , · · · , v~k in a vector space V are said to form a basis for V if (a) v~1 , v~2 , · · · , v~k span V and (b) v~1 , v~2 , · · · , v~k are linearly independent.
Example Show that S = {t2 + 1, t − 1, 2t + 2} is a basis for the vector space P2 . If S = {v~1 , v~2 , · · · , v~k } is a basis for a vector space V , then every vector in V can be written in one and only one way as a linear combination of the vectors in S.
Subspaces of lRn
Vector Spaces
Basis
The vectors v~1 , v~2 , · · · , v~k in a vector space V are said to form a basis for V if (a) v~1 , v~2 , · · · , v~k span V and (b) v~1 , v~2 , · · · , v~k are linearly independent.
Example Show that S = {t2 + 1, t − 1, 2t + 2} is a basis for the vector space P2 . If S = {v~1 , v~2 , · · · , v~k } is a basis for a vector space V , then every vector in V can be written in one and only one way as a linear combination of the vectors in S. Find a basis for the subspace W of P2 , consisting of all vectors of the form at2 + bt + c, where c = a − b.
Subspaces of lRn
Vector Spaces
Basis
The vectors v~1 , v~2 , · · · , v~k in a vector space V are said to form a basis for V if (a) v~1 , v~2 , · · · , v~k span V and (b) v~1 , v~2 , · · · , v~k are linearly independent.
Example Show that S = {t2 + 1, t − 1, 2t + 2} is a basis for the vector space P2 . If S = {v~1 , v~2 , · · · , v~k } is a basis for a vector space V , then every vector in V can be written in one and only one way as a linear combination of the vectors in S. Find a basis for the subspace W of P2 , consisting of all vectors of the form at2 + bt + c, where c = a − b. Let S = {v~1 , v~2 , · · · , v~k } be a set of nonzero vectors in a vector space V and let W = spanS. Then some subset of S is a basis for W .
Subspaces of lRn
Vector Spaces
Finite-Dimensional Vector Space
The dimension of a nonzero vector space V is the number of vectors in a basis for V .
Subspaces of lRn
Vector Spaces
Finite-Dimensional Vector Space
The dimension of a nonzero vector space V is the number of vectors in a basis for V . If S is a linearly independent set of vectors in a finite-dimensional vector space V , then there is a basis T for V , which contains S.
Subspaces of lRn
Vector Spaces
Finite-Dimensional Vector Space
The dimension of a nonzero vector space V is the number of vectors in a basis for V . If S is a linearly independent set of vectors in a finite-dimensional vector space V , then there is a basis T for V , which contains S.
Example Find a basis for lR4 that contains the vectors v1 = (1, 0, 1, 2), v2 = (0, 1, 1, 2) and v3 = (1, 1, 1, 3).
Subspaces of lRn
Vector Spaces
Finite-Dimensional Vector Space
Let V be an n-dimensional vector space, and let S = {v~1 , v~2 , 路 路 路 , v~n } be a set of n vectors in V . (a) If S is linearly independent, then it is a basis for V . (b) If S spans V , then it is a basis for V .
Subspaces of lRn
Vector Spaces
Finite-Dimensional Vector Space
Let V be an n-dimensional vector space, and let S = {v~1 , v~2 , 路 路 路 , v~n } be a set of n vectors in V . (a) If S is linearly independent, then it is a basis for V . (b) If S spans V , then it is a basis for V .
Exercise P3 is a 4-dimensional vector space. Show that t3 + t2 + t + 1, t3 + 2t2 + t + 3, 2t3 + t2 + 3t + 2, t3 + t2 + 2t + 2 form a basis for P3 .
Subspaces of lRn
Vector Spaces
Vector Spaces and Linear Systems
Vector Spaces Basis and Dimension
Subspaces of lRn Row(Column) Space of a Matrix Orthonormal Bases in lRn
Subspaces of lRn
Vector Spaces
Definition
Let
a11 a21 : am1 be an m × n matrix.
a12 a22 : am2
··· ··· ··· ···
··· ··· ··· ···
··· ··· ··· ···
a1n a2n : amn
Subspaces of lRn
Vector Spaces
Definition
Let
a11 a21 : am1
a12 a22 : am2
··· ··· ··· ···
··· ··· ··· ···
··· ··· ··· ···
a1n a2n : amn
be an m × n matrix. The rows of A, v1 v2 : vm
= (a11 , a12 , · · · , a1n ) = (a21 , a22 , · · · , a2n ) = (am1 , am2 , · · · , amn ),
considered as vectors in lRn , span a subspace of lRn , called the row space of A.
Subspaces of lRn
Vector Spaces
Definition
Similarly, the columns of A,
a11 a12 a21 a22 w1 = : , w2 = : am1 am2
a1n , · · · , wn = a2n , : amn
considered as vectors in lRm , span a subspace of lRm , called the column space of A.
Vector Spaces
Subspaces of lRn
If A and B are two m Ă— n row equivalent matrices, then the row spaces of A and B are equal.
Subspaces of lRn
Vector Spaces
If A and B are two m × n row equivalent matrices, then the row spaces of A and B are equal.
Example
1 3 A= 2 −1
−2 0 3 −4 2 8 1 4 3 7 2 3 2 0 4 −3
Subspaces of lRn
Vector Spaces
If A and B are two m × n row equivalent matrices, then the row spaces of A and B are equal.
Example
1 3 A= 2 −1
1 0 2 0 −2 0 3 −4 0 1 1 0 2 8 1 4 ;B= 0 0 0 1 3 7 2 3 0 0 0 0 2 0 4 −3
1 1 −1 0
Subspaces of lRn
Vector Spaces
Finding a basis
The procedure for finding a basis for the subspace V of lRn given by V = span S, where S = {v1 , v2 , · · · , vk } is a set of vectors in lRn that are given in row form, is as follows. Step 1. Form the matrix
v1 v2 A= : vk whose rows are given vectors in S. Step 2. Transform A to reduced row echelon form, obtaining the matrix B. Step 3. The nonzero rows of B form a basis for V.
Subspaces of lRn
Vector Spaces
Example Let S = {(1, −2, 0, 3, −4), (3, 2, 8, 1, 4), (2, 3, 7, 2, 3), (−1, 2, 0, 4, −3)} and let V be the subspace of lR5 given by V = span S. Find a basis for V .
Subspaces of lRn
Vector Spaces
Example Let S = {(1, −2, 0, 3, −4), (3, 2, 8, 1, 4), (2, 3, 7, 2, 3), (−1, 2, 0, 4, −3)} and let V be the subspace of lR5 given by V = span S. Find a basis for V . 1 −2 0 3 −4 3 2 8 1 4 A= 2 3 7 2 3 −1 2 0 4 −3
Subspaces of lRn
Vector Spaces
Example Let S = {(1, −2, 0, 3, −4), (3, 2, 8, 1, 4), (2, 3, 7, 2, 3), (−1, 2, 0, 4, −3)} and let V be the subspace of lR5 given by V = span S. Find a basis for V . 1 −2 0 3 −4 1 0 2 0 1 3 2 8 1 4 1 ;B= 0 1 1 0 A= 2 3 7 2 3 0 0 0 1 −1 −1 2 0 4 −3 0 0 0 0 0
Subspaces of lRn
Vector Spaces
Example Let S = {(1, −2, 0, 3, −4), (3, 2, 8, 1, 4), (2, 3, 7, 2, 3), (−1, 2, 0, 4, −3)} and let V be the subspace of lR5 given by V = span S. Find a basis for V . 1 −2 0 3 −4 1 0 2 0 1 3 2 8 1 4 1 ;B= 0 1 1 0 A= 2 3 7 2 3 0 0 0 1 −1 −1 2 0 4 −3 0 0 0 0 0 Hence, w1 = (1, 0, 2, 0, 1), w2 = (0, 1, 1, 0, 1) and w3 = (0, 0, 0, 1, −1) form a basis for V .
Subspaces of lRn
Vector Spaces
Example Let S = {(1, −2, 0, 3, −4), (3, 2, 8, 1, 4), (2, 3, 7, 2, 3), (−1, 2, 0, 4, −3)} and let V be the subspace of lR5 given by V = span S. Find a basis for V . 1 −2 0 3 −4 1 0 2 0 1 3 2 8 1 4 1 ;B= 0 1 1 0 A= 2 3 7 2 3 0 0 0 1 −1 −1 2 0 4 −3 0 0 0 0 0 Hence, w1 = (1, 0, 2, 0, 1), w2 = (0, 1, 1, 0, 1) and w3 = (0, 0, 0, 1, −1) form a basis for V . Given that the vector v = (5, 4, 14, 6, 3) is in V , we can write v as a linear combination of the basis obtained above, as follows: v = 5w1 + 4w2 + 6w3
Vector Spaces
Subspaces of lRn
The dimension of the row space of A is called the row rank of A, and the dimension of the column space of A is called the column rank of A.
Vector Spaces
Subspaces of lRn
The dimension of the row space of A is called the row rank of A, and the dimension of the column space of A is called the column rank of A. The row rank and column rank of the m Ă— n matrix A are equal.
Subspaces of lRn
Vector Spaces
Computing for the rank
The procedure for computing the rank A is as follows. Step 1. Using elementary row operations, transform A to matrix B in reduced echelon form. Step 2. Rank A = the number of nonzero rows of B.
Subspaces of lRn
Vector Spaces
Nullity A
If A is an m Ă— n matrix, the nullity of A as the dimension of the null space of A, that is, the dimension of the solution space of A~x = ~0.
Subspaces of lRn
Vector Spaces
Nullity A
If A is an m × n matrix, the nullity of A as the dimension of the null space of A, that is, the dimension of the solution space of A~x = ~0.
Example
1 3 Given A = 2 −1
−2 0 3 −4 2 8 1 4 . 3 7 2 3 2 0 4 −3
Subspaces of lRn
Vector Spaces
Nullity A
If A is an m × n matrix, the nullity of A as the dimension of the null space of A, that is, the dimension of the solution space of A~x = ~0.
Example
1 3 Given A = 2 −1
−2 0 3 −4 2 8 1 4 . 3 7 2 3 2 0 4 −3
If A is an m × n matrix, then rank A+ nullity A = n.
Subspaces of lRn
Vector Spaces
Rank and Singularity
An n Ă— n matrix A is nonsingular if and only if rank A = n.
Subspaces of lRn
Vector Spaces
Rank and Singularity
An n Ă— n matrix A is nonsingular if and only if rank A = n. Let S = {v1 , v2 , . . . , vn } be a set of n vectors in lRn and let A be the matrix whose rows(columns) are the vectors in S.
Subspaces of lRn
Vector Spaces
Rank and Singularity
An n Ă— n matrix A is nonsingular if and only if rank A = n. Let S = {v1 , v2 , . . . , vn } be a set of n vectors in lRn and let A be the matrix whose rows(columns) are the vectors in S. Then S is linearly independent if and only if det(A) 6= 0.
Subspaces of lRn
Vector Spaces
Rank and Singularity
An n Ă&#x2014; n matrix A is nonsingular if and only if rank A = n. Let S = {v1 , v2 , . . . , vn } be a set of n vectors in lRn and let A be the matrix whose rows(columns) are the vectors in S. Then S is linearly independent if and only if det(A) 6= 0. The homogeneous system A~x = ~0 of n linear equations in n unknowns has a nontrivial solution if and only if rank A < n.
Subspaces of lRn
Vector Spaces
Definition
A set S = {u1 , u2 , 路 路 路 , uk } in lRn is called orthogonal if any two distinct vectors in S are orthogonal, that is, if ui 路 uj = 0 for i 6= j.
Subspaces of lRn
Vector Spaces
Definition
A set S = {u1 , u2 , · · · , uk } in lRn is called orthogonal if any two distinct vectors in S are orthogonal, that is, if ui · uj = 0 for i 6= j. An orthonormal set of vectors is an orthogonal set of unit vectors. That is, S = {u1 , u2 , · · · , uk } is orthonormal if ui · uj = 0 for i 6= j, and ui · ui = 1 for i = 1, 2, · · · , k.
Subspaces of lRn
Vector Spaces
Definition
A set S = {u1 , u2 , · · · , uk } in lRn is called orthogonal if any two distinct vectors in S are orthogonal, that is, if ui · uj = 0 for i 6= j. An orthonormal set of vectors is an orthogonal set of unit vectors. That is, S = {u1 , u2 , · · · , uk } is orthonormal if ui · uj = 0 for i 6= j, and ui · ui = 1 for i = 1, 2, · · · , k. Let S = {u1 , u2 , · · · , uk } be an orthogonal set of nonzero vectors in lRn . Then S is linearly independent.
Subspaces of lRn
Vector Spaces
Definition
A set S = {u1 , u2 , · · · , uk } in lRn is called orthogonal if any two distinct vectors in S are orthogonal, that is, if ui · uj = 0 for i 6= j. An orthonormal set of vectors is an orthogonal set of unit vectors. That is, S = {u1 , u2 , · · · , uk } is orthonormal if ui · uj = 0 for i 6= j, and ui · ui = 1 for i = 1, 2, · · · , k. Let S = {u1 , u2 , · · · , uk } be an orthogonal set of nonzero vectors in lRn . Then S is linearly independent. An orthonormal set of vectors in lRn is linearly independent.
Subspaces of lRn
Vector Spaces
Dot Product
Definition The dot product or inner product of the n-vectors a = (a1 , a2 , · · · , an ) and b = (b1 , b2 , · · · , bn ) is defined as a · b = a1 b1 + a2 b2 + · · · + an bn =
n X i=1
ai bi
Subspaces of lRn
Vector Spaces
Gram-Schmidt Process Let W be a nonzero subspace of lRn with basis S = {u1 , u2 , · · · , um }. Then there exists an orthonormal basis T = {w1 , w2 , · · · , wm } for W.
Subspaces of lRn
Vector Spaces
Gram-Schmidt Process Let W be a nonzero subspace of lRn with basis S = {u1 , u2 , · · · , um }. Then there exists an orthonormal basis T = {w1 , w2 , · · · , wm } for W. Step 1. Let v1 = u1 . Step 2. Compute the vectors v2 , v3 , · · · , vm , successively, one at a time, by the formula „ « „ « „ « ui · v 1 ui · v2 ui · vi−1 vi = ui − v1 − v2 −· · ·− vi−1 v1 · v1 v2 · v2 vi−1 · vi−1 The set of vectors T ∗ = {v1 , v2 , · · · , vm } is an orthogonal set. Step 3. Let wi =
vi , 1 ≤ i ≤ m, kvi k
√ where kvi k = vi · vi Then T = {w1 , w2 , · · · , wm } is an orthonormal basis for W.
Subspaces of lRn
Vector Spaces
Exercises 1. Consider the basis S = {(1, 1, 1), (−1, 0, −1), (−1, 2, 3)} for lR3 . Use the Gram-Schmidt process to transform S to an orthonormal basis for lR3 . 2. Use the Gram-Schmidt process to find an orthonormal basis for the subspace of lR4 with basis {(1, −1, 0, 1), (2, 0, 0, −1), (0, 0, 1, 0)}
Subspaces of lRn
Vector Spaces
Questions? See you next meeting!