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# Class Note for MATH 215 at UA

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This 12 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 19 views.

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Date Created: 02/06/15

THE UNIVERSITY a OF ARIZONA Math 215 Introduction to Linear Algebra Section 35 Subspaces Basis Dimension and Rank Paul Dostert September 2 2008 112 A Subspaces The motivation for subspaces comes from the question What does the space spanned by 1 0 0 and 010 look like Is this in R2 or R3 A subspace of R is any collection S of vectors in R st 1 The zero vector 0 is in S 2 If uv e S then u v e S S is closed under addition 3 If u e S and c is a scalar then cu e S S is closed under scalar multiplication Ex Show that the space spanned by 1 0 0 and 010 is a subspace of R3 Thm Let v1 vk be vectors in Rquot Then span v1 vk is a subspace of R We say span v1 vk is the subspace spanned by v1 vk Ex Do vectors of the form 16 form a subspace of R2 Do vectors of the a form y x 2y form a subspace of R3 A Matrix subspaces 96 Ex Determine whether the set of all vectors y with z y 4x is a Z x subspace of R3 What about vectors y with c y z 1 2 Let A be an m gtlt n matrix 1 The row space of A is the subspace row A of R spanned by the rows of A 2 The column space of A is the subspace 01 A of R spanned by the columns of A 1 2 Ex Consider A 1 1 5 1 a Is b 17 47 5T in the column space of A b Is w 27 3 in the row space of A c Describe row A and col A A Matrix subspaces amp null space Thm Let B be any matrix that is row equivalent to a matrix A Then row B row Thm Let A be an m gtlt n matrix and let N be the set of solutions of the homogeneous linear system Ax 0 Then N is a subspace of Rquot Ex Prove this theorem Let A be an m gtlt n matrix The null space of A is the subspace of R consisting of solutions of the homogeneous linear system Ax 0 It is denoted by null A The idea of null space comes into the proof of the following VERY important theorem Thm Let A be a matrix whose entries are real numbers For any system of linear equations Ax b exactly one of the following is true a There is no solution b There is a unique solution c There are infinitely many solutions Pf The idea is that we show if we cannot have a or b then c must be true To show c we show the null space has infinitely many vectors A Basis A basis for a subspace S of R is a set of vectors in S that are linearly independent and span S We call the standard unit vectors e17 en e R the standard basis for Rquot 1 1 1 Ex Find a basis for span 2 7 1 7 2 1 0 2 8 1 2 4 Ex Find a basis for span 2 7 3 7 1 1 2 3 Ex Find a basis for the row space of A i Oii Di i i Di ii I l A Basis Here we briefly summarize the procedure to find the row column and null space of A 1 Find the reduced row echelon form R of A 2 Use the nonzero row vectors of R to form a basis for row A 3 Use column vectors of A corresponding to the columns of R containing the leading ls to form a basis for col A 4 Set the free variables to parameters Solve for the leading variables of Rx 0 in terms of these free variables ie write x in terms of only free parameters t s Write the resulting x as a linear combination of vectors with the free parameters as coefficients These vectors then span the null space Ex Find a basis for the row column and null space of 1 1 2 1 2 1 0 1 3 1 6 3 1 1 3 4 A A Dimension Thm The Basis Thm Let S be a subspace of Rquot Then any two bases for S have the same number of vectors Ex Describe how you would attempt to prove this If S is a subspace of Rquot then the number of vectors in a basis for S is called the dimension of S denoted dimS Note We define dim0 0 1 1 2 1 2 1 0 1 Ex For the matrix on the prevuous slide A 3 1 6 3 what are 1 1 3 4 the dimensions of the row column and null spaces How are these related Thm The row and column spaces of a matrix A have the same dimension Ex Look at the proof The idea is dim row dim row nonzero rows ofR leading 1s ofR dim col dim col All Rank The rank of a matrix A is the dimension of its row and column spaces and is denoted by rank A The nullity of a matrix A is the dimension of its null space and is denoted by nullity A Cor For any matrix A rank AT rank A Ex Prove this Thm The Rank Thm If A is an m gtlt n matrix then rank A nullity A n 1 2 1 2 3 4 Ex Find the nullity of A 2 3 and B 1 1 2 6 by using 3 4 1 5 4 2 the rank theorem and NOT by solving Ax 0 A Fundamental Thm of lnvertible Matrices With linear independence spans basis dimension and rank all defined we can now add to the previous Fundamental Thm The Fundamental Theorem of lnvertible Matrices Version 2 Let A be an n gtlt n matrix The following statements are equivalent a b C 0 6 f g h i J k I m A is invertible Ax b as a unique solution for every b e Rquot Ax 0 has only the trivial solution The rref of A is In A is a product of elementary matrices rank A n nullity A 0 The column vectors of A are linearly independent The column vectors of A span Rquot The column vectors of A form a basis for R The row vectors of A are linearly independent The row vectors of A span R The row vectors of A form a basis for Rquot Pf Show A Using the Fundamental Thm Thm Let A be an m gtlt n matrix Then rank ATA rank A and ATA is invertible iff rank A n Ex Prove using the rank theorem and the fund thm 1 0 1 Ex Show that the vectors 2 3 and 4 form a basis for R3 Recall that is B is a basis for a subspace S of R then we can write any vector in S as a linear combination of basis vectors The following takes this idea one step further Thm Let S be a subspace of R and B v17 vk a basis for S Every v e S can be written as a unique linear combination of basis vectors in B v clvl ckvk with unique 3239 Ex Prove this At Coordinate Representation Let S be a subspace of R and let l3 v17 7vk be a basis for S Let v 01V1 Ckvk be a vector in S Then the 3239 are called the coordinates ofv with respect to B and the column vector C1 Vlg Ck is called the coordinate vector ofv with respect to 8 Ex Show that w is in span 8 and find the coordinate vector WB for Mi ili ilwliil Ah Matlab Examples Let us attempt to find a basis for the column and null spaces of 1 1 0 1 0 1 1 1 0 1 1 1 We start by entering A We then calculate the rref which actually returns 2 arguments the rref matrix R and the vector jb which gives the index of columns for a basis of A Le A7jb is a basis for the range of A We also have a function null which returns a normalized orthonormal actually basis for the null space of A So in Matlab we could do A1 1 0101 1 101 1 1 R jb rrefA Ajb nullA Ajb would indicate that the 152 2quotd and 4th columns form a basis nullA gives us 057747 057747 057747 00000 as a basis which can be rescaled to 1717170

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