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Applied Linear Algebra

by: Mary Veum

Applied Linear Algebra MATH 2210

Mary Veum
GPA 3.97

Sarah Glaz

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Sarah Glaz
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This 12 page Class Notes was uploaded by Mary Veum on Thursday September 17, 2015. The Class Notes belongs to MATH 2210 at University of Connecticut taught by Sarah Glaz in Fall. Since its upload, it has received 21 views. For similar materials see /class/205804/math-2210-university-of-connecticut in Mathematics (M) at University of Connecticut.

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Date Created: 09/17/15
22 The Inverse of a Matrix The inverse ofa real number a is denoted by a l For example 7 1 17 and 77 17171 An n x n matrixA is said to be invertible if there is an n x n matrix C satisfying CA AC In where In is the n x n identity matrix We call C the inverse ofA FACT lfA is invertible then the inverse is unique Proof Assume B and C are both inverses ofA Then BBB 1 C So the inverse is unique since any two inverses coincide The inverse ofA is usually denoted by A l We have 71147114 1quot Not all n x n matrices are invertible A matrix which is not invertible is sometimes called a singular matrix An invertible matrix is called nonsingular matrix Theorem 4 LetA a Z If adi bc 0 thenA is invertible and C d 7b A71 adibc C a If adi bc 0 thenA is not invertible Assume A is any invertible matrix and we wish to solve Ax b Then Ax b and so x orx Suppose w is also a solution to Ax b Then Aw b and Aw 7b which means w A 1 b 80 w A lb which is in fact the same solution We have proved the following result Theorem 5 lfA is an invertible n x n matrix then for each b in Rquot the equation Ax b has the unique solution X A lb 3xz2 72x2139 Solution Matrix form ofthe linear system 7 Z x1 I x2 EXAMPLE Use the inverse ofA 77 3 to solve 77x1 5 72 5x1 Theorem 6 Suppose A and B are invertible Then the following results hold a A 1 is invertible and A 1 1 A Le A is the inverse ofA l b AB is invertible and AB 1 B IA 1 c AT is invertible and Alf1 A 1T Partial proof of pan b ABXB IA I A A 1 A A71 Similarly one can show that B lA IXAB 1 Theorem 6 part b can be generalized to three or more invertible matrices ABCYI Earlier we saw a formula for nding the inverse of a 2 x 2 invertible matrix How do we find the inverse of an invertible n x n matrix To answer this question we first look at elementary matrices Elementary Matrices Definition An elementary matrix is one that is obtained by performing a single elementary row operation on an identity matrix 100 100 100 EXAMPLE LetEl 020 E2 001 E3 010 and 001 010 301 abc A def ghz39 E1 E2 and E3 are elementary matrices Why Observe the following products and describe how these products can be obtained by elementary row operations on A E114 1 0 0 0 2 0 0 0 1 Q d g Q 139 abc 2d222f ghz39 abc ghz39 def a b c d e f 3ag 3bh 3ci If an elementary row operation is performed on an m x n matrix A the resulting matrix can be written as EA Where the m x m matrix E is created by performing the same row operations on 1quot Elementary matrices are invertible because row operations are reversible To determine the inverse of an elementary matrix E determine the elementary row operation needed to transform E back into 1 and apply this operation to 1 to nd the inverse For example WOl I Ol O l OO Eg1 1 X m 3 395 539 39 SP I ll l l 39 l o le l I O O O NIH O H l 3 D 3 100 100 100 EA 020 70 7301 001 010 010 100 100 0 E2E1A 001 7301 010 010 010 7301 100 00 100 E3E2E1A 010 10 010 301 7301 001 E3E2E1A 3 Then multiplying on the right by A l we get E3E2E1A 3 So 3E2E113 A71 The elementary row operations that row reduce A to In are the same elementary row operations that transform n into A l Theorem 7 An n x n matrixA is invertible if and only ifA is row equivalent to In and in this case any sequence of elementary row operations that reducesA to n will also transform n to A l Algorithm for finding A 1 Place A and 1 sidebyside to form an augmented matrix A I Then perform row operations on this matrix which will produce identical operations onA and 1 So by Theorem 7 A I will row reduce to I A l orA is not invertible 2 0 0 EXAMPLE FindtheinverseofA 73 01 ifitexists 0 10 Solution 2 0 010 o 10 0 o 0 A1 730 o10010001 0 10 0 01 0 01 0 8014 Ixle O NIH l i Order of multiplication is important EXAMPLE Suppose ABC and D are invertible n x n matrices and A BD 7quotC Solve forD in terms ofABC and D Solution 7A7 731 JACi D 7quot B IAC 1 D 7quot B lAC 1 D 42 Null Spaces Column Spaces amp Linear Transformations The null space of an m x n matrix A written as Nul A is the set of all solutions to the homogeneous equation Ax 0 NulA X X is in Rquot and Ax 0 set notation THEOREM 2 The null space of an m x n matrixA is a subspace of Rquot Equivalently the set of all solutions to a system Ax 0 of m homogeneous linear equations in n unknowns is a subspace of Rquot Proof Nul A is a subset of Rquot sinceA has n columns Must verify properties a b and c ofthe definition of a subspace Property a Show that 0 is in NulA Since 0 is in Property b If u and v are in NuIA show that u v is in NuIA Since u and v are in NuIA and Therefore Au v Property c Ifu is in NulA and c is a scalar show that cu in NulA Acu 714w 00 0 Since properties a b and c hold A is a subspace of Rquot Solving Ax 0 yields an explicit description of Nul A 3 6 6 3 9 EXAMPLE FInd an explICIt descrIptIon of Nul A where A 6 12 13 0 3 Solution Row reduce augmented matrix corresponding to Ax 0 366390 12013330 61213030 001767150 x1 72m 713x11 7 33xs X2 x2 X3 6X4 15xs x4 X4 965 x5 72 713 733 1 0 0 x2 0 x4 6 x5 15 0 1 0 0 0 1 Then NulA spanuvw Observations 1 Spanning set of NulA found using the method in the last example is automatically linearly independent 72 713 733 0 1 0 0 0 C1 0 02 6 63 15 0 2617 02 03 0 1 0 0 0 0 1 0 2 If Nul Aqt 0 the the number of vectors in the spanning set for Nul A equals the number of free variables in Ax 0 The column space of an m x n matrixA Col A is the set of all linear combinations of the columns ofA IfA a1 anthen CoIA Spana1 an THEOREM 3 The column space of an m x n matrixA is a subspace of R quot Why Theorem 1 page 221 Recall that ifo b then b is a linear combination ofthe columns ofA Therefore lColA b b AX for some X in Rquot x 7 2y EXAMPLE Find a matrixA such that W ColA where W 3y xy in R x y Solution x 7 2y 1 72 3y x 0 y 3 x y 1 1 Therefore A By Theorem 4 Chapter 1 The column space of an m x n matrixA is all of R if and only ifthe equation Ax b has a solution for each b in R quot The Contrast Between Nul A and Col A 1 2 3 EXAMPLE LetA 2 4 7 3 6 10 0 0 1 a The column space ofA is a subspace of Rk where k b The null space ofA is a subspace of Rk where k c Find a nonzero vector in Col A There are infinitely many possibilities OOJRN d Find a nonzero vector in NulA Solve Ax 0 and pick one solution 12 3 0 12 0 0 2 4 7 0 0 010 row reduces to 3 610 0 0 0 0 0 0 0 10 0 0 0 0 x172m xzisfree X30 Letx27and then 961 X x2 x3 Contrast Between NuIA and CoIA whereA is m x n see page 232 Review A subspace of a vector space Vis a subset H of Vthat has three properties a The zero vector of Vis in H b For each u and v in H u v is in H In this case we sayH is closed under vector addition 0 For each u in H and each scalar c cu is in H In this case we say H is closed under scalar multiplication lfthe subset H satis es these three properties then H itself is a vector space THEOREM 12 and 3 Sections 41 amp 42 lfv1vp are in a vector space V then Spanv1 vp is a subspace of V The null space of an m x n matrixA is a subspace of Rquot The column space of an m x n matrixA is a subspace of R quot EXAMPLE Determine whether each of the following sets is a vector space or provide a counterexample aH x I x 7y 4 Solution Since is not in H H is not a vector y space 7 0 Solution Rewrite x y as y Z 0 y z 0 b V PO 710 1 So V NulA whereA 0 1 1 Since NulA is a subspace of R2 Vis a vector space xy c S 2x7 3y xyz are real 3y One Solution Since x y 1 1 2x 7 3y x 2 y 73 3y 0 3 1 1 S span 2 73 therefore S is a vector space by Theorem 1 3 Another Solution Since x y 1 2x 7 3y 7 x y 73 3y 3 1 1 S Col A whereA 2 73 therefore S is a vector space since a column space is a vector 0 3 space Kernal and Range of a Linear Transformation A linear transformation T from a vector space Vinto a vector space W is a rule that assigns to each vector X in Va unique vector TX in W such that i Tu v Tu Tv for all uv in V ii TcucTu for all u in tin Vand all scalars c The kernel or null space of T is the set of all vectors u in Vsuch that Tu 0 The range of T is the set of all vectors in Wof the form Tu where u is in V 80 if TX AX colA range of T


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