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## Linear Alg&Diff Eq

by: Mrs. Preston Lehner

8

0

6

# Linear Alg&Diff Eq MATH 214

Mrs. Preston Lehner
UM
GPA 3.87

Staff

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COURSE
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KARMA
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## Popular in Mathematics (M)

This 6 page Class Notes was uploaded by Mrs. Preston Lehner on Thursday October 29, 2015. The Class Notes belongs to MATH 214 at University of Michigan taught by Staff in Fall. Since its upload, it has received 8 views. For similar materials see /class/231511/math-214-university-of-michigan in Mathematics (M) at University of Michigan.

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Date Created: 10/29/15
Math 214 Winter 708 Review for Midterm 2 March 14 Note These notes are intended to summarize the essential material for Midterm 2 You still have to read the textbook and to review the recent HWs Chapter 4 0 Linear Space Def 411 To have a linear space V you need to know how to add two elements of V and how to multiply by a scalar constant The addition and scalar multiplication satisfy the usual properties Essential 0 0 E V 1 closed under addition Ly E V gt z y E V 2 closed under scalar multiplication z E V gt c z E V Concretely to say that a certain set V is a linear subspace you have to explain that V is closed under addition and scalar multiplication in our examples the meaning of addition and the scalar multiplication is obvious Concrete examples polynomials of certain degree 0 square matrices V Mn functions With certain properties etc To tes your understanding explain Why the set V of upper diagonal 2 X 2 matrices form a linear space but ifl require in addition that the matrices in V are invertible then V is no longer a linear space 0 Linear Transformation Def 421 To check that T V A W is linear you have to check that T 1 behaves as expected W1 t addition TI y Tz Ty 2 behaves as expected Wrt scalar multiplication Tc z c z E V In particular one also have T0 0 Typical examples derivatives multiplications by constant matrices etc et 2 A M2 an 3 11 Explain Why TX AX A is linear but TX XAX is not Span Linear independence Basis Dimension Def 413 The key points to remember 7 basis span and linear independent 7 dimension the number of elements in a basis 7 in practice to check that something is a basis for V it is enough to check that you have the right numbers of elements and either they span V or that they are linearly independent 7 see 416 and the related examples for how to nd a basis Concrete examples nd a basis for the upper triangular matrices for degree 2 polynomials passing through 722 etc Important examples nd a basis for the image of a linear transformation or for the kernel Example let T P2 7gt P2 be given by f nd a basis for KerT and ImT Note the fundamental result dim KerT dim ImT dim V where V is the domain of T o Isommphism 422 and 424 In practice the easiest way to see that T V 7gt W a linear transformation is an isomorphism is to check 7 dimV dimW 7 and K TT 0 ie solve the equation Tz 0 and deduce that the only solution is z Question is it T P2 7gt P2 de ned by If an isomorphism What if Tf zf f o Coordinates If we have given a basis 3 111 vn for V then we can transform any element 1 of V in vector Ills and work with as if we have V R the coordinate isomorphism The role of this is to make the abstract notions of chapter 4 very concrete as in the previous chapters A few important points Ills an simply means I a1v1anvn 7 with respect to the standard basis77 is quite obvious what is the coordinate vector Say that V P2 3 12zl and f 512 31 7 7 Then clearly 7 if 1 try to do the same example as above but With respect to the exotic basis A I2 z 1 z 11 the situation becomes trickyi 1 need to nd a b c such that f5123177a12z1b11c1 After expanding this becomes a linear system of equations After solving 1 get a b c and then 7 Alternatively since 1 can easily nd f5 1 can use the formula of change of basis lflA 55H fle to compute flgi The matrix B 7gt A is obtained by taking the coor dinates of the basis 3 With respect to the basis A ie 55 llvllxw x 7 lvnlAl Concretely in our example we express the basis 3 12z1 in terms of the basis A 12 z 1 z 11 ie 12 12z17z1 z 1171 1 1 Thus 1 0 0 58M 71 1 0 0 71 0 AndWeget O 1 0 0 5 5 lflASBHA39lle1 1 3 2 J 0 71 0 77 710 compare With the direct approach to compute 7 Note the important formula 58 SAAB71 7 Note also that SAAB it is easy to nd compute itli o The matrix of a linear transformation You have T V 7gt V7 and B v17 71177 a basis Then T is given as in the case V R by a n X n matrix B st Tz5 B MB The key formula is 7 B Tv157 7 Tvn57 ie the columns of the matrix B are the coordinate vectors of the transforms by T of the basis vectors Example compute the matrix of the linear transformation T P2 7gt P27 where z f f with respect to the standard basis 3 127171 Now do the same with respect to the nonstandard basis A I2 z 171 171 you should get another matrix A representing T This can be done and should be done for practice in two ways 7 the same type of computations as for B 7 by using the base change formula A S B 5717 where S 55 NB 5 1 can be computed as the inverse of S or by 571 SAAB Chapter 5 o Orthogonal vectors Orthonormal basis A few important points 7 orthonormalortho ie ulnuj 0 ifi j and normal ie lluz ll 1 7 if u17 7 on are orthonormal7 then they are linearly independent Thus to check that u17 7 an orm an orthonormal basis for some linear subspace V it is enough to check that the vectors are or thonormal and the right number of vectors7 ie n dim V 0 With respect to orthonormal bases it is easy to compute coordinates and projections Let V a linear subspace in some RN and B u17 7 an an orthonormal basis The following two formulas are essential 7 if z 6 V7 then Ills una the coordinates are computed by taking the dot product 7 For any I not necessary in V7 the orthogonal projection on V is computed by zll projvz zu1u1 Note also that z Ill Ii In the GramSchmidt algorithm you need Ii Which is computed by z z 7 Ill and the previous formula 0 GramSchmidt see 52 You are given an arbitrary basis vb i i vn for V and want to nd an orthonormal basis uh i un The idea is quite simple it easy to get norm 1 just normalize the vectors the hard part is to obtain ortho this is done by computing the orthogonal projection Speci cally Step 1 Normalize v1 iiel ul H27 Step 2 Compute 1 as explained above iiei 1 U2 v2 u1u1 i 2 HUM Step 3 Normalize 11 ie ug Step 4 Compute v3i erlti ul ug vgi v3 7 vglu1u1 7 vgiu2u2 Step 5 Normalize v3i etc 0 To compute an orthogonal basis for a linear space V eigi K TA ImA you rst need to nd a basis vb i i vn as we did in the previ ous chapters and then apply GramSchmidt to get an orthonormal basis Recall that to nd a basis for ImA you simply have to remove the re dundant vectors among the columns of A To nd a basis for Ker A you have to nd the relations between the columns vectors of Al 0 Suppose that you are given a linear subspace V then the orthogonal complement Vl is de ned as the set of vectors orthogonal on V To nd Vi you have to solve the linear system zivl 0 111 0 Where v1 i 1 is a basis for Vi The orthogonal complement has com plementary dimension to V Lei dimVdimVi N Where N is the dimension of the ambient space V is a subspace in some RN Example Let V Span 2 7 5 Then Vi is l 3 6 dimensional since dim V 2 and dim Vi dimV 3 and Vi contains 1 4 a all the vectors 1 orthogonal to c 2 and 5 This means 3 6 a 212 Sc 0 4a5b6c 0 You have to solve for a 127 5 ie express a b in terms of the free variable 0 Since Vi is ldimensional7 a basis is given by a nontrivial solution of the previous linear system just set the free variable 5 to some random nonzero value In fancy words7 in the previous example V ImA and Vi Ker AT see 5417 Where 14 A 2 5 and AT 3 36 This means that V is spanned by the column vectors of A7 and Vi is ob tained by solving the linear system AT 1 ie the linear system considered a feW lines above Orthogonal TransformationsMathew A linear transformation T R 7gt R is orthogonal if it preserves the norm An orthogonal transformation automatically preserves also the angles7 ie I L y gt Tz L This gives an alternative characterization of orthogonal transformations 7 T is orthogonal if and only if Tel7 Ten forms an orthonormal asis The same thing in terms of the matrix A representing T is 7 A is orthogonal if and only if the columns of A form an orthogonal basis The important fact about orthogonal matrices is that it is easy to invert 7 A 1 AT7 Where AT is the transpose of A the columns of AT are the rows of A

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