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## Linear Algebra

by: Miss Noel Mertz

19

0

7

# Linear Algebra MATH 2270

Miss Noel Mertz
The U
GPA 3.95

Nicholas Korevaar

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COURSE
PROF.
Nicholas Korevaar
TYPE
Class Notes
PAGES
7
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 7 page Class Notes was uploaded by Miss Noel Mertz on Monday October 26, 2015. The Class Notes belongs to MATH 2270 at University of Utah taught by Nicholas Korevaar in Fall. Since its upload, it has received 19 views. For similar materials see /class/229924/math-2270-university-of-utah in Mathematics (M) at University of Utah.

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Date Created: 10/26/15
Math 22702 Tuesday September 18 Finding bases for imaged and kerf We work with the vectors from problem 24 in section 33 page 13 l gt restart gt withlinalg Warning the protected names norm and trace have been redefined and unprotected 39gt Atmatrix55l232l36963324l284915 O455l this is the quottransposequot of the matrix I want ie the columns and rows have been reversed l 2 3 2 l 3 6 9 6 3 An 3 2 4 l 2 8 4 9 l 5 0 4 5 5 l gt AtransposeAt this is the matrix I want the image of fxZ x is the subspace of 24 page 131 13380 2 6 2 4 4 A 3 9 4 9 5 2 6 l l 5 l 3 2 5 1 If I was doing 24 I am looking for a basis of subspace imagef where fXAX In 24 I want my basis to be a subset ofthe original 5 column vectors These 5 vectors in the quotcolumn spacequot may be dependent in which case I would like to throw away redundant vectors until I am left with a basis I can discover dependencies from the kemel i which I get from rrefA gt rref A l 3 0 l 3 0 0 l 3 l 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Backsolve to find a basis for keff Think of A and rrefA augmented by a zero column for the right hand side Check that you have a basis for kerf because this is something you need to know how to do in lots of applications How does your work compare to Maple gt nullspaceA basis for the kernel of f also called the nullspace of A P30101P31000Lamp 0013 Actually since column dependencies are really quotthe samequot as solutions to Ax0 they are unchanged as you do elementary row operations Therefore since in the reduced matrix I have the the second column equal to 3 times the first the same fact is true in the original matrix Since in the reduced matrix the 4th column is 3 times the third minus the rst the same is true in the original matrix Similarly the 5th column is dependent on the rst and third columns Thus an acceptable basis for the span of my 5 column vectors is gt imbasiscolAl colA3 imbasis 1 2 3 2 1 3 2 4 1 2 A nicer basis would have vectors with more zero entries We can obtain this by doing elementary column operations to A since these do not change the span of the columns This is the same as computing gt transpose rref At l 0 0 0 0 0 l 0 0 0 l 2 0 0 0 2 4 l 5 39 0 0 0 2 4 N blH This is how maple does it gt colspaceA basis for column space of A ie for the image of f lolile l rT2 4 It s always easy to tell that the basis you get this way is actually a basis the independence follows easily and for example it is also easy to express your basis obtained by deletion in terms of this col op derived basis Do it MATH 22702 Symmetric matrices and quadrics Decmnber42001 Quadric surfaces 22 page 511 of Kolman gt restart gt withlinalgwithplotsfor computations and pictures withstudentto do algebra computations like completing the square 3 342 224 2 228 Bmatrixl36 lO2 l l94 2 2 8 B6 10 2 3929 fmatv gtevalmtransposevampAampv Bampv C this gives a l by 1 matrix V fxyz gtsimplifyfmatmatrix3lxyz11 this extracts the entry simplifies it and defines f fxyz this is fhopefully 9 4x24xy 4xz4y2 4yz8226x 10y22 Z gt eigenvaluesA compute eigenvalues i 2410 Thus we have an ellipsoid a point or the empty set gt dataeigenvectsA i data4llll2lL l0lOlll2 We can read of the axes of our ellipsoid from the eigenvectors and adjust so that S is a rotation matrix gt wllsqrt3matrix3llll w2lsqrt2matrix3ll lO w3lsqrt6matrix3lll 2 Saugmentwlw2w3 16128 l l l 35 52 E6 l l l S 3 3 2 2 6 6 l l 3 0 6 1 gram gt evalmtransposesampAampS just checking 4 0 0 0 2 0 0 0 10 gt guvw gtsimplifyfmatevalmSampmatrix3luvw11 change of coordinates guvw 4u22v dOw2 J u8JEV JJw gt completesquare completesquare v completesquarew 2 4 2v210w28Ev Ew 2 2v2 2 4u 10 wZ g w 2 2 lop mzj 2V2 34u1 12 3 So it really will be an ellipsoid 39gt implicitplot3dfxyz0x 5ly l5z 22 axesboxedI adjusted the ranges to get a good picture dgi Ez smis gai quotES7 quot 28 7gt Amatrix338l6 2l6 8 2 2 2lO Bmatrixl3000 Cmatrixll 24 8 16 2 2 2 10 B 0 0 0 Cf2 gt fmatv gtevalmtransposevampAampv Bampv C this gives a l by 1 matrix fxyz gtsimplifyfmatmatrix3lxyz11 this extracts the entry simplifies it and defines f fxyz this is fhopefully 8x232xy 4xz 8y2 4yz1022 24 gt eigenvectsA 69 1 1 1 1 3923924919391919 A one or two sheeted hyperboloid Note there are no linear terms here so you can tell it s going to be 1sheeted why gt implicitplot3dfxyzx 33y 33z 33 axesboxed

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