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# LINR ALG & NUM ANLY AMATH 352

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This 8 page Class Notes was uploaded by Mossie Pfannerstill V on Wednesday September 9, 2015. The Class Notes belongs to AMATH 352 at University of Washington taught by Staff in Fall. Since its upload, it has received 29 views. For similar materials see /class/192279/amath-352-university-of-washington in Applied Mathematics at University of Washington.

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Date Created: 09/09/15

AMATH 352 LECTURE 9 JULY 14 2008 1 Recap of last lecture Here are the main points from the last lecture For a linear transformation T Rd a R7 the nullspace of T is NullspaceT V 6 Rd TV 0 1 This is the set of all vectors that T maps to 0 and its a subspace of Rd lt7s dimension is called the nullity of T nullityT dimNullspaceT 2 The range of T is the set of all possible outputs from T ln set notation RangeT W 6 RT W TV for some V 6 Rd 3 This is a subspace of RT and its dimension is called the rank of T rankT dimRangeT 4 We also saw that RangeT spancolumns of T 5 2 HOW do I nd NullspaceT l7ve de ned the null space of a transformation but I havent actually shown how to nd it Conveniently this boils down to Gaussian elimination And with a little bit of care along the way it also steers me towards my next goal nding a basis of the range Take the transformation given by the matrix 4 12 32 T 1 43 756 6 3 71 52 To nd NullspaceT7 I need to know which vectors this maps to 0 So l7ll take a generic vector V and set TV 0 4 12 32 0 1 11 12 13 0 3 71 52 0 l7m about to turn this into a linear system of equations7 but remember that we started out with this equation based on the columns 7 it7ll show up later Looking at the three components individually7 I get the 3 gtlt 3 linear system 411 02 113 0 11 12 7 213 0 311 7 12 13 0 And now I solve this using painfully detailed Gaussian elimination 1 3 1 3 4111 Uz 1130 Ui 02 030 4 5 scale 4 5 111 11275030 7 111 11275030 9 311 7 12 213 311 7 12 213 0 11 02 113 0 03 0 10 11 11 7 812 81370 pivot 29 7 aw i 2 113 0 pg 0 11 11 7 7312 313 7 0 scale 7 12 Ui 023030 Ii 02 7 v3 0 12 0 0 This system has one degree of freedom From here I have a couple of options The rst is to decide that 113 will be undetermined Conventionally7 the rightmost variable possible is the 2 undetermined one In that case7 the second equation gives me 12 13 And then the rst equation gives me 1 3 i 0 gt i 1 14 11 803 803 7 11 7 213 Then my generic vector in the null space is yg V 113 15 113 For the second option7 l7ll nd vectors that span the null space Since there7s only one degree of freedom in the linear systern7 this means that NullspaceT has dimension 1 So nding one vector will do the trick This means I can pick speci c values If I set 113 17 then I end up with 112 1 and 01 712 lend up with 712 NullspaceT span 1 16 1 This also tells me that nullityT 1 17 3 How do I nd a basis for RangeT In the last lecture7 we saw that RangeT spancolurnns of T 18 Sticking with the same T from the previous section7 this tells me that 4 12 32 RangeT span 1 7 43 7 756 19 3 71 52 However7 it7s cornpletely possible that l have more information than I need here 7 ie7 that the vectors on the right are linearly dependent 3 To check for linear dependence7 I set up the equation 4 12 32 0 1 51 43 52 756 03 0 20 3 41 52 0 with the goal of determining 01 02 and 03 Now7s the part where I go Whoaaaaaa wait a second here I already did tha 77 Look back at Sure it has 07S instead of os7 but otherwise its the exact same thing That means I can use 111 112 and 03 that I found before as my 04 values here So for any 113 I have 4 1 12 32 0 1 759 43 53 756 53 0 21 3 41 52 0 To break this dependence7 I need to throw out one of the vectors in this relationship I7ll get rid of the rst one death to integers7 and I end up with 12 32 RangeT span 43 7 756 7 22 71 52 with the vectors on the right forming a basis of RangeT This also tells me that rankT 2 23 4 An example with nullity2 Now I7ll walk through the same process for another transforrnation7 this one given by 12 43 1 U 756 20 452 24 1 424 3 To nd the null space7 the rst thing I7ll do is set UV 0 for a generic vector V This gives me 13 43 1 0 756 51 20 52 452 53 0 25 1 424 3 0 Like last time7 I turn this into a linear system7 which I simplify using Gaussian elimination 11812130 01724023ilg0 73m 2002 i 3713 0 73 2002 i 3713 0 26 01724023ilg0 11812130 11 7 2402 313 0 IL 0 0 27 0 0 This system has two degrees of freedom One way to characterize the null space is by all vectors of the form 2402 7 313 V 112 for all 11203 28 Z13 Or7 I can pick a couple of spanning vectors by taking speci c values for 112 and 113 24 11217030 gt 1 0 73 12 07 13 1 gt 0 1 In this case7 24 73 NullspaceU span 1 7 0 7 31 1 with the vectors on the right forming a basis This also tells me that nullityU 2 32 Now l7ll move on to nd a basis for RangeU Just like the last time7 each vector in the null space indicated a dependence between the columns of 12 is 1 0 756 24 20 1 752 0 0 33 1 724 3 0 12 78 1 0 756 i3 20 0 752 1 0 34 1 724 3 0 For each dependency l have to throw out one of the columns of U One way to do this is by throwing out the second column based on 33 and the third column based on 34 This leaves me with 12 RangeU span 756 35 1 where the right side is a basis for RangeU This also tells me that rankU 1 36 You might notice though that things would be a little screwy if I made these choices differently Say I threw out the rst column based on 33 Then it would seem that this breaks the second dependency in 34 But in fact the second and third columns are dependent on one another as well Column 3 78 gtlt Column How can I avoid this problem Well l7ll de nitely have to throw out one column for every element in the basis of NullspaceU The trick is to throw out the columns that correspond to the free variables in the linear system I got while nding the null space In this case the free variables were 02 and 113 so throwing out the second and third columns will do the trick 5 RankNullity Theorem In the past two examples we7ve seen a connection between the null space and the range of a transformation This translates into a really nice link between the rank and the nullity of a transformation which is called the Rank Nullity Theorem I wont actually prove the theorem but we7ve already seen the reasoning behind it For a transformation T Rd a R I know that RangeT is spanned by the columns of T This means that rankT is potentially as large as the number of columns which is d However the value of nullityT tells me how many dependencies there are among the columns of T l have to throw out this many ofthe columns to get a basis for RangeT The number of columns left over tells me rankT To summarize of columns 7 of dependencies rankT 37 And this boils down to d 7 nullityT rankT 38 Just for sake of reference here7s the theorem as concisely as possible RankNullity Theorem Let T Rd 7 R be a linear transformation Then d 7 nullityT rankT 39 6 Geometric interpretation You can make pictures of the null space an range of a transformation using the m le nullrangeplotm in the Lecture 9 section of the course page To use this enter the transformation T and then pass it to the m le For example gtgt T 4 1 3 12 43 1 32 56 521 Will input the matrix from the rst example l7m entering it by columns which is a little extra work but l7m trying to be consistent with the think in terms of the columns77 Vibe Then gtgt nullrangeplot T will show you Rd and RT side by side with NullspaceT in Rd and RangeT in R For this example here7s what I get Rd blue 2 null space R39 red 2 range If you like you can also use VectorSdim from the course page to plot some Vectors For instance say I Want to plot 92 in the left gure and T92 in the right picture To plot 92 in the left picture gtgt subplot121 gt 92 0 1 0 gtgt hold on gtgt Vector3de2 gtgt light The command light isn7t necessary but it adds some shading Which makes the Vectors look nicer Then to plot T92 in the right picture gtgt subplot122 gtgt hold on gtgt Vect0r3dTeZ gtgt light You can also use nullrangeplot for 2 X 2 2 X 3 and 3 X 2 matrices Give it a try

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