Math 416, Spring 2010 Coordinate systems and Change of Basis February 16, 2010 COORDINATE SYSTEMS AND CHANGE OF BASIS. 1.

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1 Math 46 Spring Coordinate systems and Change of asis February 6 COORDINAE SYSEMS AND CHANGE OF ASIS Announcements Don t forget that we have a quiz on hursday and test coming up the following hursday Finishing the last lecture In the previous class we discussed the notion of dimension We ended class by talking about ways for easily computing bases for the image and the kernel of a given matrix We started class today by reviewing an example of this calculation Example Suppose that A efore we started we pointed out that ima R 3 and that kera R 7 o get started on computing bases for the image and the kernel of A we first needed to compute the reduced row echelon form for A We did this in class and got rrefa Now we can see that the second and fourth columns are the pivot columns of rrefa his means that the second and columns of A form a basis for the image So we get basis for ima o find a basis for the kernel we needed to compute linear relations amongst the comlumns of A We got one such linear relation for each non-pivot column of rrefa For instance the fifth column of rrefa is not a pivot column he fact that the fifth column of rrefa is records the linear relation c5 first pivot column + second pivot column c + c 4 his can be reexpressed as c + c 4 c 5 and so pulling off the coefficients in this relation gives the vector acs@mathuiucedu acs/w/math46 Page of 7

2 Math 46 Spring Coordinate systems and Change of asis February 6 I can follow a similar procedure for all the non-pivot columns and each gives me a vector in the kernel of A ogether these vectors form a basis for kera: basis for kera One simple observation you can make from this procedure is the following heorem If A is an r c matrix then dimima # of pivots of rrefa rka dimkera c # of pivots of rrefa c rka he first statement follow because each pivot column of rrefa corresponds to a basis vector for ima Since the dimension of ima counts the number of vectors in a basis for ima we see that the dimension of A is the same as the number of pivots for A For the second statement: each non-pivot column of rrefa corresponds to a vector in the kernel of A Hence the dimension of kera a count on the number of vectors in a basis for kera is just the number of non-pivot columns in rrefa y adding up these two relations we get the famed heorem Rank-Nullity theorem If A is an r c matrix then dimima + dimkera c 3 ases and Coordinates he next topic we re moving into is extrememly powerful but is grounded on a fairly simple idea: instead of preferring the standard basis e e n of R n as the right basis for R n we will examine coordinate systems for an arbitrary basis of R n y allowing ourselves this flexibility we will be able to write the matrix of seemingly complicated linear operators with ease he following theorem is the cog which turns the whole machine heorem 3 If { v v k } is a basis of a subspace V inside of R n then for any v V there are unique c i R so that v c v + + c k vk Proof We already know that is a basis for V and hence forms a spanning set for V So there exist some choice of c c k so that v c v + + c k vk What we need to show is there is only one way of choosing c i For this suppose that we can write v as a linear combination in two ways: v c v + + c k vk d v + + d k vk Our goal is to show that c i d i for all i acs@mathuiucedu acs/w/math46 Page of 7

3 Math 46 Spring Coordinate systems and Change of asis February 6 Using our two expressions for v we have v v c v + + c k vk d v + + d k vk c d v + + c k d k v k ut since v v k is a basis for V it is linearly independent his means that the only way to write as a combination of the v i is by having each coefficient be Applying this to the equation above we have c i d i for all i and hence c i d i he previous theorem justifies the following Definition 3 If v V and is a basis for V then the c i in the previous theorem are called the coordinates of v the vector c v Example is called the -coordinate vector of v c k he vectors 4 3 form a basis for a -dimension space in R3 We ll call this basis What are the -coordinates for the vector o answer this question we simply need to find values of c and c which make the following equation true: c c c Once we ve done this then -coordinates will be c o find these numbers we compute Hence we get rref 3 4? 3 3 heorem 3 For any v V the vector representing v in the standard coordinates is given by v v vk [ v ] acs@mathuiucedu acs/w/math46 Page 3 of 7

4 Math 46 Spring Coordinate systems and Change of asis February 6 Notice that if V R n then the matrix we ve written down is in fact invertible his means if we have a vector v written in standard coordinates we can find the vector [ v ] as [ v ] v v n v Example In our previous example we saw that with Hence this means that we have his theorem is probably best explained graphically In the figure below we see that the way from move be- tween the two coordinate systems is by applying the matrices S and S where of course S v v n S basis standard basis S Figure Moving between coordinate systems 4 Operators in new coordinate systems We said the motivation of looking at new coordinate systems is that sometimes a linear operator can be understand very easily in one coordinate system If we can write down the linear operator as a matrix in the new coordinate system then we might be able to write down the matrix of the linear transformation in the standard basis more easily o make these notions precise we begin by introducing the following Definition 4 Let be a linear transformation : R n R n and let be a basis of R n he -matrix for is the n n matrix so that [ x ] [ x ] his means that the -matrix of is the matrix which acts on -coordinate vectors and spits out -coordinate vectors according to the action of the linear operator acs@mathuiucedu acs/w/math46 Page 4 of 7

5 Math 46 Spring Coordinate systems and Change of asis February 6 heorem 4 he -matrix of is given by [ v ] [ v n ] Proof We didn t prove this theorem in class but the proof is analagous to how we showed that the column vectors of a matrix in the standard basis are given by the vectors e i again in the standard basis Example Let L be the line spanned by Find a basis of R so that the -matrix of proj L is nice Solution Projection onto the line L acts really simply on the vector and on the vector : since the first vector is in L we know proj L keeps this vector fixed and since the second vector is orthogonal to L we know that proj L kills this vector his means that the action of proj L on the basis { } is given by the -matrix [ ] [ ] [ ] [ ] 5 Moving between coordinate systems Now that we know how to write linear operators in two coordinate systems we would like to be able to move between these two worlds After all the point of doing all this coordinate stuff was that some operators are easier to see in coordinate systems which are not given by the standard basis vectors If we can t find the standard matrix representation for an operator given the -matrix representation for the operator what good would all this work be? heorem 5 Let be an operator and let A be the matrix of in standard coordinates while is the matrix of in -coordinates hen A SS where S is the matrix whose columns are the basis vectors in We also have the equality S AS Proof by sketch he second statement follows from the first so this is the only fact we need to justify he sketch below is much like the sketch we used to visualize transformations between coordinates before except now we ve included the action of the operator on R n in the two different coordinate systems o see that this is a proof of the theorem just notice that the action of on the left given by A is equivalent to changing coordinates to multiplying by S acting by in the basis which means the matrix and then switching back to standard coordinates multiplying by S acs@mathuiucedu acs/w/math46 Page 5 of 7

6 Math 46 Spring Coordinate systems and Change of asis February 6 S basis standard basis multiply by multiply by A S Figure ransforming between matrix under standard basis and -basis Example Let be the operator which rotates vectors through an angle of θ degrees around the line spanned by /3 /3 /3 o write this matrix out in standard coordinates we ll first write out this matrix in a convenient basis for R 3 then change basis First note that the vector v /3 /3 doesn t get moved by the operator For this reason the action of /3 is especially simple on this vector so we want it in our basis So we get /3 /3 v v 3 /3 where it s our job to choose nice vectors v and v 3 he way I ll find them is by realizing that in the plane perpendicular to v vectors rotate in precisely the same they would rotate in two dimensional space So if we can find two vectors v and v 3 which form a suitably nice basis for this plane then I should be set What does suitably nice mean in this case? Well in the case of R rotations we know that a rotation by θ degrees is represented by the matrix cosθ sinθ sinθ cosθ he basis which represents this matrix is the standard basis so we need a basis for this plane which behaves like the standard basis In this problem that will mean we need our basis to consist of two perpendicular vectors with the same length First I ll pick one vector in the plane simply by eyeballing it: we saw in class that the vector v 4 is perpendicular to v so it lies on the plane o find the next vector v 3 we want it to be perpendicular to v so it sits in the appropriate plane plus perpendicular to v so we can easily determine what rotation does to v and v 3 For this we ll let v 3 v v his is the cross product of v and v and you can see how this is defined in the book When you do the computations you wind up getting v Now that I ve compiled my basis I need to determine what this linear operator does to the basis We ve already said that the operator fixes the vector v : v v acs@mathuiucedu acs/w/math46 Page 6 of 7

7 Math 46 Spring Coordinate systems and Change of asis February 6 Now for the vectors v and v 3 we said the rotation performed by the operator simply acts like a typical rotation in R In R rotation by θ does the following rotation by θ sends to the vector cosθ rotation by θ sends to the vector sinθ + sinθ + cosθ Our basis for the plane in R 3 will now just play the role of the standard basis in R : the operator sends v to cosθ v + sinθ v 3 the operator sends v 3 to sinθ v + cosθ v 3 All in all then the -matrix for our operator is given by [ v ] [ v ] [ v 3 ] [ v ] [cosθ v + sinθ v 3 ] [ sinθ v + cosθ v 3 ] cosθ sinθ sinθ cosθ Once we have this we can write the matrix for this operator in the standard basis as v v v3 cosθ sinθ v v v3 sinθ cosθ his example brings up an important concept: similarity Definition 5 o n n matrices A and are similar if there exists an invertible matrix S so that SAS One can show that similarity is an equivalence relation which means it is a weakened version of equivalence between matrices asically two matrices being similar means that their actions are geometrically the same but just defined on different bases acs@mathuiucedu acs/w/math46 Page 7 of 7