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1 BASES, COORDINATES, INEAR MAPS, AND MATRICES Math 223 Bases and Coordinates et be a vector space, and suppose thatb = [v ;:::; v n ] is a basis for This basis induces a linear map from to, namel, given ( ;: ::; ) T in we use ;:::; as coe±cients of v ;::: ; v n in a linear combination In smbols the map is ()! ; A7! v + + v n : This map de nes a linear one-to-one correspondence between and Although the map going from back to doesn't have a nice formula in general, it's a familiar process Namel, given v in, write v as a linear combination of v ;:::; v n The coe±cients de ne an element of These coe±cients are called the coordinates of v relative to the basis B, and are denoted as [v] B More precisel, [v] B is the unique element of that maps to v in () Note that the formula in () can be written as (2) v + + v n = [v ;:::; v n ] A; where we can think of [v ;:::; v n ] as a row matrix whose entries are elements of (as opposed to numbers) If is a subspace of R k and we write v ;:::; v n as column vectors, then [v ;:: :; v n ] is a k n matrix, and the product in (2) is simpl the usual matrix multiplication (Recall that if A is a matrix with n columns and c is a vector in, then Ac is the linear combination of the columns of A using the entries of c as coe±cients) Using this notation, we get that (3) v = [v ;:: :; v n ][v] B for all v2 and 2 3 4[v ;:::; v n ] A5 B = A for all A2 : In this handout, all vectors in R k will be denoted b column vectors

2 (Pause for a moment to think about what these equations mean!) The relationship between the basisband the coordinate sstem it induces on can be summarized b the following diagram x mult b [v;:::;vn] Example et be the subspace of R 3 given b 3x 2 + 4z = The vectors v = (2; 3; ) T and v 2 = (; 2; ) T are in and are independent Since dim is 2, then B = [v ; v 2 ] is a basis for The vector w = (2; 5; 6) T is in, and so is a linear combination of v and v 2 In fact, µ µ w = v + 6v 2 = [v ; v 2 ] ; and so [w] 6 B = : 6 Example 2 et =fp2p 3 j p() = g This is a two-dimensional subspace ofp 3, and a basis for is B = [x ;x(x )] The polnomial q(x) = 2x 2 + x 3 is in, and so is a linear combination of x and x(x ) We have µ µ 3 3 q(x) = 3(x ) + 2x(x ) = [x ;x(x )] ; and so [q] B = : Change of Basis Now suppose thatb = [v ; :::; v n ] and ~ B = [~v ;:: :; ~v n ] are two bases for We then have two coordinate sstems on, and it's good to have a wa to get from one to the other, that is, if ou have the coordinates of v relative to one basis, ou should be able to compute the coordinates relative to the other basis The advantage of representing coordinate sstems as linear maps from to is that these maps will do the work for us The two coordinate maps from to are :! ; [v] B = ~ A ; and ~ :! Rn ; [v] B ~ = A ; ~ where ;:: :; are the coordinates of v relative tob and ~ ;::: ; ~ are the coordinates of v relative to ~ B We can arrange these maps into a diagram as follows & [ ] ~ B! 2

3 The map across the bottom is de ned b going around the diagram: a vector ( ;:::; ) T in the left is taken up to via [ ] B, which is the map in (), and then down to the right via Now ( ;:::; ) T is the coordinate vector relative to B of some vector v in The result of going up to via [ ] B is just v itself! Going down to the right Rn results in the coordinates of v relative to B ~ Thus, the map across the bottom changes the coordinates of v relative tob into the coordinates of v relative to B! ~ We denote it b C B B ~ This map is linear, since it's the composition of linear maps, and so it is given b multiplication b some n n matrix, also denoted b C B B ~ Here is the completed diagram (4) & [ ] ~ B! C B ~ B In this diagram there are two was to take a vector v in to the right, namel go directl there, or go the long wa through the left B the wa the bottom map is de ned, it follows that it doesn't matter which wa ou go, ou get the same thing The equation that sas this is (5) C B ~ B [v] B = [v] ~B : Note that the equation sas that mulitpling the coordinates of v relative to B b the matrix C B ~ B results in the coordinates of v relative to ~ B We know that if A is a matrix with n columns and if e j is the j th vector in the standard basis for, then Ae j is the j th column of A Since [v j ] B = e j, equation (5) gives us (j th column of C B ~ B ) = C B ~ B e j = C B ~ B [v j ] B = [v j ] ~B : In words, the j th column of the matrix C B ~B consists of the coordinates of the jth basis vector ofb relative to the basis ~ B We can write C B ~ B = ([v ] ~B ;:::; [v n ] ~B ) : Here's another wa to see this If v2, then v = [v ;:::; v n ][v] B = [~v ;:::; ~v n ][v] ~B = [~v ;:::; ~v n ]C B ~ B [v] B ; where the last equalit follows from (5) If we let [v] B = ( ;:::; ) T, then we have for all ( ;:::; ) T in It follows that [v ;:::; v n ] A = [~v ;:::; ~v n ]C B B ~ A [v ;::: ; v n ] = [~v ;:::; ~v n ]C B ~ B : 3

4 This last step is similar to the fact that if A and B are matrices of the same dimension and that Ax = Bx for all vectors x, then A = B) Each side of this equation is a list of n vectors in The j th vector on the left is v j The j th vector on the right is [~v ;:::; ~v n ] times the j th column of C B ~ B Thus v j = [~v ;: ::; ~v n ](j th column of C B ~ B ): But this sas that the j th column of C B ~ B is the coordinate vector of v j relative to ~ B b the ~ B version of (3) Finall, note that C ~BB, the matrix that converts coordinates relative to ~ B into coordinates relative tob, should be the inverse of C B ~ B, since it represents the linear transformation along the bottom of (4) from right to left Example (cont) A second basis for =f(x;;z)j 3x 2 + 4z = g is B ~ = [~v ; ~v 2 ], where ~v = (2; 5; ) T and ~v 2 = ( 4; ; 3) T You can show that v = 3 5 ~v 5 ~v 2 and ³ µ v 2 = 2 5 ~v + 5 ~v 2 It follows that C B B ~ = 3=5 2=5 = 3 2 =5 =5 5 Recall that the coordinate vector of w = (2; 5; 6) T relative tob is [w] B = (; 6) T Its coordinate vector relative to B ~ should then be [w] ~B = C B B ~ [w] B = µ µ = : 5 µ 6 Indeed, we have [~v ; ~v 2 ] µ 3 = 3~v + ~v 2 = w: It's important to remember here that the vector w doesn't change What changes is the wa we refer to it The situation is similar to measuring a distance in feet and in meters You get di erent results, not because the distance changes, but because ou are measuring it di erentl The matrix converting coordinates relative to B ~ into coordinates relative tob is C ~BB = ³ C B B ~ = 2 3 Example 2 (cont) A second basis for = fp2 P 3 j p() = g is B ~ = [x ;x 2 ] Since x is the rst element of both bases, its coordinate vector relative to either basis is (; ) T We also have that x(x ) = (x ) + (x 2 ) = [x ;x 2 ] µ ; and so [x(x )] ~B = ( ; ) T The change of coordinates matrix is then C B B ~ = µ 3 From above we have [q] B =, and so [q] ~ B = C B ~B [q] B = µ µ µ 3 = : ³ 4

5 Checking this, µ [x ;x 2 ] = (x ) + 2(x 2 ) = 2x 2 +x 3 = q(x); as needed inear Maps and Matrices et W be another vector space, and suppose ^B = [w ;:: :; w m ] is a basis for W As above, we get the linear map [ ] ^B : W! R m so that the entries of [w] ^B are the coordinates of w2 W relative to ^B Now suppose :! W is linear We get the following diagram of spaces and maps! W [ ] ^B! R m The map across the bottom is de ned b going around the diagram from up to via [ ] B, which is the map in (), then across to W via, then down to Rm via [ ] ^B This map is linear, and so is given b multiplication b some matrix, namel the matrix of relative to the basesb and ^B, denoted b [] B ^B This can actuall be used as the de nition of this matrix! The diagram then becomes (6)! W [ ] ^B! [] B ^B R m With the bottom map de ned b going around the other three sides of the diagram, it follows that if ou start with a vector v in and follow the maps to R m, it doesn't matter which wa ou go, ou get the same thing The equation that sas this is (7) [] B ^B [v] B = [(v)] ^B : The left side of this is the result of going from down to b taking the coordinates of v relative to B, and then across to R m b multipling b the matrix [] B ^B The right side is the result of going across to W b, and then down to R m b taking coordinates of (v) relative to ^B We know that if A is a matrix with n columns and if e j is the j th vector in the standard basis for, then Ae j is the j th column of A Since [v j ] B = e j equation (7) gives us [] B ^B e j = [] B ^B [v j] B = [(v j )] ^B ; 5

6 which sas that the j th column of [] B ^B is [(v j )] ^B In words, the j th column of the matrix of consists of the coordinates of (v j ) relative to the basis ^B This is, of course, how we have computed matrices all along The point is that this form for the matrix is forced upon us as soon as we sa that the map across the bottom in diagram (6) is de ned b going around the other three sides In man applications we work with a linear map from a vector space to itself, that is, W = in (6) In this case we generall take the bases B and ^B to be the same, and the diagram becomes (6 )!! [] B Note that we write [] B instead of [] BB, as the extrab is now redundant The equation that goes around the diagram both was is (7 ) [] B [v] B = [(v)] B ; which replaces equation (7) Example (cont) et P : R 3! R 3 be orthogonal projection onto the x-plane, and let Q: R 3! R 3 be orthogonal projection onto the subspace As ou computed in a homework problem, the matrix for Q± P is 29 µ We de ne to be this composition restricted to, that is, :!, (v) = (Q±P)(v) Appling this to the basis vectors of B we get 2 (v ) = (Q± P )(v ) =µ 3 = v + v 2 = 2 29 (3v + 8v 2 ) and (v 2 ) = (Q± P)(v 2 ) = 29µ = (3v + 8v 2 ) = 6 29 v v 2: (Note that v is in both and the x-plane, so it is unchanged b either projection) The matrix for relative to B is then [] B = ³ 6=29 6=29 Example 2 (cont) De ne :! b (p) = ( x)p (x) You should verif that this is linear Appling to the elements of the basis B we get µ (x ) = ( x)(x ) = x = (x ) + x(x ) = [x ;x(x )] and x(x ) = ( x)(x 2 x) = 2x 2 +3x = (x ) 2x(x ) = [x ;x(x )] The matrix for relative tob is then [] B = : ³ 2 µ : 6

7 How the Matrix for a inear Map Changes Under a Change of Basis If B = [v ;:: :; v n ] and B ~ = [~v ;:::; ~v n ] are bases for, and if :!, we get matrices C B B ~, C ~BB, [] B, and [] ~B How are these matrices related We alread know that the rst two are inverses of each other, so the real question is how the matrix for changes when the basis is changed from B to B ~ The answer lies in putting together a complicated but natural diagram of spaces and maps b combining the previous diagrams (4) and (6 ) along with their analogs for C BB ~ and [] ~B Here the are!! (8) & [ ] ~ B! CB B ~ &! C ~BB! []B! [] ~B The latter two diagrams can be attached along their common edge! []B [ ] ~ Ã! B R n [] ~B Ã! R n We now consider the map across the top from the right to the left It is C BB ~! Similarl, the map across the bottom from left to right is C B ~B Filling these in, the resulting diagram contains all four of those in (8)! C ~BB x (9) Ã! R n [] B [] ~B Ã x! R n C B ~ B Multipling b the matrix [] ~B goes from the upper right to the lower right Another, equivalent, wa to go between these corners is to go around the outside of the diagram counter clockwise, taking, in turn, the arrows marked C ~BB, [] B, and C B ~ B The product of these matrices, in the appropriate order, must equal [] ~B We have [] ~B = C B ~ B [] B C ~BB = C B ~ B [] B C B ~ B : 7

8 We could have discovered this formula b combining the formulas (5) and (7 ) with their analogs for C ~BB and [] ~B, since these formulas contain the same information as the corresponding diagrams In retrospect it's not too di±cult, and it emphasizes that, despite the complexit of the forest of arrows in (9), the matrix multiplications [] B and [] ~B down the left and right sides of the diagram are reall just representations of the map down the middle If v2, then [] ~B [v] ~B = [(v)] ~B = C B ~ B [(v)] B = C B ~ B [] B [v] B = C B ~ B [] B C ~BB [v] ~B : We then have [] ~B [v] ~B = C B ~ B [] B C ~BB [v] ~B for all v2 This is the same as [] ~B A = C B B ~ [] B C ~BB A for all ( ;:::; ) T 2 It follows that [] ~B = C B ~ B [] B C ~BB This should make sense What it sas is that computing (v) b using the coordinates of v and matrix of relative to ~ B is the same as converting to the B coordinate sstem, doing the computation in B coordinates, and then switching back to ~ B coordinates Can ou gure out which corners of the diagram the coordinate vectors [v] ~B, [(v)] ~B [(v)] B, and [v] B in this computation belong to Example (cont) The matrix for relative to ~ B is [] ~B = C B ~ B [] B C B ~ B = 5 µ µ Example 2 (cont) The matrix for relative to ~ B is [] ~B = C B ~ B [] B C B ~ B = µ µ 2 µ 2 = 3 45 µ : 9 88 µ µ 2 = : 2 Robert Foote, November 997 8