THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Summer Shool Math26 28 Tutorial week s You are given the following data points: x 2 y 2 Construt a Lagrange basis {p p p 2 p 3 } of P 3 using the x values from the data set Hene find the unique ubi polynomial px that fits the data exatly Estimate the value of y when x 2 By definition the vetors in the Lagrange basis of P 3 are as follows: xx x 2 p x 2 xx x 2 6 x + x x 2 p x + 2 x + x x 2 2 x + x x 2 p 2 x + 2 xx + x 2 2 x + x x p 3 x 2 + 2 2 xx + x 6 Hene the polynomial that fits the data is px p x + p x 2p 2 x + p 3 x This simplifies to px x + When x 2 we estimate that y p 2 8 2 + 7 8 875 2 The matrix J is a redued row ehelon form of the matrix A A 2 5 2 5 3 2 7 8 3 8 9 9 3 7 7 9 5 8 J 29 8 2 a Find a basis for the olumn spae of A b Find a basis for the null spae of A Verify that the sum of the rank of A and the nullity of A is the number of olumns of A a The first three olumns of J are a basis for the olumn spae of J so the first three olumns of A are a basis for the olumn spae of A So a basis for the olumn spae of A is } { 2 8 7 25 9 7 3 7 9 9
Summer Shool Math26 Tutorial week s Page 2 b Let x x be in the null spae of A As NullA NullJ looking at J we see that and are free variables and that x 29 8 + 2 Hene x x 29 8 +2 8 x x 5 for all R Consequently a basis for the null spae of A is { 8 29 2 } 29 2 + The rank of A is the dimension of the olumn spae of A so ranka 3 The nullity of A is the dimension of the null spae of A so nullitya 2 Hene ranka + nullitya 3 + 2 5 whih is the number of olumns of A as it must be 3 a Let A Show that geometrially the null spae of A orresponds to the points on the z axis and that the olumn spae orresponds to the points on the xy plane b Find a 3 3 matrix whose null spae orresponds to the x axis and whose olumn spae orresponds to the yz plane Find the null spae and olumn spae of the 3 3 identity matrix x a The null spae of A onsists of all olumn vetors suh that x and an take any real value Therefore x is in the null spae if and only if x where R Geometrially this orresponds to all the points on the z-axis in three dimensional spae R 3 A basis for the olumn spae of A is { } ab So a typial element of the olumn spae has the form where a b R Geometrially the olumn spae orresponds to all points on the plane z ; that is the points on the xy-plane b The null spae of the matrix is the subspae of R 3 onsisting of the points on the x-axis The olumn spae is the subspae of points on the yz-plane Sine Ix x for all x R 3 the null spae of the identity matrix I is the zero vetor subspae {} of R 3 As I the standard basis vetors of R 3 are a basis for the olumn spae of I Hene the olumn spae of I is the whole of R 3 Whih of the following funtions are linear transformations a T : P 2 R 2 ; a + bx +
Summer Shool Math26 Tutorial week s Page 3 b T : P 2 P ; px px 2 T : R 2 a R; b a 2 + b 2 d T : P 2 P 2 ; px px + e T : P 2 R; px p3 a Let a + bx + and d + ex + f be two polynomials in P 2 Let k R and T a + bx + + d + ex + f T a + d + b + ex + + f a+d+b+e +f + d+e f T a + bx + + T d + ex + f T ka + bx + T ka + kbx + k ka+kb k k kt a + bx + Hene T is a linear transformation as it preserves addition and salar multipliation b T is not a linear transformation beause T 2x 2 2T T is not a linear transformation beause T 2 but T + T + 2 That is T T + T + T d Let p q be any two polynomials in P 2 and let k R T p + qx p + qx + T kpx px + + qx + T px + T qx kpx + kt px Hene T is a linear transformation e Let px qx P 2 k R T px + qx p3 + q3 T px + T qx T kpx kp3 kt px Hene T is a linear transformation 5 Suppose that V and W are vetor spaes and that T : V W is a linear transformation a Using indution on n show that T r v + r 2 v 2 + + r n v n r T v + r 2 T v 2 + + r n T v n for any r r n R and any v v n V b Now suppose that S : V W is another linear transformation and that {v v n } is a basis of V Show that S T if and only if Sv i T v i for i n
Summer Shool Math26 Tutorial week s Page a Sine T is a linear transformation T r v r T v Hene the statement is true when n Now assume by indution that Then T r v + + r n v n r T v + + r n T v n T r v + + r n+ v n+ T r v + + r n v n + r n+ v n+ T r v + + r n v n + T r n+ v n+ by linearity T r v + + r n v n + r n+ T v n+ r T v + + r n T v n + r n+ T v n+ where the last line follows by indution b If S T then Sv T v fort all v V ; in partiular Sv i T v i for i n Conversely suppose that Sv i T v i for i n Let v be an arbitrary vetor in V Sine {v v n } is a basis of V we an write v r v + + r n v n for some r i R Then using part a twie we have Hene S T as required Sv Sr v + + r n v n r Sr v + + r n Sv n r T r v + + r n T v n T r v + + r n v n T v 6 The matries below are the augmented matrix and a redued row ehelon form of the augmented matrix for a system of linear equations Ax b A b 3 2 2 7 2 9 25 3 7 7 and J 2 3 a Find the null spae of A and solve the system Ax b b Write down two other onsistent systems of linear equations whih have A as their same oeffiient matrix That is find olumn vetors b and b 2 suh that the two systems of equations Ax b and Ax b 2 both have solutions Using parts a and b and without row reduing write down the solution of the two systems of linear equations that you gave in b a Now if x x x is in the null spae of A then + The null spae of A has dimension 3 and basis + + { }
Summer Shool Math26 Tutorial week s Page 5 If x satisfies Ax b then x x 2 + 3 x5 2 3 + + + where an have any real value b Ax b is a onsistent{ system if and } only if b is in the olumn spae of A A basis for the 2 37 olumn spae of A is sine the first and third olumns of J form a basis 3 for the olumn spae of J We need only selet two different values of b in the olumn spae of A say b 2 + 37 3 9 7 and b 2 2 37 3 2 5 9 Then Ax b and Ax b 2 are onsistent systems In the augmented matrix A b the olumn vetor b is the sum of olumns and 2 of A Therefore the same dependeny relationship exists in the olumns of the redued row ehelon matrix of A b Therefore a row redued form of the Augmented matrix A b is The solution is therefore x + + + The olumn vetor b 2 is the differene of the first and third olumns of A Therefore a row redution of A b 2 is the augmented matrix The solution is therefore x + + Note that there are infinitely many values of b and b 2 that ould be hosen inluding any of the olumns of A itself