Name Michelle Yun AMATH 352 Final Review 1 1st Half Concepts Vocabulary Essential vocabulary 1 Linear combination a linear combination of v1 vn is any weighted sum 1v1 nvn for some 1 n R 2 Linear dependence v1 vn are linearly dependent if 1v1 nvn 0 for some 1 n 3 Linear independence v1 vn are linearly independent if 1v1 nvn 0 only if 1 R n 0 4 Subspace a subset S of vectors is a subspace if S satis cid 28 es the following conditions a u v S for all u v S b u S for all R and u S where 1 n R spans S 5 Span the span of a set of vectors v1 vn is the set of all linear combinations 1v1 nvn 6 Basis a set of vectors v1 vn is a basis for the subspace S if the set if linearly indepdent and 7 Norm a norm is a function f Rn R that satis cid 28 es the following conditions a cid 107 x cid 107 0 for all x Rn b cid 107 x cid 107 0 only when x 0 c cid 107 x cid 107 cid 107 x cid 107 for all R and x Rn d cid 107 x y cid 107 cid 107 x cid 107 cid 107 y cid 107 for all x y Rn a f x y f x f y for all x y U b f x f x for all R and x U 8 Linear function a linear function f U V must satisfy the following conditions 9 Dimension the dimension of a subspace S is the number of vectors in any basis of S 10 Rank the dimension of the column row space of a matrix The rank of a matrix can be found by row reducing the matrix down to echelon form and counting the number of non zero rows 11 Range the span of a matrix s columns Ax x Rn 12 Null space the set of all x such that Ax 0 ie x Rn Ax 0 13 Elementary vectors ei is a vector whose ith entry is 1 with all other entries equal to 0 Concepts and skills 1 Solving systems of linear equations a Put the equations into a matrix and row reduce 1 b There are three possibilities i Inconsistent system row of zeros non zero entry ii Exactly one solution iii In cid 28 nitely many solutions free parameter s 2 Matrix multiplication of the vector x b Dot product of each of A s rows with x 3 Finding the basis for a subspace a Can be thought of as a weighted sum of the columns of A with weights provided by the entries a If given a set of vectors put the vectors into a matrix with the vectors as the rows of the matrix and row reduce the remaining rows form a basis b If given an equation solve the equation for one of the variables and express the solution in parametrized form the number of vectors will be the same as the number of free variables c If given polynomials plug in special values to a general form of the polynomial and di cid 27 erentiate if needed solve for the restricted variables and plug these back into the general form Collect like terms factor and the polynomials multiplying with each variable form a basis for the subspace 4 Finding the basis for the range of a matrix a Row reduce the matrix b The columns with a pivot position in the reduced form correspond to the columns in the original matrix that form a basis just choose these vectors and these will be a basis for the range 5 Finding the basis for the null space of a matrix a Row reduce the matrix b Solve for the restricted variables starting from the bottom by setting each row equal to 0 c Express solutions in terms of the free variables d Remember that the dimension of R A the dimension of N A n 2 2 Least Squares We need least squares because there sometimes aren t exact solutions to Ax b due to measurement 2 cid 107 A x b cid 107 2 by taking the gradient and setting it equal errors or noise In this case we can instead minimize 1 to zero Note that In other words cid 107 cid 70 cid 107 2 2 cid 70 Taking the gradient of the 2 norm of a vector squared is just the vector 2 cid 107 x cid 107 2 then times 2 Ax b is a vector so this same principle applies If g x f Ax b where f x 1 Therefore solving the least squares problem is equivalent to setting AT Ax b 0 and solving This gives the new linear system cid 107 x cid 107 2 x2 cid 107 x cid 107 2 2x1 2xn 1 x2 n 2x g x AT f Ax b cid 107 Ax b cid 107 2 AT 1 2 AT Ax b AT Ax b 0 AT Ax AT b 0 AT Ax AT b In MATLAB this system can be solved with a few lines of code Atb A b AtA A A sol AtA Atb Or equivalently sol A A A b It is also important to realize that if x minimizes 1 2 cid 107 Ax b cid 107 2 then so does x w for any w N A Also if x1 and x2 both minimize 1 2 cid 107 Ax b cid 107 2 then x1 x2 N A Subtracting the two above equations from each other gives AT Ax b AT A x w b AT A x Aw b AT b 0 b AT 0 0 AT Ax1 b 0 AT Ax2 b 0 3 AT Ax1 b AT Ax2 b 0 AT Ax1 b Ax2 b 0 AT Ax1 Ax2 0 AT A x1 x2 0 A x1 x2 0 Since A x1 x2 0 by de cid 28 nition x1 x2 N A 2 1 Applications 2 1 1 Linear Regression In a linear regression setting we have some vector of responses y some vector of predictors x and a matrix of data A Our goal is to predict the values of y based on given values of x 1 a11 1 am1 y Ax cid 15 a1n amn x0 xn cid 15 1 cid 15 m y yi x0 ai1x1 ainxn where cid 15 is a vector with small values that correspond to some noise or variation in the data If yi represents the response of one individual in the study then It is important to append a column of ones to A because we must account for the constant term x0 Here is an example in MATLAB from homework 4 M csvread SwissFertility csv 1 1 b M 1 A M 2 6 A ones 47 1 A Atb A b AtA A A x AtA Atb the first column has responses the rest of the columns have data appending …