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# Midterm 1 Study Guide AMATH 352

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This 8 page Study Guide was uploaded by Jeremy Dao on Monday February 1, 2016. The Study Guide belongs to AMATH 352 at University of Washington taught by Aleksandr Y. Aravkin in Winter 2016. Since its upload, it has received 278 views. For similar materials see APPLIED LINEAR ALEGRA AND NUMERICAL ANALYSIS in Applied Mathematics at University of Washington.

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Date Created: 02/01/16

AMATH 352 Midterm Review Practice problems and concept list. (1) Vocabulary (a) Linear combination • A Linear Combination of vectors v ,v ,v ,...,v in V is any vector w 1 2 3 n such that w = ↵1 1+ ↵ 2 2 ··· + ↵ n n where ↵ 1↵ 2...,↵ nin R (any real number). Basically, w is a sum or combinationof vectors in V . 1 2 1 2 • Example: v1= 3 ,v2= 5 , then 6 3 2 5 is a linear combination of 1 ,2 . • Note that not all vectors in V need to be used, i.e some ↵’s can be equal to zero. • Note that if we take 1 = ↵2= ··· = ↵n=0 , we will get 0v +0 v + ··· +0 v =0 for ANY v ,v ,...,v 1 2 n 1 2 n (b) Linear dependence • v1,v2,...,n are called linearly dependent if there exists any linear combination besides all 0’s to make 1 1 + ↵2 2+ ··· + ↵n n=0 , meaning ↵ 1 ↵ =2··· = ↵ =0ncannot be the only linear combination that makes the ove staemet true. 1 1 1 1 0 • Example: v 1 2 ,v2= 2 , then 11 +1 v2= 2 + 2 = 0 . In this case,1↵ = 2 =0 . Thus, v1,v2are linearly dependent. • Note that this implies that one vector is a copy or some scalar multiple of another. In the case above,2v = 1 ,sothe v1and v2were not unique. • If 2 or more vectors in a set are linearly dependent, then the whole set is linearly dependent as well (c) Linear independence • v ,v ,...,v are called linearly independent if they are not dependent, 1 2 n or precisely the only linear combination such that ↵ v + ↵ v + ··· + ↵ v =0 is ↵ = ↵ = ··· = ↵ =0 1 1 2 2 n n 1 2 n • Example: v = 1 ,v = 1 1 2 2 0 ↵ 1 + 1 = 0 ! ↵ + 2↵ =0 ! ↵ =0 ,↵ + =0 ! =0 2 2 0 2↵ 0 Thus only ↵ = =0 works andv ,1 2 are linearly independent. • We can also look at the linear dependence of functions. Ex. are ,ex independent? x x We want↵e +e =0 for all x. So we can simply plug in some points to get equations and solve for ↵ and . 1 2 (x = 1) ↵ + =0 =0 1 2 (x = 1)↵e + e =0 ! ↵ + e =0 ↵ =0 Since ↵ = =0 ,ex and e are linearly independent functions. • Theorem: If functions f 1x),f 2x),...,f n(x) are linearly independent, then so are their derivatives1f (x)2f (x),...n(x). So to solve the example we can also take the derivative: x x x ↵e + e =0 . Combining the two equations we obtain 2e =0 ! =0 . Thus ↵ =0 as well and the two functions are linearly independent. (d) Subspace (of a linear space) • V is a subspace of W is V is a subset of W (V is in W)and V is a linear space. • V is a linear space if it satisﬁes the follow 2 properties: (i) Closure under addition: If 1 ,2 are in V , then1v +2v is also in V . (ii) Closure under scalar multiplication: I1 v is in V , then1↵v is also in V for any real ↵ • These two properties imply some simpler ones that are useful for easily showing that a space is not a linear space. For example, notice that if for closure under scalar multiplication we let ↵ =0 , we get the zero vector. Thus the zero vector MUST be in every linear space. This is a good ﬁrst check when determining if a space is a linear space or not. • Important Note: Every subspace of a linear space is itself a linear space • Functions from R ! R form a linear space (e) Span (of a set of vectors) • The span of vectors v 1v 2...,v n is the set of all linear combinations ↵1 1+ ↵ 2 2 ··· + ↵ vn nr any ↵ ,↵1,.2.,↵ n2 R • In English, basically the span is every possible combination of the vectors. • Note that the span of any vector(s) in R is a subspace of R (f) Basis (of a subspace) • Absi B = {v 1...,v k} for a subspace S is a minimal set of vectors v1,...,vk so that S =span{ v1,...,vk}. • "Minimal" means v ,1..,v kare linearly independent. 8 2 3 2 3 2 3 9 2 3 2 3 < 1 2 0 = 1 0 • Example, let S = span 4 1 4,524 , 0 , then 415 , 0 form a basis : 0 0 1 ; 0 1 2 3 2 3 1 2 for S because they are linearly independent. Notice that4 1 and 4 2 0 0 were multiples of each other and were thus linearly dependent, or not "minimal". So the original set was not a basis. (g) Norms • A norm is an abstract notion of a certain kind of distance. 3 • kxk is a function from R ! R and is a norm when: n (i) kxk 0 for all x 2 R (ii) kxk =0 if and only if x =0 (iii) k↵xk = |↵|kxk. If we scale x by ↵, the norm is also scaled by ↵. Notice that we take the absolute value of ↵ because the norm(length) is independent of direction. (iv) kx+ykk xk+kyk for all x,y 2 R called the "triangle inequality". • The 1,2,3,..., 1 norms (called the p-norms) are given by the equation P n p p kxk p= ( i=1|xi| ) . P • For p =1 we get the 1-norm, which is just kxk 1 n |xi|,thesumof i=1 the absolute value of all the elements of x. • For p =2 we get the 2-norm, which is our traditionally understanding of the norm in that ip gives the Euclidean distance (or the magnitude of the vector). kxk2= x + x + ··· + x . Note that if the subscript is 1 2 n omitted from the norm (kxk) this is generally taken to be the 2-norm. We can also deﬁne the 2- norm in terms of the inner dot product: p kxk = x · x • The 1-norm is deﬁned as kxk 1 = max(|x |1..., |xn|) the maximum of the absolute value of the elements. • Note that these are NOT the only norms. Any function that satisﬁes the above 4 properties is considered a norm. Look at the review sheet problems for some examples of this. • There is also the notion of norm balls: The p-norm ball B = {x : kxk 1} is the surface which contains all vectors whose p p 2 p-norm is less than 1. So the 2-norm ball in R is a circle of radius 1 (sphere of radius 1 if in R ). Here are some pictorial depictions of the p-norm balls: (h) Linear functions • A linear function should really be considered as a Linear Map. A linear map f from V to W satisﬁes f(↵v +v )= ↵f(v )+ f(v ) (the linear 1 2 1 2 map has linearity, meaning that it is closed under addition and scalar multiplication. 4 • Consider the maps from Rn to R . We can deﬁne them as an n ⇥ m m⇥n matri2 A, where A 2 R 3 as a rectangular array from real numbers. a11 a12 ··· a1n 6 .. .. 7 6a21 . . a2n7 th th A = 4 . .. .. . 5 aijis the entry in the row and the j . . . . am1 ··· ··· amn column of A. • We can use the idea of linear maps to deﬁne matrix-vec2or3multiplica- 2 3 v1 1211 6 7 tion in 2 ways. For example, let A =42130 5,v = 6v27 with 4v35 1001 v4 m⇥n 4 A 2 R ,v 2 R We can write Av to mean: 2 3 2 3 v1 2 3 1211 6 7 r1· v (i)4 2130 5 6v27 = 4 r · v, where r denotes the i row of A. 4v35 2 i 1001 v r3· v 4 This is expressing Av in terms of the dot product of the rows of A and v. 2 3 2 3 2 3 2 3 2 3 2 3 1211 v1 1 2 1 1 4 5 6v27 4 5 4 5 4 5 4 5 (ii) 2130 4v 5 = v1 2 + v2 1 + v3 3 + v4 0 1001 3 1 0 0 1 v4 = v 1 1 v 2 2 v 3 3 v 4 4 where cidenotes the i column of A. This is expressing Av as a linear combination of columns of A with weights given by entries of v. • Any linear function F from R to R can be written as F(x)= Wx for some matrix W. • Never forget to think of A 2 Rm⇥n as a map from R ! R . m (i) Dimension • The dimension of8a2 3 2 3 2 3 t9e numbe82 3 2 3to9s in any basis of S < 1 2 0 = < 1 0 = 4 5 4 5 4 5 4 5 4 5 • Thus if S = span: 1 , 2 , 0 ; ,nd : 1 , 0 ; forms a basis 0 0 1 0 1 for S, then the dimension of S is 2. (j) Range (of a matrix) • The range of a matrix A 2 Rn is the span of the columns of A. • To state this explicitly: For a matrix A, b is in its range if and only if there exists a x such that Ax = b • Note that since A 2 R⇥n, the columns and thus the range is in R 5 • The range of A is a subspace of Rm • The range of A is denoted range(A) (k) Rank (of a matrix) • The rank of a matrix A is the dimension of range(A) • The rank of A is denoted rank(A) • The row rank of A is equal to the column rank of A (l) Nullspace (of a matrix) m⇥n • The nullspaceof A 2 R is the set of vectors x satisfying Ax =0 . This set is denotes nullspace(A). • The nullspace of A is a subspace of R . • Note: Since Ax =0 , by the row deﬁnition of matrix-vector multiplication (see Linear Functions above), x is orthogonal to the rows of A (since the dot product of the rows of A and x is 0). • null(A) is the dimension of the nullspace of A • Rank-Null Theorem: Let A 2 R m⇥n , then: n (the number of columns) = rank(A) + null(A)) n (m) Elementary vectors e 1...,e n in R • The elementary vectors e 1...,e n in R are vectors in R with 1 in the ithspot and zeroes everywhere else. 2 3 2 3 1 0 6 0 617 6 0 607 • For example, e1= 6 7 ,e2= 6 7 4 . 4 . 0 0 • An important thing to notice is that the elementary vectors form a basis n for R ⇥ ⇤ • The matrix whose columns are the elementary vectors, me2ning e1 e23 ··· en 10 ··· 0 6 7 601 ··· 07 is called the identity matrix. This is denoted I n = 6. . .. . 4. . . . 00 ··· 1 where there are n 1’s along the diagonal and zeroes everywhere else. – The main property of the identity matrix is that any square matrix A times the identity produces A (hence the name identity). Explic- n⇥n itly, this means An = InA = A (remember that A 2 R in order for the multiplication to make sense. See matrix multiplication be- low for more detail) (2) Concepts and skills n m (a) Equivalence of linear functions from R to R and matricies • We can write a system of linear functions as a function of matrices. 6 • For example, take the system: 2x + y + z =2 x y z =1 x +2 y +0 z =3 2 32 3 2 3 21 1 x 2 This is equivalent to:1 1 54 y = 415,or Ax = b, where 12 0 z 3 2 3 2 3 2 3 21 1 x 2 41 1 1 5 4 y 415 A = ,x = ,b = 12 0 z 3 (b) Solving linear systems of equations • An easy way to solve linear systems is put them into matrix form then perform row operations. • For example, take the system: 2x + y + z =2 x y z =1 x +2 y +0 z =3 Putting this into an augmented matrix (only take the coeﬃcients and 2he solutions) 3e get: 21 1 2 4 5 1 1 1 1 which we can then perform row operations on to get 12 0 3 the matrix to be triangular. 2 3 2 1 1 3 2 1 1 3 21 1 2 R1= R1 1 2 2 1 R =R R 1 2 2 1 4 1 1 1 1 5 ! 4 1 1 1 1 5 ! 4 01 1 0 5 R3=R3R1 3 1 12 0 23 312 0 3 0 2 2 2 1 1 1 1 x + y + z =1 R3=R31.5R2 4 2 2 5 2 2 ! 01 1 0 ! y + z =0 00 2 2 z = .25 We can then just back-substitute to solve for x,y,z. • This process is called Gaussian Elimination, or row reduction. • When performing Gaussian elimination, there is the possibility that we would end up with free variables, which we can choose to be anything. For example, take the system: x x + x =0 2 1 11 0 3 1 2 3 2x1+ x2 x 30 ! 4 21 1 0 5 x +2 x 2x =0 12 2 0 1 2 3 7 2 3 1 11 0 After performing Gaussian elimination we obtain: 4 03 3 0 5 00 0 0 Notice that the last row is simply 0=0, so we only have 2 meaningful equations. We don’t have an equation for just x 3 (as we did in the triangular case above), so we will make x a free variable. If we let 3 x 31 , we get 3x 2 3x ! x3=1 an1 2 3 x x 10 . 2 3 0 0 So a solution is 415 ,utlo c 415 for any c 2 R is a solution as well. 1 ⇢ 0 Thus, the solution space is span 1 frm [o] • We can tell which variables are free and which are ﬁxed by looking at the columns of the matrix. In the above example, we can see that x a1d x 2re the ﬁrst non-zero elements in rows 1 and 2 respectively. Thus, x 1nd x a2e ﬁxed variables. Whatever is left (in this case x ) 3re free variables. • Note: Row operations or even column operations do not aﬀect the row space, column space, or the dimensions of either of the matrix. • A system of equations is consistent if it has at least one solution, i.e b is in the range of A. • We can get a hint about the solutions of Ax =0 by looking at the columns of A. – If rank(A) = n, then the columns of A are linearly independent, thus the only solution is x = 0. – If m<n ,r kA) m<n , so there must be at least one other non-zero solution. (c) Matrix multiplication • Using our deﬁnitions of matrix-vector multiplication (see Linear Maps above) we can deﬁne matrix-matrix multiplication. • Let A 2 R m⇥n2 and B 2 R n⇥k,3then2AB is: 3 2 3 . . . . . . . . 6 . . . .7 6 . . . . 7 4 A 5 6 b b ··· b 7 = 6 Ab Ab ··· Ab 7 4 .1 .2 . .5 4 .1 .2 . .n 5 . . . . . . . . • Note that not all matrices can be multiplied by each other! Matrix- matrix multiplication only makes sense if the inner dimensions of the matrices are equal. In the above example, A had dimensions m ⇥ n and B and dimensions n ⇥ k. Notice that m ⇥ n ! n ⇥ k the inner dimensions (n) are equal. The solution will have the outer dimensions of the 2 matrices (m ⇥ k) (d) Finding a basis for a subspace 8 • First ﬁnd a set of vectors that span the subspace, then put them into a matrix A. Row reduce the matrix, then take the ﬁrst columns containing the ﬁrst non-zero elements of each row. The corresponding columns in the original matrix A forms a b⇢ pace. 1 2 1 1 • Example. Find a basis for S = spa2 , 4 , 3 , 2 A = 1211 ! 1211 The ﬁrst and second columns 2434 reduce 0012 have the ﬁrst non-zero element of each row, thus the ﬁrst and second ⇢ column of A form a basis for the subspace. A basis for , is 2 3 • Note that we can do this because row operations do not change the space or even the dimension. (e) Finding a basis for the range of a matrix • Since the range of a matrix A is a subspace, we can follow the above procedure. • Just row reduce A, take the columns containing that ﬁrst non-zero ele- ment of each row, and the corresponding columns in A form a basis for the range. (f) Finding a basis for the nullspace of a matrix • Solve the equation Ax =0 and examine the result. The number of free variables in the solution will correspond to the number of vectors in the basis. • xamle: o fre variables 12 10 ! Thus, the vectors are linearly independent, and only 34 01 the zero vector is in the nullspace • Example: 1 free variable ⇢ 12 ! 12 . Thus 1 = 2x 2 so the nullspace = span 1 . 12 00 2 1 free variable means 1 vector in the span. • 2xample3 2 f2ee vari3bles 2 3 2 3 123 123 2 3 4123 5 ! 4000 5,so x1= 2x 2 3x 3 4 1 5x 2 4 0 5x3, 123 000 0 1 2 3 2 3 2 3 so4 1 5,4 0 5 forms a basis. 2 free variables, 2 vectors in the basis. 0 1

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