线性代数第一章英文题目

22 Chapter 1. Introduction to Vectors 1.3 Matrices This section is based on two carefully chosen examples. They both start with three vectors. I will take their combinations using matrices. The three vectors in the first example are u;v, and w: First example u D 2 4 1�1 0 3 5 v D 2 4 01 �1 3 5 w D 2 4 00 1 3 5 . Their linear combinations in three-dimensional space are cuC dv C ew: Combinations c 2 4 1�1 0 3 5 C d 2 4 01 �1 3 5 C e 2 4 00 1 3 5 D 2 4 cd � c e � d 3 5 . (1) Now something important: Rewrite that combination using a matrix. The vectors u;v;w go into the columns of the matrix A. That matrix “multiplies” a vector: Same combination is now A times x 2 4 1 0 0�1 1 0 0 �1 1 3 5 2 4 cd e 3 5 D 2 4 cd � c e � d 3 5 . (2) The numbers c; d; e are the components of a vector x. The matrix A times the vector x is the same as the combination cuC dv C ew of the three columns: Matrix times vector Ax D 2 4 u v w 3 5 2 4 cd e 3 5 D cuC dv C ew . (3) This is more than a definition of Ax, because the rewriting brings a crucial change in viewpoint. At first, the numbers c; d; e were multiplying the vectors. Now the matrix is multiplying those numbers. The matrix A acts on the vector x. The result Ax is a combination b of the columns of A. To see that action, I will write x1; x2; x3 instead of c; d; e. I will write b1; b2; b3 for the components of Ax. With new letters we see Ax D 2 4 1 0 0�1 1 0 0 �1 1 3 5 2 4 x1x2 x3 3 5 D 2 4 x1x2 � x1 x3 � x2 3 5 D 2 4 b1b2 b3 3 5 D b . (4) The input is x and the output is b D Ax. This A is a “difference matrix” because b contains differences of the input vector x. The top difference is x1 � x0 D x1 � 0. 1.3. Matrices 23 Here is an example to show differences of numbers (squares in x, odd numbers in b): x D 2 414 9 3 5 D squares Ax D 2 41 � 04 � 1 9 � 4 3 5 D 2 413 5 3 5 D b . (5) That pattern would continue for a 4 by 4 difference matrix. The next square would be x4 D 16. The next difference would be x4 � x3 D 16 � 9 D 7 (this is the next odd number). The matrix finds all the differences at once. Important Note. You may already have learned about multiplying Ax, a matrix times a vector. Probably it was explained differently, using the rows instead of the columns. The usual way takes the dot product of each row with x: Dot products with rows Ax D 2 4 1 0 0�1 1 0 0 �1 1 3 5 2 4 x1x2 x3 3 5 D 2 4 .1; 0; 0/ � .x1; x2; x3/.�1; 1; 0/ � .x1; x2; x3/ .0;�1; 1/ � .x1; x2; x3/ 3 5 : Those dot products are the same x1 and x2 � x1 and x3 � x2 that we wrote in equation (4). The new way is to work with Ax a column at a time. Linear combinations are the key to linear algebra, and the output Ax is a linear combination of the columns of A. With numbers, you can multiply Ax either way (I admit to using rows). With letters, columns are the good way. Chapter 2 will repeat these rules of matrix multiplication, and explain the underlying ideas. There we will multiply matrices both ways. Linear Equations One more change in viewpoint is crucial. Up to now, the numbers x1; x2; x3 were known (called c; d; e at first). The right hand side b was not known. We found that vector of differences by multiplying Ax. Now we think of b as known and we look for x. Old question: Compute the linear combination x1uC x2v C x3w to find b. New question: Which combination of u;v;w produces a particular vector b? This is the inverse problem—to find the input x that gives the desired output b D Ax. You have seen this before, as a system of linear equations for x1; x2; x3. The right hand sides of the equations are b1; b2; b3. We can solve that system to find x1; x2; x3: Ax D b x1 D b1 �x1Cx2 D b2 � x2 Cx3D b3 x1 D b1 Solution x2 D b1 C b2 x3 D b1 C b2 C b3: (6) Let me admit right away—most linear systems are not so easy to solve. In this example, the first equation decided x1 D b1. Then the second equation produced x2 D b1 C b2. The equations could be solved in order (top to bottom) because the matrix A was selected to be lower triangular. 24 Chapter 1. Introduction to Vectors Look at two specific choices 0; 0; 0 and 1; 3; 5 of the right sides b1; b2; b3: b D 2 400 0 3 5 gives x D 2 400 0 3 5 b D 2 413 5 3 5 gives x D 2 411C 3 1C 3C 5 3 5 D 2 414 9 3 5 : The first solution (all zeros) is more important than it looks. In words: If the output is b D 0, then the input must be x D 0. That statement is true for this matrix A. It is not true for all matrices. Our second example will show (for a different matrix C ) how we can have Cx D 0 when C ¤ 0 and x ¤ 0. This matrix A is “invertible”. From b we can recover x. The Inverse Matrix Let me repeat the solution x in equation (6). A sum matrix will appear! Ax D b is solved by 2 4 x1x2 x3 3 5 D 2 4 b1b1 C b2 b1 C b2 C b3 3 5 D 2 4 1 0 01 1 0 1 1 1 3 5 2 4 b1b2 b3 3 5 . (7) If the differences of the x’s are the b’s, the sums of the b’s are the x’s. That was true for the odd numbers b D .1; 3; 5/ and the squares x D .1; 4; 9/. It is true for all vectors. The sum matrix S in equation (7) is the inverse of the difference matrix A. Example: The differences of x D .1; 2; 3/ are b D .1; 1; 1/. So b D Ax and x D Sb: Ax D 2 4 1 0 0�1 1 0 0 �1 1 3 5 2 4 12 3 3 5 D 2 4 11 1 3 5 and Sb D 2 4 1 0 01 1 0 1 1 1 3 5 2 4 11 1 3 5 D 2 4 12 3 3 5 Equation (7) for the solution vector x D .x1; x2; x3/ tells us two important facts: 1. For every b there is one solution to Ax D b. 2. A matrix S produces x D Sb. The next chapters ask about other equations Ax D b. Is there a solution? How is it computed? In linear algebra, the notation for the “inverse matrix” is A�1: Ax D b is solved by x D A�1b D Sb: Note on calculus. Let me connect these special matrices A and S to calculus. The vector x changes to a function x.t/. The differences Ax become the derivative dx=dt D b.t/. In the inverse direction, the sum Sb becomes the integral of b.t/. The Fundamental Theorem of Calculus says that integration S is the inverse of differentiation A. Ax D b and x D Sb dx dt D b and x.t/ D Z t 0 b: (8) 1.3. Matrices 25 The derivative of distance traveled .x/ is the velocity .b/. The integral of b.t/ is the distance x.t/. Instead of adding CC , I measured the distance from x.0/ D 0. In the same way, the differences started at x0 D 0. This zero start makes the pattern complete, when we write x1 � x0 for the first component of Ax (we just wrote x1). Notice another analogy with calculus. The differences of squares 0; 1; 4; 9 are odd numbers 1; 3; 5. The derivative of x.t/ D t2 is 2t . A perfect analogy would have produced the even numbers b D 2; 4; 6 at times t D 1; 2; 3. But differences are not the same as derivatives, and our matrix A produces not 2t but 2t � 1 (these one-sided “backward differences” are centered at t � 1 2 ): x.t/� x.t � 1/ D t2 � .t � 1/2 D t2 � .t2 � 2t C 1/ D 2t � 1: (9) The Problem Set will follow up to show that “forward differences” produce 2t C 1. A better choice (not always seen in calculus courses) is a centered difference that uses x.t C 1/ � x.t � 1/. Divide x by the distance t from t � 1 to t C 1, which is 2: Centered difference of x.t/ D t2 .t C 1/ 2 � .t � 1/2 2 D 2t exactly. (10) Difference matrices are great. Centered is best. Our second example is not invertible. Cyclic Differences This example keeps the same columns u and v but changes w to a new vector w�: Second example u D 2 4 1�1 0 3 5 v D 2 4 01 �1 3 5 w� D 2 4 �10 1 3 5 : Now the linear combinations of u, v, w� lead to a cyclic difference matrix C : Cyclic Cx D 2 4 1 0 �1�1 1 0 0 �1 1 3 5 2 4 x1x2 x3 3 5 D 2 4 x1 � x3x2 � x1 x3 � x2 3 5 D b: (11) This matrix C is not triangular. It is not so simple to solve for x when we are given b. Actually it is impossible to find the solution to Cx D b, because the three equations either have infinitely many solutions or else no solution: Cx D 0 Infinitely many x 2 4 x1 � x3x2 � x1 x3 � x2 3 5 D 2 4 00 0 3 5 is solved by all vectors 2 4 x1x2 x3 3 5 D 2 4 cc c 3 5 : (12) 26 Chapter 1. Introduction to Vectors Every constant vector (c; c; c) has zero differences when we go cyclically. This undeter- mined constant c is like the CC that we add to integrals. The cyclic differences have x1 � x3 in the first component, instead of starting from x0 D 0. The other very likely possibility for Cx D b is no solution at all: Cx D b 2 4 x1 � x3x2 � x1 x3 � x2 3 5 D 2 4 13 5 3 5 Left sides add to 0Right sides add to 9 No solution x1; x2; x3 (13) Look at this example geometrically. No combination of u;v; and w� will produce the vector b D .1; 3; 5/. The combinations don’t fill the whole three-dimensional space. The right sides must have b1 C b2 C b3 D 0 to allow a solution to Cx D b, because the left sides x1 � x3, x2 � x1, and x3 � x2 always add to zero. Put that in different words. All linear combinations x1u C x2v C x3w� D b lie on the plane given by b1 C b2 C b3 D 0. This subject is suddenly connecting algebra with geometry. Linear combinations can fill all of space, or only a plane. We need a picture to show the crucial difference between u;v;w (the first example) and u;v;w�. Figure 1.10: Independent vectors u;v;w. Dependent vectors u;v;w� in a plane. Independence and Dependence Figure 1.10 shows those column vectors, first of the matrix A and then of C . The first two columns u and v are the same in both pictures. If we only look at the combinations of those two vectors, we will get a two-dimensional plane. The key question is whether the third vector is in that plane: Independence w is not in the plane of u and v. Dependence w� is in the plane of u and v. The important point is that the new vector w� is a linear combination of u and v: uC v Cw� D 0 w� D 2 4 �10 1 3 5 D �u� v: (14) 1.3. Matrices 27 All three vectors u;v;w� have components adding to zero. Then all their combinations will have b1 C b2 C b3 D 0 (as we saw above, by adding the three equations). This is the equation for the plane containing all combinations of u and v. By including w� we get no new vectors because w� is already on that plane. The original w D .0; 0; 1/ is not on the plane: 0 C 0 C 1 ¤ 0. The combinations of u;v;w fill the whole three-dimensional space. We know this already, because the solution x D Sb in equation (6) gave the right combination to produce any b. The two matrices A and C , with third columns w and w�, allowed me to mention two key words of linear algebra: independence and dependence. The first half of the course will develop these ideas much further—I am happy if you see them early in the two examples: u;v;w are independent. No combination except 0uC 0v C 0w D 0 gives b D 0. u;v;w� are dependent. Other combinations (specifically uC v Cw�) give b D 0. You can picture this in three dimensions. The three vectors lie in a plane or they don’t. Chapter 2 has n vectors in n-dimensional space. Independence or dependence is the key point. The vectors go into the columns of an n by n matrix: Independent columns: Ax D 0 has one solution. A is an invertible matrix. Dependent columns: Ax D 0 has many solutions. A is a singular matrix. Eventually we will have n vectors in m-dimensional space. The matrix A with those n columns is now rectangular (m by n). Understanding Ax D b is the problem of Chapter 3. REVIEW OF THE KEY IDEAS 1. Matrix times vector: Ax D combination of the columns of A. 2. The solution to Ax D b is x D A�1b, when A is an invertible matrix. 3. The difference matrix A is inverted by the sum matrix S D A�1. 4. The cyclic matrix C has no inverse. Its three columns lie in the same plane. Those dependent columns add to the zero vector. Cx D 0 has many solutions. 5. This section is looking ahead to key ideas, not fully explained yet. WORKED EXAMPLES 1.3 A Change the southwest entry a31 of A (row 3, column 1) to a31 D 1: Ax D b 2 4 1 0 0�1 1 0 1 �1 1 3 5 2 4 x1x2 x3 3 5 D 2 4 x1�x1 C x2 x1 � x2 C x3 3 5 D 2 4 b1b2 b3 3 5 : Find the solution x for any b. From x D A�1b read off the inverse matrix A�1. 28 Chapter 1. Introduction to Vectors Solution Solve the (linear triangular) system Ax D b from top to bottom: first x1 D b1 then x2 D b1 C b2 then x3 D b2 C b3 This says that x D A�1b D 2 4 1 0 01 1 0 0 1 1 3 5 2 4 b1b2 b3 3 5 This is good practice to see the columns of the inverse matrix multiplying b1; b2, and b3. The first column of A�1 is the solution for b D .1; 0; 0/. The second column is the solution for b D .0; 1; 0/. The third column x of A�1 is the solution for Ax D b D .0; 0; 1/. The three columns of A are still independent. They don’t lie in a plane. The combi- nations of those three columns, using the right weights x1; x2; x3, can produce any three- dimensional vector b D .b1; b2; b3/. Those weights come from x D A�1b. 1.3 B This E is an elimination matrix. E has a subtraction, E�1 has an addition. Ex D b � 1 0 �` 1 � � x1 x2 � D � b1 b2 � E D � 1 0 �` 1 � The first equation is x1 D b1. The second equation is x2 � `x1 D b2. The inverse will add `x1 D `b1, because the elimination matrix subtracted `x1 : x D E�1b � x1 x2 � D � b1 `b1 C b2 � D � 1 0 ` 1 � � b1 b2 � E�1 D � 1 0 ` 1 � 1.3 C Change C from a cyclic difference to a centered difference producing x3 � x1: Cx D b 2 4 0 1 0�1 0 1 0 �1 0 3 5 2 4 x1x2 x3 3 5 D 2 4 x2 � 0x3 � x1 0 � x2 3 5 D 2 4 b1b2 b3 3 5 : (15) Show that Cx D b can only be solved when b1 C b3 D 0. That is a plane of vectors b in three-dimensional space. Each column of C is in the plane, the matrix has no inverse. So this plane contains all combinations of those columns (which are all the vectors Cx). Solution The first component of b D Cx is x2, and the last component of b is �x2. So we always have b1 C b3 D 0, for every choice of x. If you draw the column vectors in C , the first and third columns fall on the same line. In fact .column 1/ D �.column 3/. So the three columns will lie in a plane, and C is not an invertible matrix. We cannot solve Cx D b unless b1 C b3 D 0. I included the zeros so you could see that this matrix produces “centered differences”. Row i of Cx is xiC1 (right of center) minus xi�1 (left of center). Here is the 4 by 4 centered difference matrix: Cx D b 2 664 0 1 0 0 �1 0 1 0 0 �1 0 1 0 0 �1 0 3 775 2 664 x1 x2 x3 x4 3 775 D 2 664 x2 � 0 x3 � x1 x4 � x2 0 � x3 3 775 D 2 664 b1 b2 b3 b4 3 775 (16) Surprisingly this matrix is now invertible! The first and last rows give x2 and x3. Then the middle rows give x1 and x4. It is possible to write down the inverse matrix C�1. But 5 by 5 will be singular (not invertible) again : : : 1.3. Matrices 29 Problem Set 1.3 1 Find the linear combination 2s1 C 3s2 C 4s3 D b. Then write b as a matrix-vector multiplication Sx. Compute the dot products (row of S ) �x: s1 D 2 4 11 1 3 5 s2 D 2 4 01 1 3 5 s3 D 2 4 00 1 3 5 go into the columns of S . 2 Solve these equations Sy D b with s1; s2; s3 in the columns of S :2 4 1 0 01 1 0 1 1 1 3 5 2 4 y1y2 y3 3 5 D 2 4 11 1 3 5 and 2 4 1 0 01 1 0 1 1 1 3 5 2 4 y1y2 y3 3 5 D 2 4 14 9 3 5 . The sum of the first n odd numbers is . 3 Solve these three equations for y1; y2; y3 in terms of B1; B2; B3: Sy D B 2 4 1 0 01 1 0 1 1 1 3 5 2 4 y1y2 y3 3 5 D 2 4 B1B2 B3 3 5 . Write the solution y as a matrix A D S�1 times the vector B. Are the columns of S independent or dependent? 4 Find a combination x1w1 C x2w2 C x3w3 that gives the zero vector: w1 D 2 4 12 3 3 5 w2 D 2 4 45 6 3 5 w3 D 2 4 78 9 3 5 : Those vectors are (independent) (dependent). The three vectors lie in a . The matrix W with those columns is not invertible. 5 The rows of that matrix W produce three vectors (I write them as columns): r1 D 2 4 14 7 3 5 r2 D 2 4 25 8 3 5 r3 D 2 4 36 9 3 5 : Linear algebra says that these vectors must also lie in a plane. There must be many combinations with y1r1 C y2r2 C y3r3 D 0. Find two sets of y’s. 6 Which values of c give dependent columns (combination equals zero)?2 4 1 3 51 2 4 1 1 c 3 5 2 4 1 0 c1 1 0 0 1 1 3 5 2 4 c c c2 1 5 3 3 6 3 5 30 Chapter 1. Introduction to Vectors 7 If the columns combine into Ax D 0 then each row has r � x D 0:2 4 a1 a2 a3 3 5 2 4 x1x2 x3 3 5 D 2 4 00 0 3 5 By rows 2 4 r1 � xr2 � x r3 � x 3 5 D 2 4 00 0 3 5 : The three rows also lie in a plane. Why is that plane perpendicular to x? 8 Moving to a 4 by 4 difference equation Ax D b, find the four components x1; x2; x3; x4. Then write this solution as x D Sb to find the inverse matrix S D A�1: Ax D 2 664 1 0 0 0 �1 1 0 0 0 �1 1 0 0 0 �1 1 3 775 2 664 x1 x2 x3 x4 3 775 D 2 664 b1 b2 b3 b4 3 775 D b: 9 What is the cyclic 4 by 4 difference matrix C ? It will have 1 and �1 in each row. Find all solutions x D .x1; x2; x3; x4/ to Cx D 0. The four columns of C lie in a “three-dimensional hyperplane” inside four-dimensional space. 10 A forward difference matrix  is upper triangular: z D 2 4 �1 1 00 �1 1 0 0 �1 3 5 2 4 z1z2 z3 3 5 D 2 4 z2 � z1z3 � z2 0 � z3 3 5 D 2 4 b1b2 b3 3 5 D b: Find z1; z2; z3 from b1; b2; b3. What is the inverse matrix in z D �1b? 11 Show that the forward differences .t C 1/2 � t2 are 2tC1 D odd numbers. As in calculus, the difference .t C 1/n � tn will begin with the derivative of tn, which is . 12 The last lines of the Worked Example say that the 4 by 4 centered difference matrix in (16) is invertible. Solve Cx D .b1; b2; b3; b4/ to find its inverse in x D C�1b. Challenge Problems 13 The very last words say that the 5 by 5 centered difference matrix is not invertible. Write down the 5 equations Cx D b. Find a combination of left sides that gives zero. What combination of b1; b2; b3; b4; b5 must be zero? (The 5 columns lie on a “4-dimensional hyperplane” in 5-dimensional space.) 14 If .a; b/ is a multiple of .c; d/ with abcd ¤ 0, show that .a; c/ is a multiple of .b; d/. This is surprisingly important; two columns are falling on one line. You could use numbers first to see how a; b; c; d are related. The question will lead to: The matrix A D � a b c d � has dependent columns when it has dependent rows.