10 CHAPTER 2. ALGEBRAIC LINEAR SYSTEMS Pause for a minute to verify that this formulation is equivalent. Theorem 2.2.2 Two different bases for the same vector space have the same number of vectors. Proof. Let a a and a a be two different bases for . Then each a is in (why?), and can therefore be written as a linear combination of a a . Consequently, the vectors of the set a a a must be linearly dependent. We call a set of vectors that contains a basis for a generating set for . Thus, is a generating set for . The rest of the proof now proceeds as follows: we keep removing a vectors from and replacing them with a vectors in such a way as to keep a generating set for . Then we show that we cannot run outof a vectors before we run out of a vectors, which proves that . We then switch the roles of a and a vectors to conclude that . This proves that . From lemma 2.1.2, one of the vectors in is a linear combination of those preceding it. This vector cannot be a , since it has no other vectors preceding it. So it must be one of thea vectors. Removing the latter keeps a generating set, since the removed vector depends on the others. Now we can add a to , writing it right after a : a a is still a generating set for . Let us continue this procedure until we run out of either a vectors to remove or a vectors to add. The a vectors cannot run out first. Suppose in fact per absurdum that is now made only of a vectors, and that there are still left-over a vectors that have not been put into . Since the a s form a basis, they are mutually linearly independent. Since is a vector space, all the a s are in . But then cannot be a generating set, since the vectors in it cannot generate the left-over a s, which are independent of those in . This is absurd, because at every step we have made sure that remains a generating set. Consequently, we must run out of a s first (or simultaneously with the last a). That is, . Now we can repeat the whole procedure with the roles of a vectors and a vectors exchanged. This shows that , and the two results together imply that . A consequence of this theorem is that any basis for R has vectors. In fact, the basis of elementary vectors e th column of the identity matrix is clearly a basis for R , since any vector b . . . can be written as b e and the e are clearly independent. Since this elementary basis has vectors, theorem 2.2.2 implies that any other basis for R has vectors. Another consequence of theorem 2.2.2 is that vectors of dimension are bound to be dependent, since any basis for R can only have vectors. Since all bases for a space have the same number of vectors, it makes sense to define the dimension of a space as the number of vectors in any of its bases. 2.3. INNER PRODUCT AND ORTHOGONALITY 11 2.3 Inner Product and Orthogonality In this section we establish the geometric meaning of the algebraic notions of norm, inner product, projection, and orthogonality. The fundamental geometric fact that is assumed to be known is the law of cosines: given a triangle with sides (see figure 2.1), we have where is the angle between the sides of length and . A special case of this law is Pythagoras’ theorem, obtained when . θ c b a Figure 2.1: The law of cosines states that . In the previous section we saw that any vector in R can be written as the linear combination b e (2.7) of the elementary vectors that point along the coordinate axes. The length of these elementary vectors is clearly one, because each of them goes from the origin to the unit point of one of the axes. Also, any two of these vectors form a 90-degree angle, because the coordinate axes are orthogonal by construction. How long is b? From equation (2.7) we obtain b e e and the two vectors e and e are orthogonal. By Pythagoras’ theorem, the square of the length b of b is b e Pythagoras’ theorem can now be applied again to the last sum by singling out its first term e , and so forth. In conclusion, b This result extends Pythagoras’ theorem to dimensions. If we define the inner product of two -dimensional vectors as follows: b c then b b b (2.8) Thus, the squared length of a vector is the inner product of the vector with itself. Here and elsewhere, vectors are column vectors by default, and the symbol makes them into row vectors. 12 CHAPTER 2. ALGEBRAIC LINEAR SYSTEMS Theorem 2.3.1 b c b c where is the angle between b and c. Proof. The law of cosines applied to the triangle with sides b , c , and b c yields b c b c b c and from equation (2.8) we obtain b b c c b c b b c c b c Canceling equal terms and dividing by -2 yields the desired result. Corollary 2.3.2 Two nonzero vectors b and c in R are mutually orthogonaliff b c . Proof. When , the previous theorem yields b c . Given two vectors b and c applied to the origin, the projection of b onto c is the vector from the origin to the point on the line through c that is nearest to the endpoint of b. See figure 2.2. p b c Figure 2.2: The vector from the origin to point is the projection of b onto c. The line from the endpoint of b to is orthogonal to c. Theorem 2.3.3 The projection of b onto c is the vector p c b where c is the followingsquare matrix: c cc c c Proof. Since by definition point is on the line through c, the projection vector p has the form p c, where is some real number. From elementary geometry, the line between and the endpoint of b is shortest when it is orthogonal to c: c b c 2.4. ORTHOGONAL SUBSPACES AND THE RANK OF A MATRIX 13 which yields c b c c so that p c c cc c c b as advertised. 2.4 Orthogonal Subspaces and the Rank of a Matrix Linear transformations map spaces into spaces. It is important to understand exactly what is being mapped into what in order to determine whether a linear system has solutions, and if so how many. This section introduces the notion of orthogonality between spaces, defines the null space and range of a matrix, and its rank. With these tools, we will be able to characterize the solutionsto a linear system in section 2.5. In the process, we also introduce a useful procedure (Gram-Schmidt) for orthonormalizinga set of linearly independent vectors. Two vector spaces and are said to be orthogonalto one another when every vector in is orthogonal to every vector in . If vector space is a subspace of R for some , then the orthogonal complement of is the set of all vectors in R that are orthogonal to all the vectors in . Notice that complement and orthogonal complement are very different notions. For instance, the complement of the plane in R is all of R except the plane, while the orthogonal complement of the plane is the axis. Theorem 2.4.1 Any basis a a for a subspace of R can be extended into a basis for R by adding vectors a a . Proof. If we are done. If , the given basis cannot generate all of R , so there must be a vector, call it a , that is linearly independent of a a . This argument can be repeated until the basis spans all of R , that is, until . Theorem 2.4.2 (Gram-Schmidt) Given vectors a a , the following construction for to a a q a q if a 0 q a a end end yields a set of orthonormal vectors q q that span the same space as a a . Proof. We first prove by induction on that the vectors q are mutually orthonormal. If , there is little to prove. The normalization in the above procedure ensures that q has unit norm. Let us now assume that the procedure Orthonormal means orthogonal and with unit norm. 14 CHAPTER 2. ALGEBRAIC LINEAR SYSTEMS above has been performed a number of times sufficient to find vectors q q , and that these vectors are orthonormal (the inductive assumption). Then for any we have q a q a q a q q because the term q a cancels the -th term q a q q of the sum (remember that q q ), and the innerproducts q q are zero by the inductive assumption. Because of the explicit normalization step q a a , the vector q ,if computed, has unit norm, and because q a , it follwos that q is orthogonal to all its predecessors, q q for . Finally, we notice that the vectors q span the same space as the a s, because the former are linear combinations of the latter, are orthonormal (and therefore independent), and equal in number to the number of linearly independent vectors in a a . Theorem 2.4.3 If is a subspace of R and is the orthogonalcomplement of in R , then Proof. Let a a be a basis for . Extend this basis to a basis a a for R (theorem 2.4.1). Orthonor- malize this basis by the Gram-Schmidt procedure (theorem 2.4.2) to obtain q q . By construction, q q span . Because the new basis is orthonormal, all vectors generated by q q are orthogonal to all vectors generated by q q , so there is a space of dimension at least that is orthogonal to . On the other hand, the dimension of this orthogonal space cannot exceed , because otherwise we would have more than vectors in a basis for R . Thus, the dimension of the orthogonal space is exactly , as promised. We can now start to talk about matrices in terms of the subspaces associated with them. The null space null of an matrix is the space of all -dimensional vectors that are orthogonal to the rows of . The range of is the space of all -dimensional vectors that are generated by the columns of . Thus, x null iff x , and b range iff x b for some x. From theorem 2.4.3, ifnull has dimension , thenthe space generated bytherows of has dimension , that is, has linearly independent rows. It is not obvious that the space generated by the columns of has also dimension . This is the point of the following theorem. Theorem 2.4.4 The number of linearly independent columns of any matrix is equal to the number of its independent rows, and where null . Proof. We have already proven that the number of independent rows is . Now we show that the number of independent columns is also , by constructing a basis for range . Let v v be a basis for null , and extend this basis (theorem 2.4.1) into a basis v v for R . Then we can show that the vectors v v are a basis for the range of . First, these vectors generate the range of . In fact, given an arbitrary vector b range , there must be a linear combination of the columns of that is equal to b. In symbols, there is an -tuple x such that x b. The -tuple x itself, being an element of R , must be some linear combination of v v , our basis for R : x v 2.5. THE SOLUTIONS OF A LINEAR SYSTEM 15 Thus, b x v v v since v v span null , so that v for . This proves that the vectors v v generate range . Second, we prove that the vectors v v are linearly independent. Suppose, per absurdum, that they are not. Then there exist numbers , not all zero, such that v so that v But then the vector v is in the null space of . Since the vectors v v are a basis for null , there must exist coefficients such that v v in conflict with the assumption that the vectors v v are linearly independent. Thanks to this theorem, we can define the rank of to be equivalentlythe number of linearlyindependent columns or of linearly independent rows of : rank range null 2.5 The Solutions of a Linear System Thanks to the results of the previous sections, we now have a complete picture of the four spaces associated with an matrix of rank and null-space dimension : range ; dimension rank null ; dimension range ; dimension null ; dimension The space range is called the left nullspace of the matrix, and null is called the rowspace of .A frequently used synonym for “range” is column space. It should be obvious from the meaning of these spaces that null range range null where is the transpose of , defined as the matrix obtained by exchanging the rows of with its columns. Theorem 2.5.1 The matrix transforms a vector x in its null space into the zero vector, and an arbitrary vector x into a vector in range . 16 CHAPTER 2. ALGEBRAIC LINEAR SYSTEMS This allows characterizing the set of solutions to linear system as follows. Let x b be an system ( can be less than, equal to, or greater than ). Also, let be the number of linearly independent rows or columns of . Then, b no solutions b solutions with the convention that . Here, is the cardinality of a -dimensional vector space. In the first case above, there can be no linear combinationof the columns (no x vector) that gives b, and the system is said to be incompatible. In the second, compatible case, three possibilities occur, depending on the relative sizes of : When , the system is invertible. This means that there is exactly one x that satisfies the system, since the columns of span all of R . Notice that invertibilitydepends only on , not on b. When and , the system is redundant. There are more equations than unknowns, but since b is in the range of there is a linear combination of the columns (a vector x) that produces b. In other words, the equations are compatible, and exactly one solution exists. When the system is underdetermined. This means that the null space is nontrivial (i.e., it has dimension ), and there is a space of dimension of vectors x such that x . Since b is assumed to be in the range of , there are solutions x to x b, but then for any y null also x y is a solution: x b y x y b and this generates the solutions mentioned above. Notice that if then cannot possibly exceed , so the first two cases exhaust the possibilities for . Also, cannot exceed either or . All the cases are summarized in figure 2.3. Of course, listing all possibilities does not provide an operational method for determining the type of linear system for a given pair b. Gaussian elimination, and particularly its version called reduction to echelon form is such a method, and is summarized in the next section. 2.6 Gaussian Elimination Gaussian elimination is an important technique for solving linear systems. In addition to always yielding a solution, no matter whether the system is invertible or not, it also allows determining the rank of a matrix. Other solution techniques exist for linear systems. Most notably, iterative methods solve systems in a time that depends on the accuracy required, while direct methods, like Gaussian elimination, are done in a finite amount of time that can be bounded given only the size of a matrix. Which method to use depends on the size and structure (e.g., sparsity) of the matrix, whether more information is required about the matrix of the system, and on numerical considerations. More on this in chapter 3. Consider the system x b (2.9) Notice that the technical meaning of “redundant”has a stronger meaning than “with more equations than unknowns.” The case is possible, has more equations ( ) than unknowns ( ), admits a solution if b range , but is called “underdetermined” because there are fewer ( ) independent equations than there are unknowns (see next item). Thus, “redundant” means “with exactly one solution and with more equations than unknowns.” 2.6. GAUSSIAN ELIMINATION 17 yes no yes no yes no underdetermined redundantinvertible b in range(A) r = n m = n incompatible Figure 2.3: Types of linear systems. which can be square or rectangular, invertible, incompatible, redundant, or underdetermined. In short, there are no restrictions on the system. Gaussian elimination replaces the rows of this system by linear combinations of the rows themselves until is changed into a matrix that is in the so-called echelon form. This means that Nonzero rows precede rows with all zeros. The first nonzero entry, if any, of a row, is called a pivot. Below each pivot is a column of zeros. Each pivot lies to the right of the pivot in the row above. The same operations are applied to the rows of and to those of b, which is transformed to a new vector c, so equality is preserved and solving the final system yields the same solution as solving the original one. Once the system is transformed into echelon form, we compute the solution x by backsubstitution, that is, by solving the transformed system x c 2.6.1 Reduction to Echelon Form The matrix is reduced to echelon form by a process in steps. The first step is applied to and c b. The -th step is applied to rows of and c and produces and c . The last step produces and c c. Initially, the “pivot column index” is set to one. Here is step , where denotes entry of : Skip no-pivot columns If is zero for every , then increment by 1. If exceeds stop. Row exchange Now and is nonzero for some . Let be one such value of .If , exchange rows and of and of c . Triangularization The new entry is nonzero, and is called the pivot.For , subtract row of multiplied by from row of , and subtract entry of c multiplied by from entry of c . This zeros all the entries inthe column below thepivot, and preserves the equalityof left- and right-hand side. When this process is finished, is in echelon form. In particular, if the matrix is square and if all columns have a pivot, then is upper-triangular. “Stop” means that the entire algorithm is finished. Different ways of selecting here lead to different numerical properties of the algorithm. Selecting the largest entry in the column leads to better round-off properties. 18 CHAPTER 2. ALGEBRAIC LINEAR SYSTEMS 2.6.2 Backsubstitution A system x c (2.10) inechelon formiseasilysolvedfor x. Tosee this, wefirstsolvethe system symbolically, leavingundeterminedvariables specified by their name, and then transform this solution procedure into one that can be more readily implemented numerically. Let be the index of the last nonzero row of . Since this is the number of independent rows of , is the rank of . It is also the rank of , because and admit exactly the same solutions and are equal in size. If , the last equations yield a subsystem of the followingform: . . . . . . Let us call this the residual subsystem. If on the other hand (obviously cannot exceed ), there is no residual subsystem. If there is a residual system (i.e., ) and some of are nonzero, then the equations corresponding to these nonzero entries are incompatible, because they are of the form with . Since no vector x can satisfy these equations, the linear system admits no solutions: it is incompatible. Let us now assume that either there is no residual system, or if there is one it is compatible, that is, . Then, solutions exist, and they can be determined by backsubstitution, that is, by solving the equations starting from the last one and replacing the result in the equations higher up. Backsubstitutions works as follows. First, remove the residual system, if any. We are left with an system. In this system, call the variables corresponding to the columns with pivots the basic variables, and call the other the free variables. Say that the pivot columns are . Then symbolic backsubstitution consists of the following sequence: for downto end This is called symbolic backsubstitution because no numerical values are assigned to free variables. Whenever they appear in the expressions for the basic variables, free variables are specified by name rather than by value. The final result is a solution withas many free parameters as there are free variables. Since any value given to the free variables leaves the equality of system (2.10) satisfied, the presence of free variables leads to an infinity of solutions. When solving a system in echelon form numerically, however, it is inconvenient to carry around nonnumeric symbol names (the free variables). Here is an equivalent solution procedure that makes this unnecessary. The solution obtained by backsubstitution is an affine function of the free variables, and can therefore be written in the form x v v v (2.11) where the are the free variables. The vector v is the solutionwhen all free variables are zero, and can therefore be obtained by replacing each free variable by zero during backsubstitution. Similarly, the vector v for can be obtained by solving the homogeneous system x with and all other free variables equal to zero. In conclusion, the general solution can be obtained by running backsubstitution times, once for the nonhomogeneous system, and times for the homogeneous system, with suitable values of the free variables. This yields the solution in the form (2.11). Notice that the vectors v v form a basis for the null space of , and therefore of . An affine function is a linear function plus a constant. . sides b , c , and b c yields b c b c b c and from equation (2. 8) we obtain b b c c b c b b c c b c Canceling equal terms and dividing by -2 yields the desired result. Corollary 2. 3 .2 Two nonzero. the equalityof left- and right-hand side. When this process is finished, is in echelon form. In particular, if the matrix is square and if all columns have a pivot, then is upper-triangular. “Stop”. then Proof. Let a a be a basis for . Extend this basis to a basis a a for R (theorem 2. 4.1). Orthonor- malize this basis by the Gram-Schmidt procedure (theorem 2. 4 .2) to obtain q q . By construction,