And, even if it is square, we may not be able to use EROs to get to the identity matrix. Consider
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1 .2. Echelon Form and Reduced Row Echelon Form In this section, we address what we are trying to achieve by doing EROs. We are trying to turn any linear system into a simpler one. But what does simpler mean for a linear system? Also, how do we get from this simpler linear system to a solution description. In Example..5 with variables x, x 2, x 3, we did EROs The set of equations on the right hand linear system is then x =, x 2 = 6, x 3 = 2, which just tells us the solution. If we can ever make the matrix of the linear system into ones on the main diagonal and zeros elsewhere then the augmented part tell us the unique solution. This matrix will come up many times and has a special name. The reason for this name will become clear in Section??. Definition.2.. The identity matrix I n is the n n matrix I n = Often, I is used, and n is determined by context. However, we can not always get the matrix of a linear system to be the identity matrix. First, of all the matrix does not need to be square - it could be 2 3, as in [ ] And, even if it is square, we may not be able to use EROs to get to the identity matrix. Consider [ ] [ ] 3 3 R R, whose second equation is =. Thus, this linear system has no solution. But how do we know there is not some other EROs we could do to make the matrix into the identity matrix? If we could, then this linear system would have a unique solution because EROs preserve the set of solutions. But it has no solution, so this is impossible. If we can t always get the identity matrix, what is the next best thing? The next best thing will be reduced row echelon form. We first introduce echelon form as an intermediate. Echelon form will be useful in its own right since we will be able to tell the number of free variables and thus the number of solutions in echelon form. Definition.2.2. A matrix or augmented matrix is in echelon form (EF) if the leading nonzero entries in each row proceed strictly from left to right as we go from top to bottom, and any rows of all zeros are at the bottom. A linear system is in EF if its augmented matrix is in EF. Here are some examples of EFs:, 3 2, 7 5, where each can represent any number.
2 2 Example.2.3. Why are the following augmented matrices are not in EF? Perform EROs to put them in EF Solution: This is not in EF since the leading entries in each row all lie in first column. Then, R 2 R , R 3 +2R R R which is in EF. This is not in EF since the leading entries in last row s leading entry is not left of leading entry in previous row, among other things. Then, R R 4 2 R 2 R , R R 4 which is in EF. Example.2.3 illustrates the utility of the swapping rows operation. Furthermore, by iteratively swapping two rows, we can achieve any rearrangement of the rows. Definition.2.4. A matrix or augmented matrix is in reduced row echelon form (RREF) if it satisfies the following: (i) It is an echelon form. (ii) All of the leading nonzero entries in each row are s. (iii) Each column with a leading has s in the rest of its entries. A linear system is in RREF if its augmented matrix is in RREF. Here are some examples of RREFs:,,, where each can represent any number. Example.2.5. Why are the following linear systems in EF not in RREF? Perform EROs to put them in RREF.
3 Solution: This is not in RREF since its leading entries are not s, and the columns with these leading entries have other nonzero entries. 4 2 R 2R R R R 3 4 R 4R 2, 5 5 R 3 which is in RREF. When solving a linear system, the number each represents will change along with the EROs, but we did not keep track of those changes here. However, the EROs needed to achieve RREF in this case do not depend on the augmented part of the linear system. This is not in RREF since its leading entries are not s, and the columns with these leading entries have other nonzero entries R R 4R 3 6/5 7/5 R R R 2R 2 6/5 3/5 6/5, 8 8 R 3 which is in RREF. So, why is RREF the next best thing to the identity matrix? Just like with the identity matrix, we can go from a linear system in RREF to describing its solution. But first, we need to discuss pivot and free columns and leading and free variables. Definition.2.6. In a linear system in EF, including RREF, a column is pivot if it contains a leading nonzero entry in one of the rows. Otherwise, a column is free. If the linear system has a solution, we call the variables corresponding to pivot columns pivot variables, and the variables corresponding to free columns free variables (guess why). Since the augmented column does not have a variable associated to it, it does not give us a free or pivot variable. For example, in, columns, 2, 5 are pivot columns, and columns 3, 4, 6, 7 are free columns, which makes x, x 2, x 5 leading variables, and x 3, x 4, x 6 free variables. Given a linear system in RREF, we can find a solution description for it by first noting it has no solution if it has any rows of the form [... k ] for some k. Each such row represents an equation of the form = k for k, which has no solution, so the whole linear system has no solution. Otherwise, it has a solution description by making all of the free variables free, and then writing the pivot variables in terms of the free variables using the equations. In this way, each equation with a pivot variable is used to determine that pivot variable in terms of the free variables. 3
4 4 Each equation without a pivot variable is then of the form = and thus adds no new information. Because we accounted for all of the information contained in the equations, we have a solution description. Example.2.7. Describe the set of solutions to the following linear systems in RREF: [ ] , Solution: First, this linear system has a solution since there are no rows of the form [... k ] where k. Columns 2, 3 are pivot, and columns, 4, 5 are free. Thus, this linear system has a solution description where x, x 4 are free, and then, solving for the pivot variables in the two equations x 2 2x 4 = 3, x 3 + 5x 4 = 7, we have x 2 = 2x 4 + 3, x 3 = 5x Do not ignore x even though its coefficient is in both equations. This means while x is a variable, the linear system does not care what it is, so x gets to be whatever it wants. This linear system has no solution since there is a row of the form [... k ] where k. The last equation translates to = which has no solution. Columns, 2, 4 are pivot, and column 3 is free. First, this linear system has a solution since there are no rows of the form [... k ] where k. Columns, 2, 5 are pivot, and columns 3, 4, 6, 7 are free. Thus, this linear system has a solution description where x 3, x 4, x 6 are free, and then, solving for the pivot variables in each equation, x = x 3 3x 4 5x 6 + 8, x 2 = 2x 3 4x 4 6x 6 + 9, x 5 = 7x 6 +. Keep in mind that there are usually other solution descriptions where one makes different choices for what variables are free. For example, another solution description to Example.2.7 is () x, x 3 free, x 2 = 2 5 x , x 4 = 5 x obtained by solving for x 2 and x 4 in the equations. However, the choice of free variables made by taking the free columns in RREF will always give a valid solution description, whereas an arbitrary choice of free variables may not. For example, because the coefficient of x is in both equations in Example.2.7, x is free in every solution description to Example.2.7, Problem
5 7. Furthermore, a variable is free in every solution description only when all of its coefficients are, Problem 8. We defined pivot and free columns and variables for any linear system in EF, but we only have used them on linear systems in RREF so far. We defined them for EF as well because one can convert from EF to RREF without changing which columns and are pivot or free. To see this, first scale the rows so that their leading nonzero entries are s. Then, working from right to left, use these leading nonzero entries to make every entry above them into. This process will look something like: Also, these EROs do not change whether there is a row of the form [... k ] where k. Thus, we can tell what the pivot and free columns will be in RREF from EF, and a linear system in EF has a solution if and only if it has no rows of the form [... k ] where k. In the examples above, we were always able to reduce a matrix to RREF using EROs. But is there be a matrix that could not be reduced to RREF with EROs. Fortunately, the answer is no. In addition, RREF has more significance than just allowing us to describe the set of solutions to a linear system. The RREF of any matrix is unique! Alice and Bob can do 2 completely different set of EROs to the same matrix to get to RREF, but they will always end up with the same RREF. In addition, for any matrix A, this lets us define RREF(A) to be the unique matrix in RREF that A is row equivalent to. Definition/Theorem.2.8. Every matrix A is row equivalent to a unique matrix in RREF, called RREF(A). For example, as seen in Example.2.5. RREF = 3/5 6/5 8 Proof. First, we need to argue that any matrix is row equivalent to some matrix in RREF. We will demonstrate this by explicitly describing how to use EROs to get any augmented matrix into RREF. First, if there exists a leading nonzero entry in column, swap it to the top, scale it so it leads with a, and then use EROs of the form R i + cr to make all other entries in column zero. Now, for each of the rest of the columns column from left to right, if there exists a leading nonzero entry in that column, swap it as high as it can below the leading entries for previous columns, say in row j. Then, scale row j so it leads with a, and then use EROs of the form R i + cr j to make all other entries in that column zero. By construction, when we are done, no column can have more than leading nonzero entry, all the leading nonzero entries are s, these entries will go from left to right, any rows of all zeros are at the bottom, and each column with a leading will have the rest of its 5
6 6 entries be zeros. Thus, we have achieved RREF using EROs. This process may look like:... a a b c a m... b m... c m d e d m 2... e m 3... In the above process, we found a leading nonzero entry among a,..., a m and b,..., b m and d,..., d m 2, but not among c,..., c m 2, so we skipped to the next column. To prove uniqueness, suppose to the contrary that A is row equivalent to two distinct RREFs, B and C. We start by arguing through one example for B, C and then explaining how to generalize this reasoning to any two distinct RREFs. For example, suppose A is 3 3 and row equivalent to both B = 2 3, and C = 4 5 Thus, A B and A C, so B C, (Section., Problem 7), Adding an augmented part of all zeros preserves row equivalence (Section., Problem 6), so (2) 5. Now, what is wrong with this situation? Trying to argue that no possible sequence of EROs can get us from B to C directly is very difficult, so I do not recommend trying this approach. Instead, we refer to something that is invariant under EROs, namely the set of solutions, see Theorem..2. By Theorem..2, the two linear systems in (.2) have the same set of solutions. However, we see directly that the two equations in (.2) have different solution sets, namely x 3 is free, x = 2x 3, x 2 = 3x 3, x 3 is free, x = 4x 3, x 2 = 5x 3, respectively, a contradiction. We can see these solution sets are different explicitly by noting that x = 2, x 2 = 3, x 3 = is a solution to the left linear system in (.2) but not the right system in (.2). By this contradiction, is false, which means B, C cannot both be row equivalent to A. Next, let us see how this argument generalizes to show that it is impossible for any matrix to be row equivalent to two distinct RREFs. Suppose A is an m n matrix. Suppose to the contrary
7 that A is row equivalent to two distinct RREFs, B and C. By the same reasoning as our example, B. and C. have the same set of solutions. However, the following contradicts Claim. Thus, proving Claim will give a contradiction and show that it is impossible for A to be row equivalent to two distinct RREFs, which will complete the proof. 7 Claim. If B, C are distinct m n RREFs, then (3) B. and C. have different sets of solutions. Proof of Claim: Let L, L 2 denote the two linear systems in 3, respectively. Note that both L, L 2 have at least one solution, namely x =,..., x n =. Solving for the pivot variables to obtain a solution description, each equation not of the form = in L and L 2 expresses x k in terms of x k+,..., x n for some k. This follows from definition of RREF and can be seen best in Example.2.7. Since B C and B, C have the same of rows, there is a first row where B, C differ as we go from top to bottom, say row i. Then, letting x k be the first among x,..., x n to appear in equation i of either L or L 2, we see that x k must depend on x k+,..., x n differently in these solution descriptions to L and L 2. Hence, L and L 2 have two different sets of solutions in general. The following example may clarify this argument: B = 2 3, C = 2 4, then i =, k = 2, and x 2 depends on x 3, x 4 differently in the linear systems L, L 2, namely L : x 2 = 2x 3 3x 4, L 2 : x 2 = 2x 2 4x 4. In this proof, it was useful to restrict our attention to linear systems whose augmented part was all zeros. In fact, most of the linear systems we will be interested in for the rest of the class will have an augmented part of all zeros. Such linear systems has a special name. Definition.2.9. A linear system is homogeneous if its augmented part is all zeros. For example, [ ] is homogeneous, but [ ] is not. Observe that every homogeneous linear system in x,..., x n always has a solution by setting x =,..., x n =. Exercises:
8 8. Reduce the following linear systems to EF using EROs. [ 4 7 ] Reduce the following linear systems in echelon form to RREF using EROs. [ ] Describe the set of solutions to the given linear system in RREF. Specify which variables are free, and then write the rest of the variables in terms of them. Use x,..., x n for the variable names. (d) [ 5 ]
9 4. Solve the following linear systems by reducing to RREF by EROs and then describing the set of solutions. Specify which variables are free, and then write the rest of the variables in terms of them. Use x,..., x n for the variable names. [ ] (d) Suppose the set of solutions to a linear system L is described by x 3, x 5 free, and x = 2x 3 + 3x 5, x 2 = 6x 3 x 5, x 4 = 4x 5 9 Find a linear system in RREF with this solution set. Is there a way to describe the set of solutions to L with x 2 and x 4 free instead? If so, how? If not, why not? Is there a way to describe the set of solutions with x 4 and x 5 free instead? If so, how? If not, why not? (d) For what subsets of 2 of the variables of x,..., x 5 does there exist a description of the solution with those variables free? Explain. You do not have to write out all of these descriptions. 6. Suppose the set of solutions to a linear system is described by x 3, x 4, x 5 free, and x = x 3 + 3x 4, x 2 = 2x 3 + 6x 4, Find a linear system in RREF with this solution description. For what subsets of 3 of the variables of x,..., x 5 does there exist a description of the solution with those variables free? Explain. You do not have to write out all of these descriptions unless you want to. 7. Specify which value(s) of the variables in each of the linear systems below give rise to no solutions, exactly one solutions, and infinitely many solutions. [ ] h k
10 [ ] 2 h 5 k h 3 8 k 8. Suppose a homogeneous linear system has 3 equations and 3 variables. How many free variables could it have in a solution description? Give an example for each possibility. Do the same for 3 equations and 4 variables. Do the same for 4 equations and 3 variables. (d) Do the same for 3 equations and 6 variables. Problems:. () Show that a linear system has a solution if and only if the augmented column is a free column. 2. Determine if the following statements are True or False. Explain your reasoning or give an example/counterexample. () Every linear system with 3 equations and 2 variables has a solution. () No linear system with 3 equations and 2 variables has a solution. () Every linear system with 2 equations and 3 variables has a solution. (d) () No linear system with 2 equations and 3 variables has a solution. (e) (2) Every homogeneous linear system with 2 equations and 3 variables has infinitely many solutions. (f) (2) Every homogeneous linear system with m equations and n variables and m < n has infinitely many solutions. (g) (2) A homogeneous linear system with 3 equations and 2 variables has a solution. (h) (2) A linear system with 3 equations and 2 variables can have a unique solution. (i) (2) A linear system with 2 equations and 3 variables can have a unique solution. 3. (2+) Suppose L is a linear system with the same number of equations as variables that has a unique solution. Suppose L is obtained from L by changing the augemented column. What are the possible number of solutions L can have? (2+) Answer if instead L has no solution. (2+) Answer if instead L has infinitely many solutions. 4. (2) Show that if A, B are matrices, then A B if and only if RREF(A) = RREF(B). 5. (3) Prove or give a counterexample: If A, B are m (n + ) augmented matrices with the same set of solutions, then A B. 6. (3) Prove or give a counterexample: If A, B are m (n + ) augmented matrices with the same set of solutions that includes at least one solution, then A B. 7. (2) Let L be a linear system in x,..., x n. Show that if the coefficient of x k is in every equation in L, then every solution description to L has x k free.
11 8. (3) Let L be a linear system in x,..., x n. Show that if the coefficient of x k is nonzero in at least one equation in L, then there exists a solution description to L where x k is not free. 9. (3+) (Requires Programming Experience) Write a program that takes in a matrix and outputs its RREF. I strongly recommend doing this, for it is the best way to figure out how this algorithm of reducing to RREF works. I bet you could look one up on the internet, but that is no fun. You can either figure it on your own or follow these steps. Method : You should be able to write this program with for loops. Do the necessary setup in whatever programming language you want to use to be able to store and manipulate matrices. Write programs to do each of the 3 EROs. Write a program to find the first nonzero entry in column. (d) Swap the row with this entry to row and make the rest of the entries in column zero by doing EROs of the form R i + cr (e) Try doing what you did in column to the rest of the columns from left to right. Explain why this is NOT reducing you to RREF. (f) Keep track of the number of leading nonzero entries you have so far, and then adjust what you did in column to the rest of the columns to correctly reduce to RREF. Where should you start searching for nonzero entries in each column? Where should you swap this row to? Which of the other entries in this column to you want to make? What EROs do you do to accomplish this? Method 2: Recursion. Do - (d) as before. (e) Write a base case. What should you return if the matrix is empty. (e) Now, assume your program works on smaller matrices. Where can you use this program to recursively do a lot of the work for you? This technique is called recursion, and. (e) If you used the method of recursion we expected you to, your program should now return an EF. Fix it so it returns RREF. What other entries you need to make after you recursively compute the RREF of this smaller matrix? How do you determine which columns are free?. Recall Example.2.7: [ ] Consider the following alternative way to obtain the solution description in (). Rearranging the variables in the order x, x 2, x 4, x 3, gives [ ] [ 2/5 ] 29/5 /5 7/5 Converting this RREF into a solution description by the methods of this section, x, x 3 are free, and then x 2 = 2 5 x , x 4 = 5 x Let L be a linear system with augmented matrix A. (4) Show that one can obtain any solution description to L by rearranging the columns of A and then converting its RREF into a solution description by the methods of this section. (3) Do two different rearrangements of A s columns necessarily lead to two different solution descriptions to L?
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