Linear Systems Math A4600 - Bianca Santoro September 3, 06 Goal: Understand how to solve Ax = b. Toy Model: Let s study the following system There are two nice ways of thinking about this system: x + y = 3 () x + y = 3 (). Finding a solution is equivalent of finding the intersection of these two lines. Probably, this is the way you are more used to think about this problem. How can two lines intersect in R?. ( Another ) way ( ) is to think about finding the correct linear combination of the vectors and that gives us the right hand side. ( ) 3 Natural questions that arise: does belong to the span of the vectors above? Is 3 there a unique linear combination? If there is more than one, what are all possible linear combinations? Let s now think of Case first. Instead of looking for the intersection of two crooked lines, how about we replace the second line by a horizontal line that passes through the same point? When we do that, is the point of intersection changing? What will that mean algebraically?
( ) Now, let s move to Case : I am looking for a linear combination of the vectors ( ) ( ) 3 and that gives me. 3 I would like to apply an invertible linear transformation T to R in order to make this problem easier. Namely, let T : R R be the transformation defined by ( ) 0 T (v) = v Is T invertible? Why? Back to the problem: I am going to apply the transformation T to my equation: ( ) [ ( ) ( )] ( ) ( ) 0 0 3 x + y = 3 ( ) ( ) ( ) ( ) ( ) 0 0 3 x + y = 3 ( ) ( ) ( ) 3 x + y 3 = 3 0 and this last problem is much easier to solve. Question: Given the linear system, how can I come up with such a nice linear transformation in order to symplify the problem? (3) (4) (5) Gauss-Jordan Elimination We are going to study a method to simplify a system Ax = b, by performing elementary operations on the rows of A. Of course, later we will need to justify why we are allowed to make this operations - we need to make sure that the solutions of our system will remain unchanged.
Reduction of A m n x n = b m We will learn how to reduce the original system to a sister system U m n x n = c m, where the matrix U m n will be in semi-echelon form (if A is square, U will be upper triangular). For the matrix A,. Locate the first (left-most) column that contains a non-zero entry.. If necessary, change the first row with another, in order to guarantee that the first non-zero column has a non-zero entry on the first row. 3. Let a j 0 is the leading entry ( in (row ), and ) the other coefficients below it are a j,, a mj. Replace (row i) by rowi a ij a j row 4. After doing 3) for all rows, we will get a new matrix, with only zeroes below a j. 5. Ignore row and column of this system, and go back to ) for the smaller system. Example : Solve for x, y and z, 3 x 4 5 6 y = z 3 3
Example : Solve for x, y, z and w, x 3 4 0 5 6 0 y 0 z = 0 8 9 0 w What are we really doing? For each of those steps, we are applying a linear transformation to the output space which is invertible, so it will not change the system. In Example, if we multiply both sides of the system by the two matrices E and E defined by 0 0 0 0 E = 4 0 ande = 0 0 0 0 3, we obtain the system after row reducing, as before. So, the matrix U can be written as U = E E A. Notice that both E and E are lower triangular, with ones along the diagonal. HW: Show that the product of two lower triangular matrices with ones along the diagonal is also lower triangular, with ones along the diagonal. Even though we haven t really learned how to compute the inverse of a matrix, use the definition of the inverse transformation to check that the inverse is also lower triangular, with ones along the diagonal. If U = E E A, and both E and E are invertible, we can write A = (E ) (E ) U = LU. The supernice fact is that to construct L, lower triangular with ones along the diagonal, we just need to write down... 4
HW: Solve (if possible) the system Av = b, for 3 4 x 0 0 y 0 0 z = 3 3 0 w 0 Can you decompose the matrix A in the LU form? Describe how. HW3: Look at the system a x + b y + c z = d (6) a x + b y + c z = d (7) Is it possible that the system above has exactly one solution? Give two explanations why (one inspired by the row picture, other by the column picture). HW4: Find a cubic polynomial p(x) = a+bx+cx +dx 3 such that p() = 5, p () = 5, p() = 7 and p () =. 5
Complete Solutions to Ax = b Sometimes, hyperplanes in R n may intersect in sets which are larger than just a single point. If that is the case, we want to understand the best we can how is the intersection. 3 x 6 Example: y = 3 3 4 z 9 By inspection (or by Gaussian Elimination, whatever you prefer), we can see that x = (,, ) is one solution. But you will agree that adding any element of Null(A) to this vector will also be a solution, right? So, our problem will reduce to understanding the nullspace of A. Is the nullspace of A related to the nullspace of U? Lemma 0. Let T : W W linear, invertible, and let S : V W linear. Then the nullspace of S is equal to the nullspace of the composite map SoT. 6
Back to our example: in order to describe the Nullspace of A, I will exhibit a basis for it. Recall that dim N(A) + dim R(A) = dim R 3. Also, note that the dimension of the range of A is the number of non-zero rows after elimination (why?). Hence, in our example, the dimension of the nullspace of A, which is the same as the nullspace of U, is equal to. So, the complete solution for a system is given by: x c = x p + x h, where x p is any particular solution to the original problem Ax = b, and x h denotes all the solutions to the homogeneous problem (in other words, all elements in the Nullspace of A. In our example, 7 x c = + α, for all α R. 7
HW5: Find the complete solution to the system 3 4 x y 3 4 5 z = 0 0 3 w HW6: Find a polynomial of degree 3 that passes through the points (, 6), (0, ), (, 4) and (, 5). Is your answer unique? Also, is it possible to find a polynomial of degree 3 that pass not only through the points above, but also through the point (0, 00)? Explain why/why not. (In the near future, we will see what we can do about this problem) HW7: Let A be an invertible 5 5 matrix, whose columns are the vectors a, a, a 3, a 4 and a 5, in this order. Let B be the matrix whose columns are the vectors a 5, a, a 4, a and a 3, also in this order. For a given vector b, the vector (,, 3, 4, 5) is the solution of Bx = b. complete solution to the problem Ax = b. Find the 8