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# class_3 - CS205 Class 3 1 So far we have discussed solving...

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CS205 - Class 3 1. So far we have discussed solving Ax=b for square n n × matrices A . For more general n m × matrices, there are a variety of scenarios. a. When m < n , the problem is underdetermined since there is not enough information to determine a unique solution for all the variables. Usually m<n implies that there are infinite solutions . However, in some cases, there may be contradictory equations leading to the absence of any solutions. b. When m > n , the problem is overdetermined although this in itself doesn’t tell us everything about the nature of the solution. For example, if enough equations are linear combinations of each other, there can still be a unique solution or infinite solutions. i. We can use the rank of the matrix to enumerate the possibilities. Recall that the rank of a matrix is the number of linearly independent columns that it has. Thus a n m × matrix has at most a rank of n. ii. If the rank < n some columns are linear combinations of others and we say that the matrix is rank-deficient and there may be an infinite number of solutions. iii. On the other hand, if the rank = n , i.e. all the columns are linearly independent and we are guaranteed either a unique solution or no solution. In the case of no solution, there is the notion of the “closest fit” in a least squares sense. That is, the least squares solution finds the closest possible solution. 2. When solving systems of equations, we want Ax=b . We define the residual as r=b-Ax , and note that the residual gives us some measure of the error. Of course the goal is to attain r= 0. 3. The least squares method finds the best-fit or best approximation to the solution in the sense that the least squares solution x minimizes the 2 L norm of the residual, i.e. it minimizes 2 r . 4. An example, consider interpolating a given a set of m data points, ( , ) i i t y , with a straight line 1 2 y x x t = + . Here, each data point leads to a new equation, for example with three data points ( m =3) we obtain 1 1 1 2 2 2 3 3 1 1 1 t y x t y x t y       =           . a. Here, if one gave the exact same pair ) , ( i i y t three times, then there is really only one point and there are an infinite number of lines that go through it. b. If the three points all line on the same line, then that line is a unique solution to the problem. c. If the three points do not all lie on a line, the problem is overdetermined and there is no solution, i.e. no straight line that passes through the points. In this case, we can look for the best least squares solution, or the line that passes as close as possible to the three points.

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