Chapter 5. The Inverse; Numerical Methods

Transcription

1 Vector Spaces in Physics 8/6/ Chapter. The nverse Numerical Methods n the Chapter we discussed the solution of systems of simultaneous linear algebraic equations which could be written in the form C - using Cramer's rule. There is another, more elegant way of solving this equation, using the inverse matri. n this chapter we will define the inverse matri and give an epression related to Cramer's rule for calculating the elements of the inverse matri. We will then discuss another approach, that of Gauss-Jordan elimination, for solving simultaneous linear equations and for calculating the inverse matri. We will discuss the relative efficiencies of the two algorithms for numerical inversion of large matrices.. The inverse of a square matri. Definition of the inverse. The inverse of a scalar number c is another scalar, say d, such that the product of the two is equal to : cd=. For instance, the inverse of the number is the number.. We have defined multiplication of one matri by another in a way very analogous to multiplication of one scalar by another. We will therefore make the following definition. Definition: For a given square matri, the matri B is said to be the inverse of if B B - We then write B. Notice that we have not guaranteed that the inverse of a given matri eists. n fact, many matrices do not have an inverse. We shall see below that the condition for a square matri to have an inverse is that its determinant not be equal to zero. Use of the inverse to solve matri equations. Now consider the matri equation just given, C - We can solve this equation by multiplying on both sides of the equation by : C C - C. Thus, knowing the inverse of the matri lets us immediately write down the solution to equation -. s an eample, let us consider a specific eample, where is a matri. -

2 Vector Spaces in Physics 8/6/ - C - f we knew the inverse of, we could immediately calculate C. n this simple case, we can guess the inverse matri. We write out the condition for the inverse, ij - s a first guess we try to make come out to zero one possibility is ij -6 Now we arrange for to be zero: ij -7 f we now look at the diagonal elements of, they come out to be = and = -. We can fi this up by changing the sign of the, and, elements of the inverse, and by multiplying it by /. So we have ij -7 Now that we have the inverse matri, we can calculate the values and : 9 6 C -8 So, the solution to the two simultaneous linear equations is supposed to be = -, =. We will write out the two equations in long form and substitute in.

3 Vector Spaces in Physics 8/6/ t checks out!. The inverse matri by the method of actors. Guessing the inverse has worked for a matri - but it gets harder for larger matrices. There is a way to calculate the inverse using actors, which we state here without proof: ji ij -9 ji M ji Here the minor Mpq is the determinant of the matri obtained by removing the p-th row and q-th column from the matri. Note that you cannot calculate the inverse of a matri using equation -9 if the matri is singular that is, if its determinant is zero. This is a general rule for square matrices: inverse does not eist Eample: Find the inverse of the matri - Here are the calculations of the four elements of. First calculate the determinant: - Then the matri elements: -

4 Vector Spaces in Physics 8/6/ - - So, - Check that this inverse works: - Eample: Calculate the inverse of the following matri using the method of actors: - Solution: This is getting too long-winded. We will just do two representative elements of. 6-6 B. Time required for numerical calculations.

5 Vector Spaces in Physics 8/6/ Let s estimate the computer time required to invert a matri by the method of actors. The quantity of interest is the number of floating-point operations required to carry out the inverse. The inverse of a nn matri involves calculating n actors, each of them requiring the calculation of the determinant of an n-n- matri. So we need to know the number of operations involved in calculating a determinant. Let's start with a determinant. There are two multiplications, and an addition to add the two terms. n= gives FLOPs. FLOP = Floating-Point Operation. To do a determinant, the three elements in the top row are each multiplied by a determinant and added together: FLOPs + FLOPs for addition n= requires + FLOPs. Now we can proceed more or less by induction. t is pretty clear that the determinant of a matri requires calculations of a determinant: --> FLOPs. nd for a determinant, operations. t is a pretty good approimation to say the following: No. of operations for nn determinant = n! -7 This means that calculating the inverse by the actor method n actors requires n n! FLOPs. fast PC can today do about GigaFLOPs/sec. This leads to the table given below showing the eecution time to invert matrices of increasing dimension. dim. for for inverse time sec for inverse time sec n determinant method of actors PC Gauss-Jordan PC n! n^n! n^ 8 8E-.E-9 6.E-9 8.8E-8 8.8E-8 6.6E E E E E E E E E E+.7869E E-6.77E+.97E E+.6E E E+.79E E E+.77E E-6 9.6E+7.99E E-6.9E E E+9.E+.E+ 7.7E-6.E+.6E+.6E+ 9.9E-6.8E+.677E+.677E E-6 6.E+.778E+6.778E E-6 Table -. Floating-point operations required for calculation of n n determinants and inverses of n n matrices, and computer time required for the matri inversion. Results -

6 Vector Spaces in Physics 8/6/ are given for two different numerical methods. s a useful conversion number, the number of seconds in a year is about. 7. t can be seen from the table that the inversion of a matri could take a time on a fast computer about equal to the age of the Universe. This suggests that a more economical algorithm is desirable for inverting large matrices! Teasing Mathematica: Try this calculation of a determinant. n= m=table[random[],{n},{n}] Det[m] Does this suggest that the algorithm used for Table - is not the fastest known C. The Gauss-Jordan method for solving simultaneous linear equations. There is a method for solving simultaneous linear equations that avoids the determinants required in Cramer's method, and which takes many fewer operations for large matrices. We will illustrate this method for two simultaneous linear equations, and then for three. Consider the matri equation solved above, C - This corresponds to the two linear equations -8 standard approach to such equations would be to add or subtract a multiple of one equation from another to eliminate one variable from one of the equations. f we add the first equation to the second, we get addt eq. toeq Now we eliminate from the top equation, by subtracting / the bottom equation: add eq. toeq. 9 - from eq. subtract / eq. 9 nd finally, multiply the second equation by /: add eq. toeq. 9 - subtract / eq. from eq. multiply eq. by / 9 So we have found that = - and =, as determined earlier in the chapter using the inverse. - 6

7 Vector Spaces in Physics 8/6/ Note that the same operations could have been carried out using just the coefficients of the equations, and omitting and, as follows. The assembly of the coefficients of and and the constants on the right of the equation is refered to as the augmented matri. add eq. toeq. 9 - subtract. eq. from eq. multiply eq. by / 9 The results for and appear in the column to the right. Eample: Use the Gauss-Jordan method to solve the system of linear equations represented by C - Solution: We set up the augmented matri, and then set about making the matri part of it diagonal, with ones on the diagonal. This is done in the following systematic fashion. First use the first equation to eliminate and. Net use the second equation to eliminate. subtract / from and from / / / 7 / - subtract from / / Net we work upwards, using equation to eliminate and. fter that, equation is used to eliminate. t this point the matri is diagonal. the final step is to multiply equations and by a constant which makes the diagonal elements of become unity. - 7

8 Vector Spaces in Physics 8/6/ / add./ to, add to / subtract multiply by. from and by - The solution for the unknown 's is thus =, =, = -. - SUMMRY: Work your way through the matri, zeroing the off-diagonal elements, N THE ORDER SHOWN BELOW, zeroing ONE, then TWO, then THREE, etc. f you try to invent your own scheme of adding and subtracting rows, it may or may not work. SX FVE. ONE FOUR TWO THREE D. The Gauss-Jordan method for inverting a matri. There is a very similar procedure which leads directly to calculating the inverse of a square matri. Suppose that B is the inverse of. Then B B B B -6 B B B B B B This can be thought of us three sets of three simultaneous linear equations: B B, B B B, -7 B B B, B These three sets of equations can be solved simultaneously, using a larger augmented equation, as follows: - 8

9 Vector Spaces in Physics 8/6/ - 9 / / / / / / / / / / / / / / / / / / by - and / by multiply from subtract to add to, add./ from subtract and from from / subtract -8 So, the result is B -9 The check is to multiply by its inverse: B -9 So the inverse just calculated is correct. Time for numerical inverse. Let us estimate the time to invert a matri by this numerical method. The process of zeroing out one element of the left-hand matri requires multiplying the line to be subtracted by a constant n FLOPs, and subtracting it n FLOPs. This must be done for approimately n matri elements. So the number of floating-point operations is about equal to n for matri inversion by the Gauss- Jordan method. Consulting Table - shows that, for a matri, the time required is less than a milli-second, comparing favorably with years for the method of actors. Number of operations to calculate the inverse of a nn matri. method number of FLOPs actor n n! Gauss-Jordan n PROBLEMS Problem -. a Use the method of actors to find the inverse of the matri

10 Vector Spaces in Physics 8/6/ C. b Check your result by verifying that CC. Problem -. Use the Mathematica function nverse to find the inverse of the matri C. See ppendi C for the necessary Mathematica operations. Check your result. Problem -. Prove that if an operator has both a left inverse call it B and a right inverse call it C, then they are the same that is, if B and C then B C. [Be careful to assume only the properties of B and C that are given above. t is not to be assumed that, B and C are matrices.] Problem -. Suppose that B and C are members of a group with distributive multiplication defined, each having an inverse both left-inverse and right-inverse. Let be equal to the product of B and C, that is, BC. Now consider the group member D, given by D C B. Show by direct multiplication that D is both a left inverse and a right inverse of. [Be careful to assume only the properties of B and C that are given above. t is not to be assumed that, B and C are matrices.] Problem -6. a Use the method of Gauss-Jordan elimination to find the inverse of the matri C. b Check your result by verifying that CC. -