solve EB œ,, we can row reduce the augmented matrix

PY Decomposition: Factoring E œ P Y Motivation To solve EB œ,, we can row reduce the augmented matrix Ò + + ÞÞÞ+ ÞÞÞ + l, Ó µ ÞÞÞ µ Ò- - ÞÞÞ- ÞÞÞ-. Ó " # 3 8 " # 3 8 When we get to Ò-" -# ÞÞÞ -3 ÞÞÞ-8Ó in an echelon form ( this is the forward part of the row reduction process), then we can either i) use back substitution to solve for B ii) continue on with the second ( backward ) part of the row reduction process until Ò-" -# ÞÞÞ -3 ÞÞÞ-8 Ó is in row reduced echelon form, at which point it's easy to read off the solutions of EB œ, Þ A computer might take the first approach; the second might be better for hand calculations. But both i) and ii) take about the same number of arithmetic operations (see the first paragraph about back substitution in Sec. 1.2 (p. 19) of the textbook). Whether we use i) or ii) to solve EB œ,, it is sometimes necessary in applications to solve a large system EB œ, many times for the same E but changing, each time perhaps a system with millions of variables and solving thousands of times! For example, read the description of the aircraft design problem at the beginning of Chapter 2. It's too inefficient to row reduce Ò + + ÞÞÞ+ ÞÞÞ + l, Ó each time, even if a computer is doing the work. " # 3 8 The general idea: the same EROs would be repeated to row reduce the coefficient matrix Ò + " + # ÞÞÞ + 8Ó to an echelon form Y each time, is changed. To avoid the repetitions, these necessary row operations can be recorded in a matrix. This results in a factorization of E into the form E œ ^Y. Sometimes we can choose the EROs so that ^ is a unit lower triangular matrix and, in that case, we will rename the matrix ^ as P (for lower ). The result is called an PY decomposition or PY factorization of E. Terminology: A lower triangular matrix is a square matrix that has only! entries above the main!!!!! diagonal, for example Ö Ùà the matrix is unit lower triangular if it's lower! Õ! triangular and has only "'s on the diagonal, for example Ö ÙÞ The definitions "! Õ " for upper triangular and unit upper triangular are the same except that words above the main diagonal are replaced by below the main diagonal.

Here's the more formal definition: The PY Decomposition An PY decomposition of an 7 8 matrix E is a factorization E œ P Y ÚY is an echelon form of E, where ÛP is a square unit lower triangular matrix Ü A couple of observations: ñ Y is an echelon form of E, so Y must be the same shape as E À 7 8 ñ P must be square 7 7 so that all the matrices in E PY have the right sizes. And since P is unit lower triangular, P must be invertible Ðwhy? how many pivot positions in the rref of P? Look at the examples above ). ñ If E and Y happen to be square, then Y, because it's an echelon form, will automatically be upper triangular. But even when Y is not square, Y has only!'s below the leading entry in each row, so Y still resembles an upper triangular matrix. It's not always possible to find an PY decomposition for E. But when it is possible, the number of steps to find P and Y is not much different from the number of steps required to solve EB œ, (one time!) by row reduction. And if we can factor E in this way, then it takes relatively few additional steps to solve EB œ, for B each time a new, is chosen. Why? If we can writeeb œ PYB œ, ß then we can substitute YB œ C. Then the original matrix equation EB œ, is replaced by two new ones. œ P C œ, C œ YB (*) This is an improvement (as Example 1 will show) because: i) it's easy to solve PC œ, for C because P is lower triangular; and ii) after finding Cß then solving YB œ C is also easy, because Y is already in echelon form iii) P and Y remain the same whenever we change, Þ Look at the details in the following example. Example 1 Suppose " # # EB œ, is the equation # "! B œ " Õ! % # Õ Here is an PY factorization of E ( don't worry for now about where it came from): " # # "! œ " # # % Õ! % # Õ %! " Õ ##!!

To solve " # " # # # "! B œ # % B œ ". Õ! % # Õ %! " Õ ##!! Õ Å Å Å EB œ P Y B œ, i) substitute YB œ C to get the two equations (*) œ P C œ, that is, C œ YB Ú # Ý PC œ # "! C œ " Õ %! " Õ Û " # ÝC œ YB œ! % B Ü Õ!! ## ii) First solve for C Þ To do this, we could further reduceòp, Ó to row reduced echelon form, or we can use forward substitution just like back substitution but starting with the top equation because that's where the simplest equation is. Either way we need only a few steps to get C because P is lower triangular. Using forward substitution gives: # Ú C" œ # # "! C œ ", so ÛC# œ " #C" œ and then Õ %! " Õ Ü % C œ C œ so C œ C" # C # œ ÕC Õ # iii) Substitute C into the second equation to get " # #! % B œ, which we Õ ##!! Õ can quickly solve with back substitution or further row reduction. Using back substitution we start at the bottom and work up:: B" "" So B œ B œ Ö Ù # ÕB Õ Ú * ÝB œ Ð## Ñ œ ## "& ÛB# œ %ÐB Ñ œ, so B# œ Ý ÜB œ # B #ÐB Ñ œ " # ) & "" * ## ) "" & "" ""

Example 2 For the matrix E œ " # # "! in Example 1, how could you actually find the Õ! % # PY decomposition? First, row reduce E to an echelon form Y. ( It turns out that we need to be careful about what EROs we use during the row reduction but for now, just follow along; the discussion that comes later will explain what caution is necessary.) Each elementary row operation that we use corresponds to left multiplication by an elementary matrix I 3, and we record those matrices at each step. so we get E œ " # " # " # # "! µ! % µ! % œ Y (an echelon form) Õ! % # Õ! % # Õ ##!! Å Å I " œ # "! I# œ! "! Õ!! " Õ %! " I# I" E œ Y. Then E œ ÐI# I" Ñ Y œ I" I# ÐI" I# Ñ Y Æ Æ Æ Æ Æ " # " # # "! œ # "! Y œ # % Õ! % # Õ!! " Õ % Õ %! "! " Õ ##!! In this example, I" I# turns out to be unit lower triangular, so we're done: let P œ I" I#, and E œ PY is the decomposition we want. Is it just luck that I" I# turned out to be unit lower triangular? No, it's because of the particular steps we used in the row reduction: see the discussion below.

Example 1 continued: same coefficient matrix E but different, Now that E œ PY is factored, we can solve another equation EB œ " # easily: Õ! See above: here only, has changed! Æ Ú " # "! Ý C œ # Õ %! " Õ! Û " # Ý! % B œ C Ü Õ!! ÚC " œ " By forward substitution ÛC# œ # 2Ð"Ñ œ 0 Ü % C œ Ð0Ñ œ 0 ## ß so " C œ! Þ Õ! " # " Then the second equation becomes! % B œ! and, by back substitution we get Õ ##!! Õ! ÚB œ! " ÛB# œ! %Ð!Ñ so B# œ! so B œ! Þ ÜB œ "Ð!Ñ #Ð!Ñ œ " Õ! " When does the method used in Example # produce an PY decomposition? We can always perform the steps illustrated in Example # to get a factorization E œ ^ Y, where Y is an echelon form of E and ^ is a product of elementary matrices Ðin Example #, ^ œ I" I# Ñ. But ^ won't always turn out to be lower triangular (so that ^ might not deserve the name P ). When does the method work? Suppose ( as in Example 2) that the only type of ERO used to row reduce E is add a multiple of a row to a lower row. Let these EROs correspond to the elementary matrices I" ßÞÞÞßI:. Then I ÞÞÞ I E œ Y and therefore E œ I ÞÞÞ I YÞ : " " Each matrix I3 and its inverse I3 will be unit lower triangular Ðwhy? Ñ. Then P œ I " ÞÞÞ I: is also lower triangular with "'s on its diagonal. ( Convince yourself that a product of unit lower triangular matrices is a unit lower triangular matrix.) Note that i) if a row rescaling ERO had been used, then the corresponding elementary matrix would not have only "'s on the diagonal and therefore the product I ÞÞÞ I: might not be unit lower triangular. " :

ii) if a row interchange ( swap ) ERO had been used, then the corresponding elementary matrix would not be lower triangular and therefore the product I ÞÞÞ I might not be lower triangular. " : To summarize: the method illustrated in Example 2 always produces an PY decomposition for E if the only EROs used to row reduce E to Y are of the form add a multiple of a row to a lower row. Ð** Ñ In that case: if the EROs used correspond to multiplication by elementary matrices I ßÞÞÞÞß I, theni ÞÞÞ I E œ Y is in echelon form and " : : " E œ PY, and P œ ÐI : ÞÞÞ I" Ñ œ I " ÞÞÞ I: Ð*** Ñ is unit lower triangular The restriction in (**) sounds very restrictive until to stop to think about it: ñ When you do row reduction systematically, according to the standard steps as outlined in the text, you never add a multiple of a row to a higher row. The reason for using a row replacement ERO is to create!'s below a pivot. In other words, adding multiples of rows only to lower rows doesn't actually constrain you at all: it's just standard procedure in row reductions. ñ No row rescaling can be an arithmetic inconvenience, but nothing more than an inconvenience. When row reducing by hand, rescaling a row may be helpful to simplify arithmetic, but rescaling is never necessary to get to an echelon form of E. ( Of course, row rescaling might be needed to proceed on to the row reduced echelon form because rescaling may be necessary to get "'s in the pivot positions. ) Side comment: If we did allow row rescaling, it would only mean that we could might end up with a lower triangular P in which some diagonal entries are not "'s. From the point of view of solving EB œ,, this would not be a disaster. However, the text follows the rule that P should have only "'s on its diagonal, so we'll stick to that convention.! #! But Ð Ñ really does impose some restriction on us:: for example, E œ #!! simply Õ!! # cannot be reduced to echelon form using (**) a row interchange is necessary. Here the method in Example 1 simply will not work to produce an PY decomposition. However ( see the last section of the online version of these notes), there is a work-around that is almost as good as an PY decomposition in such situations.

To find P more efficiently If E is large, then row reducing E to echelon form might involve a very large number of elementary matrices I3 in the formula Ð Ñ so that using this formula to find P is too much work. Fortunately, there is a technique, illustrated in Example 3, to write down P, entry by entry, as you row reduce E. ( The text gives a slightly different presentation, and offers an explanation of why it works. It's really nothing more than careful bookkeeping. Since these notes are just a survey about PY decompositions, we will only write down the method. ) Assume E is 7 8 and that it can be row-reduced to Y following the rule Ð** ÑÞ Then P will be a square 7 7 unit lower triangular matrix. You can write down P using these steps: ñ Write down an empty 7 7 matrix P ñ Put "'s on the diagonal of P and!'s in each position above the diagonal ñ Whenever you perform a row operation add - times row 3 to a lower row 4, then put - in the Ð4ß3Ñ position of P Careful: notice the sign change on -, and put - in the Ð4ß3Ñ position of Pß not in the Ð3ß4Ñ position ñ When E has been reduced to echelon form, stop. If there are any remaining empty positions in P, fill them with!'s. If we apply these steps while row reducing in Example 2, here's what we'd get: ñ E was, so start with P œ * "! Õ * * " ñ The row operations we used in Example 1 were: add # Ðrow 1 Ñ to row #, so put # in the Ð#ß"Ñ position in P: that is, P#" œ # % % % add Ð row #Ñ to row 3, so put in the Ðß#Ñ entry in P: that is, P œ # Now we have P œ # "! Õ * " ñ Fill in the rest of P with!'sß to get % P œ # "! ß the same as we got by multiplying elementary matrices Õ %! "

Example 3 In this example, E is a little larger and not square. But otherwise, the calculations are parallel to the ones we did earlier. If possible, "!! # " find an PY factorization of E œ #! # ) and use it to solve Õ! " EB œ, œ # Õ "!! # " "!! # " E œ #! # ) µ!! & # "# & Õ! " Õ! " "!! # " "!! # " µ!! & # "# & µ!! & # "# & œ Y Õ!! # & # Õ " "!!!! Y is in echelon form, and we used only EROs of type (**).. To find P, which must be À start with P œ "!. The row operations we used Õ " were Ú add #* Ð row "Ñ to Ð row #Ñ Ý Û add "* Ð row "Ñ to Ð row Ñ so P œ # "! Þ Then Õ # " Ý & Ü # add Ð row #Ñ to Ð row Ñ & P Y E œ "!! # " #! # ) œ "!! # " # "!!! & # "# & Õ! " Õ # " Õ " "!!!! We can solve EB œ, œ # by writing Õ & & & & &

Ú Ý Û Ý Ü PC œ,, that is C œ YBß that is C" # "! C # œ, œ # Õ # " ÕC Õ & B" B# "!! # "!! & # "# & C" B œ C# Õ " "!!!! B% & & Ö Ù ÕC B& ÕB ' The first equation gives C œ C" C # œ # #ÐC" Ñ œ!, and then the second ÕC Õ # C C Õ # " & # B" B# "!! # " B equation becomes!! & # "# & œ!. So the solution is Õ " " B%!!!! Ö Ù Õ # & & B& ÕB B " "" %B& #B ' ""! % # B# B #! "!! B #B& B' % %! # B œ œ œ B# B& B' B% B&!!! "! Ö Ù Ö Ù Ö Ù Ö Ù Ö Ù Ö Ù B& B &!! "! ÕB Õ B Õ! Õ! Õ! Õ " ' ' ' MATERIAL BEYOND THIS POINT IN THESE NOTES IS OPTIONAL

A Useful Related Decomposition Sometimes row interchanges are simply unavoidable to reduce E to echelon form. Other times, even when row interchanges could be avoided, they are desirable in computer computations to reduce roundoff errors (see the Numerical Note about partial pivoting in the textbook, p. 17). What happens if row interchanges are used? We can still get a factorization that resembles E œ PY and every bit as useful for solving equations. Here's how.. Note: as before, all row replacement operations follow the rule (**), and we do not allow rescaling of rows. 1) Look ahead ( on scratch paper!) at what you need to do to reduce E to an echelon form, using (**) and, if necessary, row interchanges. Make note what row interchanges will be used. Then go back to the beginning and rearrange rows in E that is, do all these row interchanges first. The result is E rearranged. Let's look at this in more detail: Suppose the row interchanges we use are represented by the elementary matrices I" ßÞÞÞßI5Þ Then the rearranged E is just I 5 ÞÞÞ I" E œ TE, where T œ I 5 ÞÞÞ I". Since T œ I 5 ÞÞÞ I" œ I 5 ÞÞÞ I" M we see that T is just the final result of when the same row interchanges are performed on M that is, T is just M with some rows rearranged. T is called a permutation matrix; the multiplication TE rearranges ( permutes ) the rows of E in exactly the same way that the rows of M were permuted to create T. A permutation matrix T is the same as a square matrix with exactly one " in each row and exactly one " in each column. Why? T is a product of invertible (elementary) matrices, so T is invertible and T œ I " ÞÞÞ I: is also a permutation matrix (why? what does each I3 do to M? ). ( If you know the steps to write down T, it's just as easy to actually write down the matrix T. Or, perhaps you can convince yourself that T œ T X Þ) Since T TE œ E, multiplication by T rearranges the rows of TE to restore the original matrix E. 2) T he matrix E rearranged is TE, and TE can be row reduced to an echelon form only using row replacements (**) no need for further row interchanges. Therefore we can use our previous method to find an PY decomposition of TE À TE œ PY 3) Then E œ ÐT PÑY. Here, T P is a permuted lower triangular matrix, meaning lower triangular with some rows interchanged. MATLAB Help gives T P the cute name psychologically l ower triangular matrix, a term never used anywhere else to my knowledgeþ This factorization is just as useful for equation solving as the PY decomposition. For example, we can solve EB œ ÐT PÑYB œ, by first substituting C œ YB. It's still easy to solve ÐT PÑC œ, because T P is psychologically lower triangular ( see illustration below), and

then knowing C, it's easy to solve YB œ C because Y is in echelon form.. For example suppose ÐT PÑC œ, is the equation # C "!! ÙÖC # " Ö ÙÖ Ù œ Ö Ù # # " " C " Õ" " "! ÕC Õ! % ß permuted lower triangular matrix We can simply imitate forward substation, but solve working the rows in the order that (rearranged) would make a lower triangular matrix (simplest row first, the row that involves only the variable C" ÑÞ That is, Ú ÝC" œ ", from the second row, then C# œ! #C" œ #, from the first row, then Û Ý C œ! C" C# œ ", from the fourth row, then ÜC œ " #C #C C œ #, from the third row % " # (In some books or applications, when you see E œ PY it might even be assumed that P is permuted lower triangular rather than lower triangular. Be careful when reading other books to check what the author means.) OR you could solve EB œ ÐT PÑYB œ, as follows: The equation TEB œ T, has exactly the same solutions as EB œ, because: ñ If TEB œ T, is true, then multiplying both sides by T shows that B is also a for EB œ,, and ñ If EB œ, is true, then multiplying by T shows that TEB œ T, is also true. If you think in terms of writing out the linear systems of equations: TEB œ T, is the same system of equations as EB œ, but with the equations listed in some other interchanged order determined by the permutation matrix T. Since we have an PY decomposition for TEß solving TEB œ PYB œ T, for B is easyþ " " # " " " Example 3 Let E œ Ö ÙÞ " % % Õ! " " 1) On scratch paper, do enough steps row reducing E to echelon form ( no row rescaling allowed, and, following standard procedure, only adding multiples of rows to lower rows ) to see what row interchanges, if any, are necessary. It turns out that only one, interchanging rows 2 and 4 at a certain stage, will be enough.

2) So go back to the start and perform that row interchange first, creating " " #!! "! " TE œ Ö Ù, where T œ Ö Ù " % %!! "! Õ" " " Õ! 3) Row reduce TE to an echelon form Y, keeping track of the EROs used: " " # " " # " " # " " #! " " Ð"Ñ! " " Ð#Ñ! " " ÐÑ! " " Ö Ù µ Ö Ù µ Ö Ù µ Ö Ù " % %! 2! 2!! 1 Õ" " " Õ" " " Õ!! Õ!! " " # Ð%Ñ! " " µ Ö Ù œ Y!! 1 Õ!!! The row operations were (1) add row(1) to row 3 (2) add row(1) to row 4 (3) add 3 row(2) to row Ð%Ñ add 1*row(3) to row 4 Using the method described earlier to find P more efficiently we can write down P as we row reduce TE À 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 * 1 0 0 * 1 0 0 * 1 0 0 * 1 0 0 Ö Ù Ä Ö Ù Ä Ö Ù Ä Ö Ù * * 1 0 " * 1 0 " * 1 0 " 1 0 Õ * * * 1 Õ * * * 1 Õ" * * 1 Õ" 1!!! Ä Ö Ù Ä Ö Ù œ P " "! " "! Õ" " " Õ"! " " Then TE œ P Y " " #! " " #! " "!! " " Ö Ù œ Ö Ù Ö Ù " % % "! 1 Õ" " " Õ"! " " Õ!!!

# # " ( To solve EB œ Ö Ùß write TEB œ PYB œ T, œ Ö Ù Ðrows 2,4 of, interchanged, as Õ( Õ" # ( for E earlier Ñ. We then solve PYB œ Ö Ù in the same way as before Õ" Factoring E with slight variations from E œ PY PHY decompositions Suppose we can row reduce E using only EROs described in (**) and get the factorization E œ PY. If we loosen up and then allow rescaling of the rows of Y, we can convert all the leading entries (pivot positions) in Y to "'s. This amounts to a factorization E œ PHY where, as before, P is unit lower triangular, H is a (square) diagonal matrix and Y is an echelon form of E which (by rescaling) has only "'s in its pivot positions. Since P is unit lower triangular, this produces more symmetry in appearance between P and Y. ( For example: suppose E is square and invertible, and that E œ LDU as described. Then P and Y will both be square, with P unit lower triangular and Y unit upper triangular! 's below the diagonal and only "'s on the diagonal. Why must Y have that form? ) For Y, this is an aesthetically nicer form, but frankly, it doesn't help much in solving a system of equations. To illustrate the (easy) details: first factor as E œ PY. Create a diagonal matrix H as follows: If Row of 3 Y has a pivot position: Let : 3Ð Á!Ñ be the number in the pivot position. Factor : 3 out of Row 3 of Y th (leaving a " in the pivot position) and use the number : 3 as the 3 diagonal entry in H. th Otherwise, make! the 3 diagonal element of H. w After factoring out the : 3's from the rows of Y, what's left is a new matrix Y that has only "'s in its pivot positions, and HY w œ Y (because multiplication on the left by a diagonal matrix just w w multiples each row of Y by the corresponding diagonal element of HÑÞ Therefore E œ PHY Þ At this point, we abuse the notation and write Y in place of Y w we do this only for neatness, since the rightmost matrix in this factorization is traditionally called Y : just remember that the Y in E œ PHY usually isn't the same as the Y in the original factorization E œ PY Þ E œ PHY is cleverly called an PHY matrices from Example 2. decomposition of EÞ Here's an illustration using the E œ "!! # " #! # ) œ "!! # " # "!!! & # "# & Õ! " Õ # " Õ " "!!!! & & &

P Y Ð old Y Ñ "!! # " œ # "! 0 5! # "#!! " & & " Õ # " Õ " 0 0 Õ!!! "! & & P H Y Ð new Y Ñ Note for those who use MATLAB MATLAB has a command ÒPß Yß TÓ œ lu(a) ( lu is lowercase in MATLAB ) Unless you're very comfortable with MATLAB and the material, you should probably avoid confusion by only using the Matlab command on square matrices E ( See the technical note, below.) In that case, MATLAB` should give you for which TE œ PY. 1) a square unit lower triangular P 2) Y in echelon form, and 3) a permutation matrix T If it's possible simply to factor E œ PYß then MATLAB might just give T œ M œ the identity matrix. However, sometimes MATLAB will permute some rows even when it's not mathematically necessary (because doing so, in large complicated problems, will help reduce roundoff errors). See the Numerical Note about partial pivoting in the textbook, p. 17). If E is not square, then MATLAB gives a result so that TE œ PY is still true, but the shapes of the matrices may vary from our discussion. Technical note from a department MATLAB expert: If A is 7 8, then ÒPßYßTÓ œ lu ÐEÑ should return an 7 7 unit lower triangular L, 7 8 upper triangular U, and 7 7 permutation P satisfying PA=LU. In fact, the dimensions of what MATLAB returns depends on whether 7 œ 8, 7 8, or 7 8. Evidently the algorithm performs Gaussian elimination with partial pivoting as long as it can on A, in place, then extracts relevant parts of the changed A as L and U, with 1's inserted into L. P is stored as a vector but output as a matrix. Thus U will come out as 7 8 and P will be 7 7, but L will be 7 5 with 5 œ min Ð7ß8Ñ because its zero-rows are superfluous. V. Wickerhauser)

Here's a small-scale illustration of another way in which the PY decomposition can be helpful in some large, applied computations. A band matrix refers to a square 8 8 matrix in which all the nonzero entries lie in a band that surrounds the main diagonal. For example, in the following matrix all nonzero entries are in the band that consists of one superdiagonal above the main diagonal and two subdiagonals below the main diagonal. The number of superdiagonals in the band is called the upper bandwidth ("ß in the example below) and the number of subdiagonals in the band is the lower bandwidth (2, in the example). The total number of diagonals in the band œ Ðupper bandwidth lower bandwidth Ñ is called the bandwidth (4, in the example). " "! # " %! (all other entries Ö Ù # " " " Õ " # % œ!ñ!! A diagonal matrix such as! #! is an extreme example for which bandwidth œ "Þ. Õ!! ' Use of the term band matrix is a bit subjective: in fact, we could think of any matrix as a band matrix if, say, all entries are nonzero, then the matrix is just one that has the largest possible bandwidth. In practice, however, the term band matrix is used to refer to a matrix in which the bandwidth is relatively small compared to the size of the matrix. Therefore the matrix is sparse a relatively large number of entries are!. Look at the simple problem about steady heat flow in Exercises 33-4 in Section 1.1 in the text; a more larger problem, set up the same way, then occurs in Exercise 31, Section 2.5.

Given the temperatures at the nodes around the edge of the plate, we want to approximate the temperature B" ßÞÞÞßB) at the interior nodes numbered 1,...8. This leads, as in Exercise 33, Section 1.1 to an equation EB œ, with a band matrix E À %!!! & %!!! "&! %!!! %! "! B œ!!! %!!!!! %! "! Ö Ù Ö Ù!!!!! % #! Õ!!!!! % Õ! If we can find an PY decomposition of E, then P and Y will also be band matrices ( Why? Think about how you find P and Y from E). Using the MATLAB command ÒPß Yß TÓ œ lu (A) gives an PY decomposition. Rounded to 4 decimal places, we get P œ "Þ!!!!!!!!!!!!Þ#&!! "Þ!!!!!!!!!!!Þ#&!!!Þ!''( "Þ!!!!!!!!!!!Þ#''(!Þ#)&( "Þ!!!!!!!!!!!Þ#'(*!Þ!) "Þ!!!!!!!!!!!Þ#*"(!Þ#*#" "Þ!!!!!! Ö Ù!!!!!Þ#'*(!Þ!)'" "Þ!!!!! Õ!!!!!!Þ#*%) 0.2931 "Þ!!!!

Y œ %Þ!!!! Þ!!!! Þ!!!!!!!!!! Þ(&!!!Þ#&!! Þ!!!!!!!!!! Þ( Þ!''( Þ!!!!!!!!!! Þ%#)'!Þ#)&( Þ!!!!!!!!!! Þ(!) Þ!) Þ!!!!!!!!!! Þ*"*!Þ#*#" Þ!!!! Ö Ù!!!!!! Þ(!&# Þ!)'" Õ!!!!!!! Þ)') The PY decomposition can be used to solve EB œ,, as we showed earlier in these notes. The important thing in the example is not actually finding the solution, which turns out to be Þ*&'* 'Þ&))& %Þ#*# (Þ*(" B œ &Þ'!#* )Þ('!) Ö Ù *Þ%""& Õ"#Þ!%" What's interesting and useful to notice: ñ From earlier in these notes: if we need to solve EB œ, for different vectors,, then factoring E œ PY saves time many times ñ A new observation is that for a very large 8 8 sparse band matrix E (perhaps thousands of rows and columns but with a relatively narrow band of nonzero entries), we can store E with much less computer space than is needed for a general 8 8 matrix: we only need to store the entries in the upper and lower bands, and two numbers for the upper and lower bandwidths; no need to waste space storage space for all the other matrix entries which are known to be!'s. Since P and Y are also sparse band matrices, they are also relatively economical to store. ñ Using Eß P and Y can let us avoid computations that use E. Even if E is a sparse band matrix, E may have no band structure and take up a lot more storage space than E, Pß and YÞ For instance, in the preceding example,

E œ!þ#*&!þ!)''!þ!*%&!þ!&!*!þ!")!þ!##(!þ!"!!!þ!!)#!þ!)''!þ#*&!þ!&!*!þ!*%&!þ!##(!þ!")!þ!!)#!þ!"!!!þ!*%&!þ!&!*!þ#("!þ"!*!þ"!%&!þ!&*"!þ!")!þ!##(!þ!&!*!þ!*%&!þ"!*!þ#("!þ!&*"!þ"!%&!þ!##(!þ!")!þ!")!þ!##(!þ"!%&!þ!&*"!þ#(" 0.1093!Þ!*%&!Þ!&!*!Þ! 227!Þ 0318!Þ 0591!Þ 1045!Þ 1093!Þ 3271!Þ 0509!Þ!*%& Ö Ù!Þ!"!!!Þ!!)#!Þ!")!Þ!##(!Þ!*%&!Þ!&!*!Þ#*&!Þ!)'' Õ!Þ!!)#!Þ!"!!!Þ!##(!Þ!")!Þ!&!*!Þ!*%&!Þ!)''!Þ#*&