ࡱ> \Bh^]Oh+'0 ,< `l    The MATLAB Notebook v1.5.2"Template for The MATLAB Notebook osDapartment of Mathematicsteapaapam-bookeDapartment of Mathematicste29aMicrosoft Word 8.0m@6)@@f@\/%Section 1: Dot products and Orthogonal Matrices MATLAB and Dot Products We will begin with a review of the dot product and orthogonal and symmetric matrices. Much, but not all, of this material is included in chapters 6 and 7 of Lay's book. We will also review the use of MATLAB as we proceed. We recall that the term "vector" usually denotes a column vector. This is forced on us by our desire to have the product of a matrix on the left with a vector on the right turn out to be a vector. A column vector is represented in MATLAB by a semicolon delimited list. A comma or space delimited list is a row vector, and a column vector can also be represented as the transpose of a row vector. For example [1;2;3] ans = 1 2 3 [1 2 3]' ans = 1 2 3 [1,2,3]' ans = 1 2 3 We see from the forgoing that the transpose, denoted in standard notation by a superscripted T as in AT, is represented in MATLAB by a "prime" or single quote. Let us see this for a matrix. A=[1 2 3;4 5 6;7 8 9] A = 1 2 3 4 5 6 7 8 9 A' ans = 1 4 7 2 5 8 3 6 9 We recall now that the dot product vw of two vectors v and w may be defined as the matrix product vTw. Let us see how this works in MATLAB. v=[1;2;3] v = 1 2 3 w=[4;5;6] w = 4 5 6 v'*w ans = 32 w'*v ans = 32 Notice that MATLAB does not distinguish between a 1x1 matrix (which is what vTw really is) and its single entry. Neither do we ordinarily, but we should remember that the product of a scalar with a matrix is not in general an instance of matrix multiplication. Notice also that the dot product is symmetric as regards v and w. This is important, and can be derived from two observations: vTw is a 1x1 matrrix, and therefore symmetric (equal to its transpose). The transpose of a matrix product is the product of the transposes in the opposite order. Let us see, by way of contrast, what happens if we look at the products vwT and wvT. v*w' ans = 4 5 6 8 10 12 12 15 18 w*v' ans = 4 8 12 5 10 15 6 12 18 These are 3x3 matrices and are not equal, although each is the transpose of the other. Matrix Transposes and Dot Products We now state and prove an important identity regarding transposes and the dot product: Proposition 1.1: If v is a vector of dimension n, A is an mxn matrix, and w is a vector of dimension m, then Avw=vATw. Proof: This follows from the fact that (Av)T=vTAT, so that both sides are represented by the matrix product vTATw. We will be able to prove a converse to this result as after we state another important fact regarding matrices and dot products: Proposition 1.2:If A and B are mxn matrices and Avw=Bvw for all vectors v and w of the appropriate dimensions, then A=B. The proof of Proposition 1.2 is left as an exercise. The converse we promised above now follows as a simple corollary: Corollary 1.3: If A is an mxn, matrix, B is an nxm matrix, and Avw=vBw for all vectors v of dimension n and w of dimension m, then B=AT. Orthogonal Matrices We come now to orthogonal matrices. We recall that a square matrix U is called orthogonal if UvUw=vw for all vectors v and w of the appropriate dimension. We can immediately prove: Proposition 1.4 A square matrix U is orthogonal if and only if UT=U-1. Proof: If U is orthogonal, we have vw=UvUw=vUTUw for all v and w of the appropriate dimension. It follows that UTUw=w for all w or, equivalently, that UTU=I, where I is the identity matrix of the appropriate dimension. It follows that UT=U-1. On the other hand, if we assume UT=U-1, we have UvUw=vUTUw=vw. Another important characterization of orthogonal matrices follows from this proposition. Proposition 1.5: A matrix is orthogonal if and only if its columns form an orthonormal set. Proof: For any defined matrix product AB, the ijth entry of the product is the matrix product of the ith row of A (on the left) and the jth column of B (on the right). In particular, if A and B have the same shape, then ATB is defined and the ijth entry of ATB is the dot product of the ith column of A with the jth column of B. It follows that UTU is the identity matrix if and only if the dot product of the ith and jth columns of U is 0 when i and j are distinct and 1 when they are equal. Since any matrix U is equal both to (UT)T and to (U-1)-1, it follows that if U is orthogonal, then so is UT, so that the rows of an orthogonal matrix (or, more properly, the transposes of the rows, since the dot product is only defined for column vectors) also form an orthonormal set. Orthogonal Complements and Orthogonal Projection Let V be a subspace of Rn. We recall from Section 6.1 of your text that V4% denotes the orthogonal complement of V, which is the vector subspace of Rn consisting of all vectors perpendicular to V. It follows from Theorem 3 on page 375 that V and V4% have complementary dimensions Proposition 1.6: If V is a subspace of Rn, then dim(V) + dim(V4%)=n. The proof of Proposition 1.6 is left as an exercise. An important consequence is Propostion 1.7: If V is a subspace of Rn, then V4% 4%=V. It is immediate from the definition of V4% that any vector in V is perpendicular to any vector in V4%. From this it follows that V is a subspace of V4% 4%. But by Proposition 1.6, V and V4% 4% have the same dimension. It follows that they are equal. We now proceed to amplify somewhat the discussion of orthogonal projection in Section 6.3 of Lay's text. From page 381, if y and u are vectors, the orthogonal projection of y on u, (or more properly on Span{u}) is given by  EMBED Equation.3 . Here the factor on the left is a scalar and the factor on the right is a vector and the product is not an instance of matrix multiplication. However, we can write this product differently by treating the numerator of the fraction as a 1x1 matrix and setting it on the right, so that we have  EMBED Equation.3 . This shows us that projection on u is a linear transformation whose matrix is given by  EMBED Equation.3 . Proposition 1.8: Pu is symmetric,  EMBED Equation.3 , and  EMBED Equation.3 . The proof is left to an exercise. Proposition 1.9: If u and v are perpendicular, then PuPv=PvPu=0. Proof: We have  EMBED Equation.3 , but uTv=uv=0. Similarly for the opposite product, exchanging the roles of u and v. Theorem 8 on page 390 of Lay's text can be reformulated as Proposition 1.10: If {u1, up} is an orthogonal basis for W, then orthogonal projection on W is given by the EMBED Equation.3  matrix  EMBED Equation.3 . Moreover,  EMBED Equation.3 . The proof does not require any modification. Proposition 1.11:  EMBED Equation.3 . The proof is left as an exercise. Let us use the notation we have developed to redescribe the Gram-Schmidt process as discussed in Section 6.4 of Lay's text. The computations in Theorem 11 on page 399 can be rewritten as v1=x1 v2= EMBED Equation.3 x2 v3= EMBED Equation.3 x3 etc. If some vi turns out to be 0, then the corresponding xi is a linear combination of the previous ones. Thus the Gram-Schmidt process can be used to determine the rank of a matrix. Since we are using projections systematically, I have written a MATLAB m-file called proj that computes projection matrices. proj(v) returns the Pv. This m-file, and others that I will provide, can be downloaded from the same source as this file. Problems: Prove Proposition 1.2 and Corollary 1.3 Determine which of the following matrices are orthogonal.  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  EMBED Equation.3  Deduce Proposition 1.6 from Theorem 3 on page 375 of Lay's text. Let u be the vector  EMBED Equation.3 . Use MATLAB to compute Pu. Verify that Pu is symmetric, and that  EMBED Equation.3 . Prove Proposition 1.8. Deduce Proposition 1.11 from Propositions 1.8-1.10. In each of Problems 6.3.9 and 6.3.10 in Lay's text: Let W be Span{u1,u2,u3}, and use Proposition 1.10 and MATLAB to obtain the matrix PW. Verify Propositions 1.8 and 1.11 for W. Use PW to solve the problem as stated in the text. Use MATLAB and the reformulation of the Gram-Schmidt process at the end of this section to solve Problems 6.4.11 and 6.4.12 in Lay's text. Use the Gram-Schmidt process to determine the rank of the matrix  EMBED Equation.3  EMBED Equation.3 . )*E $%bglm  ? K L e j v w   M N < = D E G O P )*.[\^_`aCDHIlmCJH*OJQJ CJOJQJH*0J0J\01IJ+, )*07>01IJ+, )*07>GHI   $%)<Odefgl~{xu  %&'%,23>?@IPW]^hi ->GHI   $%)<Odefglms> ? K L P W lms> ? K L P W ^ g h i j v w { L G yvs@A    jklu{|,W ^ g h i j v w { L  & F G H O P V i | 7 8 [ \ . & FG H O P V i | 7 8 [ \ ./&'~  ý~{xt/ KL   RSTUj}/./&'~  fg-/PR&'ikqs)*+,$&(*pv{|( ) < = > ? 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F R Y e l x  ?!A!k!{!!!!!!!!!""##7#G#V#f#u######$$#$3$B$R$k${$$$(%8%G%W%f%v%%%%&&&.&>&&&&&''!'1'J'Z'''''''(())#*(***+Dapartment of MathematicsD:\401\401-1.docDapartment of MathematicsD:\401\401-1.docDapartment of MathematicsD:\401\401-1.doc Judy GreenE:\401\401-1.doc Judy GreenE:\401\401-1.doc Judy GreenE:\401\401-1.doc Judy GreenE:\401\401-1.doc Judy GreenE:\401\401-1.doc Judy / =!"#$%Gram-Schmidt and the QR Factorization We refer to Section 6.4 of Lay's text for the detailed definitions, theorems and proofs regarding the Gram-Schmidt Process and the QR factorization. We will recall that if M is a matrix with linearly independent columns, the Gram-Schmidt process produces a matrix Q whose columns form an orthonormal basis for the column space of M and such that M=QR, where R is upper triangular with positive diagonal entries. MATLAB has a command (called QR) that directly computes QR factorizations. However, it will be instructive to ignore this command and produce a QR factorization using more fundamental MATLAB commands. Since we will be Now for a sample computation. We begin by entering a 3x4 matrix M whose columns are linearly independent. [1 2 3 4;6 5 7 8;9 11 10 12]' ans = 1 6 9 2 5 11 3 7 10 4 8 12 Q=M; We will now use the Gram-Schmidt process to modify Q until its columns form an orthonormal basis for the column space of M. The first step is to normalize the first column. Q(1:4,1)=Q(1:4,1)/norm(Q(1:4,1)) Q = 0.1826 6.0000 9.0000 0.3651 5.0000 11.0000 0.5477 7.0000 10.0000 0.7303 8.0000 12.0000 The notation Q(1:4,1) picks out the first column of Q as a vector. The effect of the command we gave is to divide the first column of Q by its norm, and leave the rest of Q unchanged. The next step is to modify the second column by subtracting off it projection on the first column. Q(1:4,2)=Q(1:4,2)-proj(Q(1:4,1))*Q(1:4,2) Q = 0.1826 3.7000 9.0000 0.3651 0.4000 11.0000 0.5477 0.1000 10.0000 0.7303 -1.2000 12.0000 Next we normalize the second column. Q(1:4,2)=Q(1:4,2)/norm(Q(1:4,2)) Q = 0.1826 0.9459 9.0000 0.3651 0.1023 11.0000 0.5477 0.0256 10.0000 0.7303 -0.3068 12.0000 Now we modify the third column by subtracting off its projections on the preceding columns. Q(1:4,3)=Q(1:4,3)-proj(Q(1:4,1))*Q(1:4,3)-proj(Q(1:4,2))*Q(1:4,3) Q = 0.1826 0.9459 -0.5098 0.3651 0.1023 3.0980 0.5477 0.0256 -1.0588 0.7303 -0.3068 -0.6275 And finally, we normalize again. Q(1:4,3)=Q(1:4,3)/norm(Q(1:4,3)) Q = 0.1826 \&]&&&U'o'''/(W(())))jlllooBqDqfqhq>r@rrrr & F & F8q&r&s&u&&&&&&&&&&&''#'$'<'='P'Q'R'S''''''',(.(\(](W)X)k)l)m)o))))))oo@r|rr0J5 jV)EHUjp\> PJUVmH j'EHUj[> PJUVmH j%EHUj> PJUVmHH* j#EHUj > PJUVmH jr"EHUj[> PJUVmH jU jbEHUj> PJUVmH1 0.9459 -0.1512 0.3651 0.1023 0.9187 0.5477 0.0256 -0.3140 0.7303 -0.3068 -0.1861 We can now check that the columns of Q are orthonormal. If they are, then the product QTQ should be the 3x3 identity matrix (why?)/ Q'*Q ans = 1.0000 0.0000 -0.0000 0.0000 1.0000 -0.0000 -0.0000 -0.0000 1.0000 Now, since we should have M=QR, then QTM=QTQR=R. R=Q'*M R = 5.4772 12.5976 19.9006 0.0000 3.9115 6.2124 -0.0000 -0.0000 3.3723 Find the QR decompositions as well.rrr(s6sttttt u^uuuxxpxrxzxxx4yvyyyz zzPzzGreenE:\401\401-1.doc Judy GreenE:\401\401-1.docW"eR)% b&eR05eR>97 7aeRh^`o(.h^`.hpLp^p`L.h@ @ ^@ `.zz{{{P|R|Z|||}V}}}}}}(~f~~~L̀΀CJOJQJkH6O16AutoInit5B* CJOJQJkH.OA.Calc<B*CJOJQJkH.@R. 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AskedForShutdown8InputCellCount 8MATLABFig16Color,8MATLABFigEmbed88MATLABFigHeightD8MATLABFigUnitsP8MATLABFigWidthd8 MATLABFmtl8MATLABFmtStyle|8MATLABStopOnError8ShowCellMarkers8ShowRecalcMarkers8no26YESYES3.5Inches4ShortLooseYES11@ll0lkC$Eƀb'FGPJmH 0J GPJmH F0J GPJmH 1F0J GPJmH 0J GPJmH 0J GPJmH |F0J GPJmH 0J GPJmH 0J GPJmH 0J GPJmH 0J GPJmH 0J GPJmH 0J GPJmH 0J GPJmH  \]^dopqrs!"#$*56789;STUV\ghijkln%0123467VWXY_jklmnp,:;<ir#$-2 0 1 2 3 4 s ?!_!`!a!b!c!!!!!!!""T"## #"#+#,#-#.#/#3################$]$|$}$~$$$%%%% %%%%%%%%%%%%%I&h&i&j&&&&&&&&&&&&;'<'['\''''''''''''''''(W+{+++@@AAAAA&Ah^`o(.h^`.hpLp^p`L.h@ @ ^@ `.h^`.hL^`L.h^`.h^`.hPLP^P`LAA,AA.A.@AjAlAnAzAAAAA@bAAAAAAAAŒAČAƌAȌAʌA֌AAAA@&A.@AAAAA*AAA0@A@BA.@AA A AA.A0A2A4AAAA6A.@AAAAA*A,A.A0AAvAtA.@AAAAA*A,A.A0A@@<@AlALl@hl@jlAllAlAlA2mApmAmAm@o@oAoAoAoAoApA:pAq@@qADqAI@@dqA@rAzrA|r@~r@rK@@rArA$s@&s@tK@A(sA.sA0sA2s@4s@vK@@6sAtAtAt@t@t@tAuAuAu@uAuA vA$vAHvAv@xAxATxAXxAjxAlx@nx@px@rx@zxA4yAry@ty@xK@Avy@y@yAyAyAyAzAz@z@z@ z@zAzA {@ {A{@:{A{AJ|AL|@N|@P|@R|A}AR}@T}@V}A}A}A}A}A}@}@}@}A~A~@~A~AAA@@AAA@@@@AAȀ@ʀAA"@,@0A2A@@B@D@FAʁA@ @ A@zK@ @|K@A@*R@@S@GTimes New Roman5Symbol3& Arial3 -3 fg?5 Courier New"V hqQQf[:$MV0d},C%b%%%%%%%% &*&I&j&&&&&&'<']'^''''''''''( (*|+++> _PID_GUIDAN{5A3D3A64-AB5C-11D0-B6F5-00A0249C4B47}  FMicrosoft Word Document MSWordDocWord.Document.89q FMicrosoft Equation 3.0 DS Equation Equation.39q_1042191004FQQ_1042191005FQQOle CompObjfObjInfoEquation Native ,_1042190969}FQQ_1042190970FQQ_1042190936iF`%Q`%Q_1042190956F`%Q`%QOle CompObj~f_1042190866Y|yF@Q@Q_1042190919uFQQ_1042190748kqF@Q@Q_1042190727mF@Q@QEquation Native kG_1041849283;^[FTTOle mCompObjZ\nf( K' K'z "q&r$-0Equation Native 6b_1041666862h3Fp~p~Ole 8CompObj249f15CG>W ."\&rz΀%')*,/24BDElG $)&(+.3Unknown Judy Green\pr"68Thj13Wkm9MOw!57;OQZnp}m(((((((((((($)8):)~)))+++++++::::::::::::::::::::::::::'INPUT_4OUTPUT_4INPUT_5OUTPUT_5INPUT_6OUTPUT_6INPUT_8OUTPUT_8INPUT_9OUTPUT_9INPUT_10 OUTPUT_10INPUT_11 OUTPUT_11INPUT_12 OUTPUT_12INPUT_13 OUTPUT_13INPUT_14 OUTPUT_14INPUT_15 OUTPUT_15INPUT_18 OUTPUT_18INPUT_19INPUT_20 OUTPUT_20INPUT_21 OUTPUT_21INPUT_22 OUTPUT_22INPUT_23 OUTPUT_23INPUT_24 OUTPUT_24INPUT_25 OUTPUT_25INPUT_26 OUTPUT_26* %gm?LjwGP 4 ?!c!#/###$ %%%&ObjInfoEquation Native 0Ole CompObjfObjInfoEquation Native ,&''+  !"#$%&(E#bkJeuN1 `!!,###|$%%%h&&['''+*-mpLOZa:=BEHKPS ! + X Z  FMicrosoft Equation 3.0 DS Equation Equation.39qVlII "" FMicrosoft Equation 3.0 DS Equation Equation.39q[ a F I Y b : = O R ` i ! * BH ,/ORhkps*,$.JLeg\g+/?A4 7 ##$$% %&&&&S)U)d)f)+΀02DFNʁ Ě ,.02 0246BD𴫙{ofTKIH*j <EHH*U#j> CJOJPJQJUVmHj]:EHH*Uj> PJUVmHj8EHH*UjG> PJUVmHj6EHH*U#jz> CJOJPJQJUVmHjM5EHH*Ujy> PJUVmHj3EHH*U#jl> CJOJPJQJUVmHj1EHH*UjX> PJUVmH jH*Uj0EHH*U=Proof: If , then (-B)v" w=0 for all v and w. In particular, setting w=(-B)v, it follows that (-B)v(-B)v=0. But the only vector whose dot product with itself is 0 is the zero vector, so we have shown that (-B)v=0 for all v. But this can only be true if A-B is the zero matrix, so that A=B.Proof: Let A be a matrix whose colums form a basis for V. Then V=Col(A), the dimension of V is the rank of A, and also of AT. We have, by the Rank theorem on page 259, rank AT + dim Nul AT=n, but by Theorem 3 on page 375, Nul AT=(Col A)= V. The result now follows. Proof: The symmetry follows from the identity (uuT)T= uTTuT= uuT. The rest follows from the fact that uTu=uu.This follows by direct computation, using Propositions 1.8 and 1.9.ĝƝĞƞȞʞОԞ֞؞02ŸH*},m6C:\Program Files\Microsoft Office\Templates\m-book.dotThe MATLAB Notebook v1.5.2!Template for The MATLAB Notebook Dapartment of MathematicsDapartment of Mathematics [4@4NormalCJOJQJkHmH 6@6 Heading 1CJOJQJkH>@> Heading 2x5CJOJQJkH@@@ Heading 3$<@& CJOJQJ<A@<Default Paragraph Font"O"NoGraphB*0O0Input5B* CJOJQJkH0O0OutputB*CJOJQJkH0O!0Error5B*CJOJQJkH6O16AutoInit5B* CJOJQJkH.OA.Calc<B*CJOJQJkH.@R. Footnote Text8&@a8Footnote ReferenceH*-l  ! " _`\]?*no2 3 !!""l"m"""""""""""#######$?$[%\%%%%%%% & &1&2&U&V&Z&y&&&&4'5'y'z'~'''''((A(B(F(e((((I)J)Q)R)X)w)))))))))*6*W*X*Z*d*e******++++,<,o,---' ' ' ' ' \\\\\??))]f7@ *-.5AHT[mpx?@LMjkwxPVLO)0HIPS[bnu Z \ J W , 1 i } <F&-s|9Akp47nuw|tw:Bsak+/  4 7 ? 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