(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 4.0, MathReader 4.0, or any compatible application. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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Actually, addition and multiplication \ by scalars is useful, too. They are simpler too, so we do them first.\ \>", \ "Text", ImageRegion->{{-0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Matrix Addition", "Subsection", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "In order to add two matrices, we insist that they be exactly the same \ size. Then we will do what we did in Calculus: coordinate by coordinate \ addition. Thus\n\nDEFINITION: If A = [a", StyleBox["i,j", FontVariations->{"CompatibilityType"->"Subscript"}], "] and B = [b", StyleBox["i,j", FontVariations->{"CompatibilityType"->"Subscript"}], "] are both m", "\[Cross]", "n, then A + B is the m", "\[Cross]", "n matrix\n given by\n\t\t A + B = [a", StyleBox["i,j", FontVariations->{"CompatibilityType"->"Subscript"}], " + b", StyleBox["i,j", FontVariations->{"CompatibilityType"->"Subscript"}], "].\t" }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a = {{1,1,1},{1,1,2},{1,1,0}}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["b = {{2,-3,1},{2,0,0},{3,2,-5}}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a + b", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "Note: the matrix of zero entries acts like an identity. It is called \ the ", StyleBox["zero", FontSlant->"Italic"], " m", "\[Cross]", "n matrix." }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["z = Table[0,{i,3},{j,3}]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a + z", "Input", ImageRegion->{{-0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Scalar Multiplication", "Subsection", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ StyleBox["Scalar multiplication", FontSlant->"Italic"], " means the multiplying of a matrix by a scalar quantity. Again, we \ imitate what was done in Calculus with vectors. Remember that ", StyleBox["scalar", FontSlant->"Italic"], " means a number. In our case it could mean a real or complex number, \ depending on the context.\n\nDEFINITION: If s is a scalar and A = [a", StyleBox["i,j", FontVariations->{"CompatibilityType"->"Subscript"}], "] an m", "\[Cross]", "n matrix, then s A is defined to be the\n m", "\[Cross]", "n matrix given by:\n\t\t s A = [s a", StyleBox["i,j", FontVariations->{"CompatibilityType"->"Subscript"}], "].\n\nREMARK: We denote (-1) A by - A, and as usual we write A - B = \ A + (- B)." }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["2 a", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["2*a", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["-1 a", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["0*a", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["3 a - 4 b", "Input", ImageRegion->{{-0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Linear Combinations", "Subsection", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "The last cell of the previous section illustrates another very useful idea \ in linear algebra:\n\nDEFINITION: A ", StyleBox["linear combination", FontSlant->"Italic"], " of the matrices (vectors) m", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ", m", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", ..., m", StyleBox["r", FontVariations->{"CompatibilityType"->"Subscript"}], " is any \n expression of the form\n\t \ a", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], " m", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], " + a", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], " m", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], " + ... + a", StyleBox["r", FontVariations->{"CompatibilityType"->"Subscript"}], " m", StyleBox["r", FontVariations->{"CompatibilityType"->"Subscript"}], ",\nwhere a", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ", a", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", ..., a", StyleBox["r", FontVariations->{"CompatibilityType"->"Subscript"}], " are scalars. If this linear combination sums to the zero matrix \ (vector), then it is called a ", StyleBox["trivial linear combination", FontSlant->"Italic"], ". Otherwise, it is a ", StyleBox["nontrivial linear combination", FontSlant->"Italic"], ".\n\nThus, we saw that 3 a - 4 b is a nontrivial linear combination of the \ matrices a and b.\nCheck the following linear combination for triviality \ and answer this question: what does this linear combination really say about \ c in terms of a and b? Test your conclusion by having ", StyleBox["Mathematica ", FontSlant->"Italic"], "do the calculation", StyleBox[".", FontSlant->"Italic"] }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["\<\ c={{1,-3,1/5},{1,-3/5,-6/5},{9/5,1,-4}} \ \>", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["3 a - 4 b + 5 c", "Input", ImageRegion->{{-0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Matrix Multiplication", "Subsection", ImageRegion->{{-0, 1}, {0, 1}}], Cell["\<\ This is the least obvious idea of today's material. Start by \ recalling how to \.aadot\.ba two vectors of the same size: line up the \ elements side by side, multiply them together and add the results:\ \>", \ "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["v = {1,2,3,4}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["w = {2,-1,1,0}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["v.w", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["w.v", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "Matrix multiplication is an extension of this idea: let A be an m", "\[Cross]", "p matrix and B a p", "\[Cross]", "n matrix. Then A B is the m", "\[Cross]", "n matrix whose ", StyleBox["(i,j)", FontSlant->"Italic"], "th entry is the result of the dot product of the ", StyleBox["i", FontSlant->"Italic"], "th row of A and", StyleBox[" j", FontSlant->"Italic"], "th column of B. Notice that these two vectors have the same length p." }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a.b", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["b.a", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "What could play the part of a \.aaone\.ba with respect to matrix \ multiplication, if anything? It turns out that there is such a matrix, \ which we call the", StyleBox[" identity matrix", FontSlant->"Italic"], ". Of course, just like zero matrices, the identity matrix can come in \ different sizes, though it is always a square matrix; that is, it has the \ same number of rows and columns:" }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["id = IdentityMatrix[3]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Let's see how it works:", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a.id", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["id.a", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["c = {{2,3,5,1},{1,2,-1,0},{1,2,1,3}}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["id . c", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["d = Transpose[c]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["d . id", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Now try this:", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["x = {x1,x2,x3}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a.x", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "Recall that two matrices are equal precisely when they are the same size \ and corresponding entries of each matrix are equal. With this out of the \ way, we arrive at a really basic idea about linear systems: the linear \ system (LS) with coefficient matrix A, right hand side vector b and column \ of unknowns x = {x", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ", x", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", ... , x", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], "}, is exactly equivalent to the matrix equation\n\t\t\t\t\tA . x = b\n\ This suggests yet another \"big idea.\" Why not try to solve (LS) by \ treating matrices like numbers? Then we could get the unique solution\n\t\t\t\ \t\tx = (1/A) . b\nBut what is it that plays the part of the \"inverse\" of \ a matrix? We know that the earlier matrix a doesn't give a system with a \ unique solution, so let's try this example (also from class):" }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a = {{1,1,1},{2,2,5},{4,6,8}}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["b = {4,11,24}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "Maybe ", StyleBox["Mathematica", FontSlant->"Italic"], " knows something about this ..." }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["ia = Inverse[a]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a . ia", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["ia . a", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Now for the real test:", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["x = ia . b", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["LinearSolve[a,b]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["\<\ It works!!! Clearly, we are going to have to explore this idea. \ Click on the earliest version of a and then click the equation defining ia. \ We have some things to think about, right?\ \>", "Text", ImageRegion->{{-0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Inverses", "Subsection", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "From class we saw how to build (right) inverses using the \ \.aasuperaugmented\.ba matrix. We saw that the m", "\[Cross]", "n matrix A had to have rank ", StyleBox["m", FontSlant->"Italic"], ", in order to have a right inverse. Here is the rest of the story for \ square matrices:\nTHEOREM: For a square n", "\[Cross]", "n matrix A, the following are equivalent (TFAE):\n (1) A.x = b has \ a unique solution ", StyleBox["x", FontSlant->"Italic"], " for some right hand side vector ", StyleBox["b", FontSlant->"Italic"], ".\n (2) A.x = b has a unique solution ", StyleBox["x", FontSlant->"Italic"], " for all right hand side vectors ", StyleBox["b", FontSlant->"Italic"], ".\n (3) The Row Echelon Form of A is I", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], ".\n (4) rank (A) = n.\n (5) A is a product of elementary \ matrices.\n (6) A is invertible.\n\nEXAMPLE. Use ", StyleBox["Mathematica", FontSlant->"Italic"], " to find a formula for the general inverse of a 2", "\[Cross]", "2 matrix by using the next cell and the Inverse command. Check your \ result." }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Clear[A,a,b,c,d]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["A={{a,b},{c,d}}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["B=Inverse[A]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Simplify[A.B]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Simplify[B.A]", "Input", ImageRegion->{{-0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Transposes", "Subsection", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "Now about transposes. We already saw this idea last time in lab:\n\n\ DEFINITION: The ", StyleBox["transpose", FontSlant->"Italic"], " of the m", "\[Cross]", "n matrix A is the n", "\[Cross]", "m matrix obtained by interchanging the rows and columns of A. It is \ denoted by A", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ".\n\nDEFINITION: A matrix is ", StyleBox["symmetric", FontSlant->"Italic"], " if it equals its transpose.\n\nExample: Verify that the following \ matrix is symmetric by using ", StyleBox["Mathematica", FontSlant->"Italic"], "." }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a={{4,2,1},{2,4,2},{1,2,3}}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a - Transpose[a]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "For complex matrices, there is another idea that is important :\n\n\ DEFINITION: The ", StyleBox["Hermitian transpose", FontSlant->"Italic"], " of a matrix A is obtained by conjugating each entry of the transpose of \ A", StyleBox[", ", FontSlant->"Italic"], "and is denoted by A* (or sometimes as A", StyleBox["H", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ").\n\nDEFINITION: A matrix is ", StyleBox["Hermitian", FontSlant->"Italic"], " (or ", StyleBox["Hermitian symmetric) ", FontSlant->"Italic"], "if it equals its Hermitian transpose", StyleBox[".", FontSlant->"Italic"] }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "EXAMPLE: Verify that the following matrix is Hermitian by using ", StyleBox["Mathematica", FontSlant->"Italic"], ":\n" }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a ={{1,I,1+I},{-I,-5,2-I},{1-I,2+I,3}}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a - Conjugate[Transpose[a]]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "There are some important properties of transposes and Hermitian transposes \ that we need to know. The main properties are summarized in the following \ theorem. Most of the parts are rather easy to verify directly from \ definitions. The statement about rank is not so obvious.\n\nKEY FACT: Let \ A and B be matrices and ", StyleBox["c", FontSlant->"Italic"], " a (complex) scalar. Then the following hold (provided A and B have \ the suitable sizes for addition or multiplication, as appropriate.)\n \t(1) \ (A", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ") ", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " = A and (A*) * = A.\n \t(2) (A + B)", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " = A", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " + B", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " and (A + B)* = A* + B*.\n \t(3) (c A)", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " = c (A", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ") and (c A) * = Conj(c) A*.\n \t(4) (A B)", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " = B", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " A", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " and (A B)* = B* A*.\n \t(5) (A", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], ")", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " = (A", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ")", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], " and (A", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], ")* = (A*)", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], ".\n \t(6) rank (A) = rank (A", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ") and rank (A) = rank (A*)." }], "Text", ImageRegion->{{-0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Exercises", "Subsection", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "Evaluate the following expressions using ", StyleBox["Mathematica", FontSlant->"Italic"], ", letting\n ", "\:f2e6", "1 5 7", "\:f2f6", " ", "\:f2e6", " 1 2 3", "\:f2f6", "\n A = ", "\:f2e7", "0 2 3", "\:f2f7", " B = ", "\:f2e7", "-1 0 6", "\:f2f7", "\n ", "\:f2e8", "9 0 1", "\:f2f8", " ", "\:f2e8", "1 0 0", "\:f2f8", "\n \n A + B\n A B\n B A (Is matrix \ multiplication commutative?)\n 5 A - 7 B\n A (B + 3 I", StyleBox["3", FontVariations->{"CompatibilityType"->"Subscript"}], ") - (A B + 3 A) \n A", StyleBox["3", FontVariations->{"CompatibilityType"->"Superscript"}], " - 4 A", StyleBox["2", FontVariations->{"CompatibilityType"->"Superscript"}], " - 58 A - 11 I", StyleBox["3", FontVariations->{"CompatibilityType"->"Subscript"}], " \n \n A", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], " B", StyleBox["-1\n ", FontVariations->{"CompatibilityType"->"Superscript"}], "B", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], " A", StyleBox["-1\n (A B)-1 ", FontVariations->{"CompatibilityType"->"Superscript"}], "(Which of the above two expressions is this equal to?)\n A", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " \n (A +", StyleBox[" i", FontSlant->"Italic"], " B) * (Use properties to write this another way, only using A", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], " and B", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], ". Check your\n answer using ", StyleBox["Mathematica", FontSlant->"Italic"], ". Hint: For real matrices R, R", StyleBox["T", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], "= R *)", StyleBox[" \n ", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}] }], "Text", ImageRegion->{{-0, 1}, {0, 1}}] }, Closed]] }, Closed]] }, FrontEndVersion->"4.0 for X", ScreenRectangle->{{0, 1152}, {0, 864}}, WindowToolbars->{}, CellGrouping->Manual, WindowSize->{520, 600}, WindowMargins->{{Automatic, 308}, {104, Automatic}}, PrivateNotebookOptions->{"ColorPalette"->{RGBColor, -1}}, ShowCellLabel->True, ShowCellTags->False, RenderingOptions->{"ObjectDithering"->True, "RasterDithering"->False}, Magnification->1.25 ] (*********************************************************************** Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. 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