(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.0' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. 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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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 40853, 1487]*) (*NotebookOutlinePosition[ 41623, 1513]*) (* CellTagsIndexPosition[ 41579, 1509]*) (*WindowFrame->Normal*) Notebook[{ Cell["Lab 4: Linear Algebra Fun", "Title", Background->RGBColor[0, 0, 1]], Cell[TextData[{ "Math 334\tSpring 2004\t", StyleBox["Chapter 3", FontFamily->"Arial", FontSize->16, FontWeight->"Bold"], " \t\t\t" }], "Subtitle", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[TextData[StyleBox["Instructions!", FontColor->RGBColor[1, 0, 0]]], "Section"], Cell[TextData[{ "Work in ", StyleBox["groups of 2 or 3", FontWeight->"Bold"], " today, going through the entire notebook. If you have questions, flag me \ down. \n\nWrite up your solutions to the lab questions and be ready to hand \ them in Next Thursday (", StyleBox["26 February", FontWeight->"Bold"], ") in class. Remember, this is your opportunity to find out how to use ", StyleBox["Mathematica", FontSlant->"Italic"], " for some matrix basics. " }], "Subsubtitle"] }, Closed]], Cell[CellGroupData[{ Cell["Matrices and Determinants", "Section", PageWidth->WindowWidth], Cell[CellGroupData[{ Cell["Some basics...", "Subsubsection"], Cell[TextData[{ "Fortunately for us, ", StyleBox["Mathematica", FontSlant->"Italic"], " will do almost everything, except perhaps cook dinner and clean the \ house... \nA matrix is entered in ", StyleBox["Mathematica", FontSlant->"Italic"], " as a list - surrounded by curly brackets {}. Each ", StyleBox["row", FontWeight->"Bold"], " of the matrix is also delimited by curly brackets. ", "Since matrices are lists, they are usually output in an ugly format, as \ seen below: " }], "Text"], Cell[BoxData[ \(\(\(Clear[A]\[IndentingNewLine] A\ = {{1, \ \(-1\), \ 1}, {2, \ \(-3\), \ 4}, \ {\(-2\), \ \(-1\), \ 1}}\)\(\ \)\)\)], "Input"], Cell[TextData[{ "You can always display the matrix using the ", StyleBox["MatrixForm ", FontWeight->"Bold"], "command: " }], "Text"], Cell[BoxData[ \(MatrixForm[A]\)], "Input"], Cell[TextData[{ "But that is a pain to type everytime you want to print a matrix... so here \ is the ", StyleBox["cool ", FontWeight->"Bold"], StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" trick of the day\[RegisteredTrademark]", FontWeight->"Bold"], ". Executing ", StyleBox["this one", FontWeight->"Bold"], " cell will make sure that all matrices are displayed in matrix format ", StyleBox["for the rest of the ", FontWeight->"Bold"], StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" session!", FontWeight->"Bold"], "..." }], "Text"], Cell[BoxData[ \($PrePrint = If[MatrixQ[#], MatrixForm[#], #] &\)], "Input"], Cell["Try it out here...", "Text"], Cell[BoxData[ \(A\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Solving Systems of Equations", "Subsubsection"], Cell[TextData[{ "Let' s look at how we can solve ", StyleBox["Ax=b", FontWeight->"Bold"], ". ", "Just to make sure we didn't forget what it was, we'll define our \ coefficient matrix ", StyleBox["A", FontWeight->"Bold"], " ", "again..." }], "Text"], Cell[BoxData[ \(\(\(Clear[A]\[IndentingNewLine] A\ = {{1, \ \(-1\), \ 1}, {2, \ \(-3\), \ 4}, \ {\(-2\), \ \(-1\), \ 1}}\)\(\ \)\)\)], "Input"], Cell[TextData[{ "Now, let's define ", StyleBox["b", FontWeight->"Bold"], ". We can enter this as a", " vector (equivalent to one row or one column of a matrix). ", StyleBox["Mathematica", FontSlant->"Italic"], " will interpret this as either a row or a column vector, depending on how \ it is used." }], "Text"], Cell[BoxData[{ \(Clear[b]\), "\[IndentingNewLine]", \(b = {0, \ \(-2\), \ 7}\)}], "Input"], Cell[TextData[{ "There are several ways to solve this in ", StyleBox["Mathematica", FontSlant->"Italic"], ". \n", StyleBox["Gaussian elimination ", FontWeight->"Bold"], "is performed if you use the ", StyleBox["LinearSolve ", FontWeight->"Bold"], "command." }], "Text", PageWidth->PaperWidth], Cell[BoxData[ \(LinearSolve[A, b]\)], "Input"], Cell[TextData[{ "The ", StyleBox["Solve", FontWeight->"Bold"], " command also works, it may use different techniques than the ", StyleBox["LinearSolve", FontWeight->"Bold"], " command." }], "Text"], Cell[BoxData[{ \(Clear[x, y, z, vars]\), "\[IndentingNewLine]", \(vars = {x, y, z}\), "\[IndentingNewLine]", \(Solve[A . vars \[Equal] b, vars]\)}], "Input"], Cell[TextData[{ "Yet another alternative is to enter the equations without matrix notation. \ Note that the equations are enclosed in curly braces, since we have a ", StyleBox["set", FontWeight->"Bold"], " of equations that we want solved." }], "Text"], Cell[BoxData[ \(Solve[{x - y + z \[Equal] 0, 2 x - 3 y + 4 z \[Equal] \(-2\), \(-2\) x - y + z \[Equal] 7}, {x, y, z}]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["A Singular System", "Subsubsection"], Cell["Let's see what happens with this system.", "Text"], Cell[BoxData[{ \(Clear[A, b]\), "\[IndentingNewLine]", \(\(A = {{1, 1, \(-1\), 2}, {1, 1, 0, 1}, {1, 2, \(-4\), 0}, {2, 1, 2, 5}};\)\), "\[IndentingNewLine]", \(\(b = {1, 2, 1, 1};\)\), "\[IndentingNewLine]", \(LinearSolve[A, b]\)}], "Input"], Cell["\<\ The error message indicates that this system has no solution, so your input \ command is sent back to you.\ \>", "Text", FontColor->RGBColor[0, 0, 1]], Cell[TextData[{ "Let's use Gaussian elimination on this matrix and see what happens. This \ is done using the command ", StyleBox["RowReduce", FontWeight->"Bold"], "." }], "Text"], Cell[BoxData[ \(RowReduce[A]\)], "Input"], Cell["\<\ Let's also check out the determinant of the coefficient matrix.\ \>", "Text"], Cell[BoxData[ \(Det[A]\)], "Input"], Cell["\<\ While we are at it, why not look at the eigenvalues for the coefficient \ matrix?\ \>", "Text"], Cell[BoxData[ \(Eigenvalues[A] // N\)], "Input"], Cell["\<\ Notice that there is an eigenvalue of 0, and that the determinant was also \ 0... \ \>", "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Another Example: 4 equations, 3 unknowns", "Subsubsection"], Cell[TextData[{ "Note the distinction between ", StyleBox["LinearSolve", FontWeight->"Bold"], " and ", StyleBox["Solve", FontWeight->"Bold"], " in this example." }], "Text"], Cell[BoxData[{ \(Clear[A, b]\), "\[IndentingNewLine]", \(\(A = {{2, 3, \(-1\)}, {\(-1\), \(-1\), 3}, {1, 2, 2}, {0, 1, 5}};\)\), "\[IndentingNewLine]", \(\(b = {3, 0, 3, 3};\)\), "\[IndentingNewLine]", \(LinearSolve[A, b]\)}], "Input"], Cell[TextData[{ "The ", StyleBox["LinearSolve", FontWeight->"Bold"], " gave us only one set of solutions, but look at what ", StyleBox["Solve", FontWeight->"Bold"], " yields." }], "Text"], Cell[BoxData[ \(Solve[A . {x, y, z} \[Equal] b, {x, y, z}]\)], "Input"], Cell["\<\ This means that z can can be assigned any value, thus yielding an entire set \ of solutions that fall along a line in 3-space.\ \>", "Text"], Cell["Let's try RowReduce on this one.", "Text"], Cell[BoxData[ \(RowReduce[A] // MatrixForm\)], "Input"], Cell[TextData[StyleBox["Question 1:\tBoth this example and the one before \ yielded a zero row after using Gaussian elimination. Explain why the previous \ one yielded no solutions and this yielded an infinite number of solutions.", FontWeight->"Bold"]], "Text", Background->RGBColor[0, 1, 1]], Cell["\<\ We might as well check out the determinant and the eigenvalues for the \ coefficient matrix.\ \>", "Text"], Cell[BoxData[{ \(Det[A]\), "\[IndentingNewLine]", \(Eigenvalues[A] // N\)}], "Input"], Cell[TextData[{ "Hmmmm.... well... ", StyleBox["that", FontWeight->"Bold"], " was interesting... ", "Take note of the pattern you are observing." }], "Text"], Cell[TextData[StyleBox["Question 2:\tSo what was the problem with finding the \ determinant or the eigenvalues of A?", FontWeight->"Bold"]], "Text", Background->RGBColor[0, 1, 1]] }, Closed]], Cell[CellGroupData[{ Cell["Another Example: 3 equations, 5 unknowns", "Subsubsection"], Cell[BoxData[{ \(Clear[A, b]\), "\[IndentingNewLine]", \(\(A = {{4, \(-8\), \(-1\), 1, 3}, {5, \(-10\), \(-1\), 2, 3}, {3, \(-6\), \(-1\), 1, 2}};\)\), "\[IndentingNewLine]", \(\(b = {0, 0, 0};\)\), "\[IndentingNewLine]", \(vars = {x\_1, x\_2, x\_3, x\_4, x\_5}\), "\[IndentingNewLine]", \(LinearSolve[A, b]\), "\[IndentingNewLine]", \(Solve[A . vars \[Equal] b, vars]\)}], "Input"], Cell[TextData[{ "Note that ", StyleBox["LinearSolve", FontWeight->"Bold"], " gave only the trivial solution, while the ", StyleBox["Solve", FontWeight->"Bold"], " command yields the relations to give us an infinite number of solutions \ (since ", Cell[BoxData[ \(TraditionalForm\`x\_2\)]], "and ", Cell[BoxData[ \(TraditionalForm\`x\_5\)]], "can take on any values)." }], "Text"], Cell[TextData[StyleBox["Question 3:\tDoes it make sense to look at the \ determinant and eigenvalues for this system? Why or why not?", FontWeight->"Bold"]], "Text", Background->RGBColor[0, 1, 1]] }, Closed]], Cell[CellGroupData[{ Cell["Inverted Thinking!", "Subsubsection"], Cell[TextData[{ "To find the inverse of a matrix, use the ", StyleBox["Inverse", FontWeight->"Bold"], " command in ", StyleBox["Mathematica", FontSlant->"Italic"], "." }], "Text"], Cell[BoxData[{ \(Clear[A]\), "\[IndentingNewLine]", \(\(A = {{18, \(-24\), 25, 27}, {\(-12\) - 2 \[Pi], 14 - \[Pi], \(-17\), \(-27\)}, {15 + 3 \@ 3, \(-18\) - 4 \@ 3, 21 + 4 \@ 3, 27}, {\(-10\), 12, \(-14\), \(-18\)}};\)\), "\[IndentingNewLine]", \(Inverse[A]\), "\[IndentingNewLine]", \(Inverse[A] // N\)}], "Input"], Cell[TextData[{ "The first output is in exact form and the second in approximate form. If \ any decimals are in the original matrix, ", StyleBox["Mathematica", FontSlant->"Italic"], " will work in approximate mode (otherwise, it will work in exact mode).\n\ What if a matrix does not possess an inverse?" }], "Text"], Cell[BoxData[{ \(\(B = {{1, 1, 0, 0, 0}, {2, 2, 2, 0, 0}, {0, 3, 3, 3, 0}, {0, 0, 4, 4, 4}, {0, 0, 0, 5, 5}};\)\), "\[IndentingNewLine]", \(MatrixForm[B]\), "\[IndentingNewLine]", \(Inverse[B]\)}], "Input"], Cell["\<\ Question 3:\tIs there any relation between no inverse and zero determinant \ and a zero eigenvalue?\ \>", "Text", FontWeight->"Bold", Background->RGBColor[0, 1, 1]] }, Closed]], Cell[CellGroupData[{ Cell["Tranposing our Thoughts...", "Subsubsection"], Cell[TextData[{ "The transpose of a matrix is the one with rows taking the place of columns \ and vice-versa. ", StyleBox["Mathematica", FontSlant->"Italic"], " uses the ", StyleBox["Transpose", FontWeight->"Bold"], " command to find it. " }], "Text"], Cell[BoxData[{ \(Clear[A]\), "\[IndentingNewLine]", \(A = {{2, 1, \(-3\), 4, 1}, {\(-2\), 1, 1, 0, 7}, {6, 2, \(-8\), 4, 9}, {\(-2\), 0, 2, \(-6\), \(-2\)}, {10, 3, \(-13\), 14, 12}}\), "\[IndentingNewLine]", \(Transpose[A]\), "\[IndentingNewLine]", \(A . Transpose[A]\), "\[IndentingNewLine]", \(Transpose[A] . A\)}], "Input"], Cell[BoxData[{ \(RowReduce[A]\), "\[IndentingNewLine]", \(RowReduce[Transpose[A]]\)}], "Input"], Cell["\<\ Note that if there is a zero row in A, there will be a zero row in its \ transpose. Compare their eigenvalues.\ \>", "Text"], Cell[BoxData[{ \(Eigenvalues[A]\), "\[IndentingNewLine]", \(Eigenvalues[Transpose[A]]\)}], "Input"], Cell["How about their determinants?", "Text"], Cell[BoxData[{ \(Det[A]\), "\[IndentingNewLine]", \(Det[Transpose[A]]\)}], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["All Together Now...", "Subsubsection"], Cell["Let's look at another matrix whose determinant is non-zero.", "Text"], Cell[BoxData[{ \(Clear[A]\), "\[IndentingNewLine]", \(A = {{4, \(-2\), 16, 27, \(-11\)}, {9, 43, 9, \(-8\), \(-1\)}, {34, 20, \(-3\), 0, 21}, {\(-5\), 4, 4, 7, 41}, {0, 12, \(-2\), \(-2\), 3}}\), "\[IndentingNewLine]", \(Ainv = Inverse[A]\), "\[IndentingNewLine]", \(Atran = Transpose[A]\)}], "Input"], Cell["\<\ Question 4:\tHow do the determinants of A, A inverse, and A transpose \ compare?\ \>", "Text", FontWeight->"Bold", Background->RGBColor[0, 1, 1]], Cell[TextData[StyleBox["Question 5:\tHow do the eigenvalues of A, A inverse, \ and A transpose compare?", FontWeight->"Bold"]], "Text", Background->RGBColor[0, 1, 1]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Vector Spaces", "Section"], Cell[CellGroupData[{ Cell["Span", "Subsubsection"], Cell[TextData[{ "One way to determine if a new vector is in the ", StyleBox["span", FontWeight->"Bold"], " of a set of vectors, ", StyleBox["V", FontWeight->"Bold"], ",", " is to see if you can write the new vector as a linear combination of the \ vectors in ", StyleBox["V", FontWeight->"Bold"], "." }], "Text"], Cell[BoxData[{ \(\(v1 = {4, \(-2\), 1, 2, 3, 1};\)\), "\[IndentingNewLine]", \(\(v2 = {2, \(-2\), 1, \(-5\), 7, 1};\)\), "\[IndentingNewLine]", \(\(v3 = {\(-17\), 22, \(-1\), 0, 2, 1};\)\), "\[IndentingNewLine]", \(\(v4 = {\(-31\), \(-3\), \(-8\), 1, 1, 0};\)\), "\[IndentingNewLine]", \(\(v5 = {17, 44, 1, \(-22\), 11, 1.9};\)\), "\[IndentingNewLine]", \(\(v6 = {1, 0, 1, 15, \(-3\), 3};\)\), "\[IndentingNewLine]", \(\(newv = {1, 2, \(-2\), 0, \(-1\), 4};\)\), "\[IndentingNewLine]", \(Solve[ newv \[Equal] c1\ v1 + c2\ v2 + c3\ v3 + c4\ v4 + c5\ v5 + c6\ v6, {c1, c2, c3, c4, c5, c6}]\)}], "Input"], Cell["\<\ If no constants can be found, the new vector is not in the space determined \ by the other six. let's see what happens if we try to solve for the \ constants when newv is the zero vector.\ \>", "Text"], Cell[BoxData[{ \(\(newv = {0, 0, 0, 0, 0, 0};\)\), "\[IndentingNewLine]", \(Solve[ newv \[Equal] c1\ v1 + c2\ v2 + c3\ v3 + c4\ v4 + c5\ v5 + c6\ v6, {c1, c2, c3, c4, c5, c6}]\)}], "Input"], Cell[TextData[StyleBox["Question 6:\tWhat does all this say about whether or \ not the vectors are linearly independent?", FontWeight->"Bold"]], "Text", Background->RGBColor[0, 1, 1]], Cell["\<\ While we are at it, let's look at the matrix formed by these six vectors and \ examine it from different perspectives.\ \>", "Text"], Cell[BoxData[{ \(M = {v1, v2, v3, v4, v5, v6}\), "\[IndentingNewLine]", \(MT = Transpose[M]\)}], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Wronskian", "Subsubsection"], Cell[TextData[{ "Although there is not a Wronskian command in ", StyleBox["Mathematica", FontSlant->"Italic"], ", it is easy to define using the command for derivatives, ", StyleBox["D[f, x]", FontWeight->"Bold"], " which represents taking the derivative of the function f with respect to \ x. If you want a second derivative, use ", StyleBox["D[f, {x,2}]", FontWeight->"Bold"], ", etc.\nRemember, we say that a set of functions is linearly independent \ if and only if the Wronskian is non-zero for ", StyleBox["some", FontWeight->"Bold"], " value of ", StyleBox["x", FontSlant->"Italic"], "." }], "Text"], Cell[BoxData[{ \(functions = {\[ExponentialE]\^\(3 x\), x\ \[ExponentialE]\^\(3 x\), \[ExponentialE]\^\(3 x\)\ Cos[ 2 x], \[ExponentialE]\^\(3 x\)\ Sin[2 x], x\ Cos[x], x\ Sin[x]}\), "\[IndentingNewLine]", \(wronmatrix = {functions, D[functions, x], D[functions, {x, 2}], D[functions, {x, 3}], D[functions, {x, 4}], D[functions, {x, 5}]}\), "\[IndentingNewLine]", \(wronskian = Det[wronmatrix] // Simplify\), "\[IndentingNewLine]", \(wronskian /. x \[Rule] 0\), "\[IndentingNewLine]", \(Solve[wronskian \[Equal] 0, x]\ \)}], "Input"], Cell[TextData[{ "The Simplify command is very helpful here, as we are able to get a \ reasonable function for the Wronskian. Note that the next to last line of \ code asks for the value of the Wronskian when x is 0, and the last line of \ code solves to see if the Wronskian is ", StyleBox["ever", FontWeight->"Bold"], " equal to 0." }], "Text"], Cell[TextData[StyleBox["Question 7:\tSo are these functions all linearly \ independent?", FontWeight->"Bold"]], "Text", Background->RGBColor[0, 1, 1]], Cell[TextData[{ StyleBox["Question 8:\tAre the functions ", FontWeight->"Bold"], Cell[BoxData[ \(\(\(x\[ExponentialE]\^x\)\(\ \)\(and\)\(\ \)\(x\^2\) \ \(\[ExponentialE]\^\(3 x\)\)\(\ \)\)\)], FontWeight->"Bold"], StyleBox["linearly independent", FontWeight->"Bold"], StyleBox["?", FontWeight->"Bold"] }], "Text", Background->RGBColor[0, 1, 1]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Finding the Basis of a Subspace", "Section", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Arial"], Cell[TextData[{ "If ", Cell[BoxData[ \(TraditionalForm\`S\)]], " is a subspace of ", Cell[BoxData[ \(TraditionalForm\`\[DoubleStruckCapitalR]\^n\)]], " that is spanned by the vectors ", Cell[BoxData[ \(TraditionalForm\`x\_1, x\_2, \[CenterEllipsis]\ x\_n\)]], ", then we can find the ", StyleBox["basis ", FontWeight->"Bold"], "of this subspace by:\n\t1. F", "orm the matrix ", Cell[BoxData[ FormBox[ RowBox[{"A", "=", RowBox[{"(", GridBox[{ {\(x\_1\), \(x\_2\), \(x\_3\), "\[CenterEllipsis]", \(x\_n\)} }], ")"}]}], TraditionalForm]]], ", this is the matrix whose columns are the vectors in the spanning set. \n\ \t2. Find the RREF of ", Cell[BoxData[ \(TraditionalForm\`A\^T\)]], ", call it ", Cell[BoxData[ \(TraditionalForm\`M\)]], ". The nonzero columns of ", Cell[BoxData[ \(TraditionalForm\`M\^T\)]], " form a basis for ", Cell[BoxData[ \(TraditionalForm\`S\)]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True, FontFamily->"Arial"], Cell[TextData[{ "Why does this work? Finding the RREF uses ", StyleBox["row operations", FontWeight->"Bold"], ", so we look at ", Cell[BoxData[ \(TraditionalForm\`A\^T\)]], ", where the original columns become rows. The row operations eliminate \ unnecessary rows (rows that are linear combinations of other rows, therefore \ not needed in the basis), and leave the remaining rows in simplified form. \ Remember, though, these rows were the ", StyleBox["columns ", FontWeight->"Bold"], "of our original matrix A. 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