(*********************************************************************** 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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Let's get right to the point. Let A be a square n ", "\[Cross]", " n matrix (these are the only kind of matrices for which eigenvalues and \ eigenvectors will be defined). An ", StyleBox["eigenvector", FontSlant->"Italic"], " of A is a NONZERO vector ", StyleBox["x", FontSlant->"Italic"], " in R", StyleBox["n", FontVariations->{"CompatibilityType"->"Superscript"}], " (or C", StyleBox["n", FontVariations->{"CompatibilityType"->"Superscript"}], ", if we are working over complex nos), such that for some scalar number \ ", "\[Lambda]", ", we have\n\n\t\t\t\t\tA x = ", "\[Lambda]", " x\n\nThe number ", "\[Lambda]", " is called an ", StyleBox["eigenvalue", FontSlant->"Italic"], " of the matrix A, and the vector ", StyleBox["x", FontSlant->"Italic"], " is called an ", StyleBox["eigenvector belonging to", FontSlant->"Italic"], " (or ", StyleBox["corresponding to", FontSlant->"Italic"], ") the eigenvalue ", "\[Lambda]", ". Observe that the eigenvalue ", "\[Lambda]", " is allowed to be the 0 scalar, but eigenvectors ", StyleBox["x", FontSlant->"Italic"], " are, by definition, never the 0 vector. (Question: what does it \ tell you about a matrix that 0 is an eigenvalue of it?) \n\nMathematica \ knows all about eigenvalues and eigenvectors. Click the following cells." }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a = {{4,0,1},{-2,1,0},{-2,0,1}}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["MatrixForm[a]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Eigenvalues[a]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Eigenvectors[a]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Curious, isn't it? Let's try another example. ", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["b = {{3,2},{-1,0}}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Eigenvalues[b]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Eigenvectors[b]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Let's check some of these proposed values:", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a . {0,1,0}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a. {-1,2,2}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a.{-1,1,1}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["b.{-1,1}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["\<\ What do you think is the number of eigenvalues of a given matrix A? \ O. K.,... try this\ \>", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["c = {{-2,-1},{5,2}}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Eigenvalues[c]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Eigenvectors[c]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["What do you say about your answer to the previous question?", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "So what is the general story? You'll find out more technical details in \ class lecture, but here are some of the basic points about eigenvalues and \ eigenvectors:\n\nA has eigenvalue ", "\[Lambda]", " iff A x = ", "\[Lambda]", " x, for some nonzero vector x\n\t\t\t\t \n\t\t\tiff A x - ", "\[Lambda]", " x = 0, x nonzero\n\t\t\t\t \n\t\t\tiff (A - ", "\[Lambda]", " I) x = 0, x nonzero\n\t\t\t\t \n\t\t\tiff ", StyleBox["x", FontSlant->"Italic"], " is a nonzero element of the Null Space of (A - ", "\[Lambda]", " I)\n\t\t\t\t \n\t\t\tiff rank (A - ", "\[Lambda]", " I) < n = size of A\n\t\t\t\t \n\t\t\tiff det (A - ", "\[Lambda]", " I) = 0.\n\nBut observe that the det above is just a polynomial of degree \ ", StyleBox[" n", FontSlant->"Italic"], " in the variable ", "\[Lambda]", ". The last equation above is called the ", StyleBox["characteristic equation", FontSlant->"Italic"], " of A. It will have exactly", StyleBox[" n", FontSlant->"Italic"], " roots, if we allow for repetitions and complex numbers.\n\nOnce we have \ found all eigenvalues of A, we can always find the appropriate Null Space \ and hence all eigenvectors belonging to each eigenvalue. By the way, this \ Null Space has a name: it is called the ", StyleBox["eigenspace corresponding to the eigenvalue ", FontSlant->"Italic"], "\[Lambda]", StyleBox[". ", FontSlant->"Italic"], "As an example, click the following cell", StyleBox[":\t\t\t\t ", FontSlant->"Italic"] }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Det[a - lambda*IdentityMatrix[3]]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Factor[Det[a-lambda*IdentityMatrix[3]]]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Det[c - lambda*IdentityMatrix[2]]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "Now fill in the next cell with one of the eigenvalues of the matrix ", StyleBox["a", FontSlant->"Italic"], " before doing the subsequent cell:" }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["eval = 1", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["NullSpace[a-eval*IdentityMatrix[3]]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "So now you see that the Eigenvector command is really calculating null \ spaces. Once you have calculated the null spaces you have a complete \ listing, in principle, of all the eigenvectors and values of the matrix. \ Actually, you specify a basis for the null space of the matrix ", StyleBox["a", FontSlant->"Italic"], " - lambda*IdentityMatrix[n]. This space, the eigenspace of the matrix ", StyleBox["a", FontSlant->"Italic"], " corresponding to the eigenvalue ", StyleBox["lambda", FontSlant->"Italic"], ", has one vector in it which is not an eigenvector. Which one is it? A \ complete listing of all eigenvalues and eigenspaces of the matrix A is \ called the ", StyleBox["eigensystem", FontSlant->"Italic"], " of the matrix A. In principle, you should be able to calculate \ eigensystems by hand. Of course, in practice this may be difficult. ", StyleBox["Mathematica", FontSlant->"Italic"], " knows about these systems and can help us out. Click the follow cells:" }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["? Eigensystem", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Eigensystem[a]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["", "Input", ImageRegion->{{-0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Why are they???", "Subsection", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "This is a good question and the answer could go on and on. We will try \ to indicate their importance for one special class of problems, namely, \ discrete dynamical systems.\n\t\nHere is how such a system is described: one \ starts with an initial state vector x", StyleBox["0", FontVariations->{"CompatibilityType"->"Subscript"}], ". Now to pass from one state vector to the next, one uses the rule\n\t\n\t\ \t\t\t\tx", StyleBox["n+1", FontVariations->{"CompatibilityType"->"Subscript"}], " = A x", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], ",\n\t\t\t\t\t\nwhere x", StyleBox["k ", FontVariations->{"CompatibilityType"->"Subscript"}], " means the kth state vector. The matrix A is fixed for the dynamical \ system. It is called the ", StyleBox["transition matrix", FontSlant->"Italic"], " of the system. You have already seen one such example, namely, Markov \ chains. There are others, from fields like biology and economics. Let's \ try to understand how state vectors change in the case of the matrix A from \ the previous section." }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["x0 = {1,1,1}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a.x0", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a.a.x0", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a.a.a.x0", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a.a.a.a.x0", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a.a.a.a.a.x0", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "What do you predict will be the next state, approximately?\n\nWhat do you \ get if you divide the above vector by its largest component? \n\nNow compare \ these to the eigenvectors and eigenvalues of A.\n\nNow do the whole sequence \ above with another random vector x", StyleBox["0", FontVariations->{"CompatibilityType"->"Subscript"}], ".\n\nNext try the same thing with the matrix ", StyleBox["b", FontSlant->"Italic"], ". Repeat the above type of sequence." }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["x0 = {1,1}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "What's going on here? Here is a start on an explanation:\n\nNotice that \ x", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], " = A x", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], " = A A x", StyleBox["0", FontVariations->{"CompatibilityType"->"Subscript"}], ",\n\n\t x", StyleBox["3", FontVariations->{"CompatibilityType"->"Subscript"}], " = A x", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], " = A A A x", StyleBox["0", FontVariations->{"CompatibilityType"->"Subscript"}], ", etc.\n\nSee the pattern? \n\t\t\t\tx", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], " = A", StyleBox["n", FontVariations->{"CompatibilityType"->"Superscript"}], " x", StyleBox["0", FontVariations->{"CompatibilityType"->"Subscript"}], ".\n\t\t\t\t\nSo, to understand how a dynamical system works, what we \ really need to know is how the powers of the transition matrix A behave. \ But in general, this is very hard!!!\n\nHere is a case we can handle: what \ if A is diagonal?" }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["d = {{1, 0, 0}, {0, 2, 0}, {0, 0, 3}}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["MatrixForm[d]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["MatrixForm[d.d]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["MatrixForm[d.d.d]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "See what's going on? It's easy to show this\n\nFACT: For a (square) \ diagonal matrix D and positive integer n the matrix D", StyleBox["n", FontVariations->{"CompatibilityType"->"Superscript"}], " is computed from D by raising each diagonal entry of D to the nth \ power. \n\nBy the way, one can use this fact to answer a question that we \ will be interested in in the next section: When do the positive powers of D \ tend to a constant matrix? Answer....?\n\nNow for a more general A. Let's \ take a 3 ", "\[Cross]", " 3 matrix A. What if we could find three linearly independent \ eigenvectors v", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ", v", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", and v", StyleBox["3", FontVariations->{"CompatibilityType"->"Subscript"}], "? We would have\n\n\t\t\t\tA v", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], " = ", "\[Lambda]", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], " v", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], "\n\t\t\t\tA v", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], " = ", "\[Lambda]", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], " v", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], "\n\t\t\t\tA v", StyleBox["3", FontVariations->{"CompatibilityType"->"Subscript"}], " = ", "\[Lambda]", StyleBox["3", FontVariations->{"CompatibilityType"->"Subscript"}], " v", StyleBox["3", FontVariations->{"CompatibilityType"->"Subscript"}], ",\n\n\t\t\t\t\t\t ", "\:f2e6\[Lambda]", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], " 0 0 ", "\:f2f6", "\nOr:\t\tA [v", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ", v", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", v", StyleBox["3", FontVariations->{"CompatibilityType"->"Subscript"}], "] = [v", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ", v", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", v", StyleBox["3", FontVariations->{"CompatibilityType"->"Subscript"}], "] . ", "\:f2f7", " 0 ", "\[Lambda]", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], " 0 ", "\:f2e7", "\n\t\t\t\t\t\t ", "\:f2e8", " 0 0 ", "\[Lambda]", StyleBox["3", FontVariations->{"CompatibilityType"->"Subscript"}], " ", "\:f2f8", "\n\nNow set P = [v", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ", v", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", v", StyleBox["3", FontVariations->{"CompatibilityType"->"Subscript"}], "] and set D = DiagonalMatrix[", "\[Lambda]", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ",", "\[Lambda]", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", ", "\[Lambda]", StyleBox["3", FontVariations->{"CompatibilityType"->"Subscript"}], "]. Then P is invertible . (Why?) Moreover, the above equation, if \ multiplied on the left by P", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], ", gives the equation\n\n\t\t\t\tP", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], " A P = D.\n\nConsider D D = (P", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], " A P ) (P", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], " A P ) =P", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], " A A P = ", "P", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], " A", StyleBox["2", FontVariations->{"CompatibilityType"->"Superscript"}], " P", ".\nIn general, the same holds for the nth powers of A and D. Finally, \ multiply on the left and right by P and P", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], " to obtain:", "\n\nKEY FACT: If there is a matrix P such that ", "P", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], " A P = D, then for all positive integers n,", "\n\n\t\t\t\tA", StyleBox["n", FontVariations->{"CompatibilityType"->"Superscript"}], " = P D", StyleBox["n", FontVariations->{"CompatibilityType"->"Superscript"}], " P", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], ".\n\nCall a matrix A ", StyleBox["diagonalizable", FontSlant->"Italic"], " if there exists an invertible matrix P such that P", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], " A P is a diagonal matrix. What is the story here? Notice that all the \ steps we performed are reversible. So what we have really shown (for the \ case n = 3, but the same argument works in general) is the following \ important fact:\n\nKEY FACT (Diagonalization Theorem): Let A be an n", "\[Cross]", "n matrix. Then A is diagonalizable if and only if there exists a basis \ of eigenvectors v", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ", v", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", ... , v", StyleBox["n ", FontVariations->{"CompatibilityType"->"Subscript"}], " of A (i.e. v", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ", v", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", ... , v", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], " are linearly independent) for R", StyleBox["n", FontVariations->{"CompatibilityType"->"Superscript"}], ". Moreover, in this case the matrix \nP = [v", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], ", v", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ", ... , v", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], "] is nonsingular and P", StyleBox["-1", FontVariations->{"CompatibilityType"->"Superscript"}], " A P is a diagonal matrix with the eigenvalues of A along its main \ diagonal.\n\nJust as an aside, here is a very handy fact worth knowing:\n\n\ FACT: A set of eigenvectors of the matrix A which correspond to distinct \ eigenvalues of A must necessarily be linearly independent.\n\nNice \ application:\n\nAPPLICATION: If the n", "\[Cross]", "n matrix A has ", StyleBox["n", FontSlant->"Italic"], " distinct eigenvalues, then A is diagonalizable.\n\t\t\t\t\nLet's \ carry out this whole program on the matrix a of the previous section." }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["s = Eigensystem[a]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["d = DiagonalMatrix[s[[1]]]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["p = Transpose[s[[2]]]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["MatrixForm[a.p]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["MatrixForm[p . d]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["d - Inverse[p].a.p", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a - p.d.Inverse[p]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["\<\ Finally, let's verify the above equation, say for the fifth power \ of A:\ \>", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["p . d.d.d.d.d . Inverse[p]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a.a.a.a.a", "Input", ImageRegion->{{-0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Foxes and Chickens???", "Subsection", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "Yes, foxes and chickens. Consider the following problem: a population of \ foxes and chickens enjoys a predator/prey relationship which is modelled as a \ linear relationship. The coefficients that model the relationship are \ determined experimentally, and one parameter, the kill rate ", StyleBox["k", FontSlant->"Italic"], " (of chickens by foxes ) varies with different populations and locales. \ If F", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], " and C", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], " are the fox and chicken populations at the beginning of year ", StyleBox["n", FontSlant->"Italic"], ", then the population in the following year is given by \n\n\t\tF", StyleBox["n+1", FontVariations->{"CompatibilityType"->"Subscript"}], " = 0.6 F", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], " + 0.5 C", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], "\n\t\tC", StyleBox["n+1", FontVariations->{"CompatibilityType"->"Subscript"}], " = -", StyleBox["k", FontSlant->"Italic"], " F", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], " + 1.2 C", StyleBox["n", FontVariations->{"CompatibilityType"->"Subscript"}], "\n\t\t\nCan you attach a meaning to each of the coefficients? In matrix \ form we can write the column vector [F", StyleBox["i", FontVariations->{"CompatibilityType"->"Subscript"}], ", C", StyleBox["i", FontVariations->{"CompatibilityType"->"Subscript"}], "] as x", StyleBox["i", FontVariations->{"CompatibilityType"->"Subscript"}], " and express the formula as\n\n\t\tx", StyleBox["i+1", FontVariations->{"CompatibilityType"->"Subscript"}], " = A x", StyleBox["i", FontVariations->{"CompatibilityType"->"Subscript"}], ",\n\nwhere the matrix A = {{0.6, 0.5}, {-k, 1.2}}. Click the \ following cell:" }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a = {{0.6, 0.5},{-k, 1.2}}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "Now here is the question: for what values of the kill rate ", StyleBox["k", FontSlant->"Italic"], " will the fox and chicken populations die out? For what values will it \ grow without bound? Is there a value for which the populations will \ stabilize?\n\nHere is where ", StyleBox["Mathematica", FontSlant->"Italic"], " really shines ! We recognize this as a discrete linear dynamical system. \ From our earlier discussion, we realize that the growth of the state vector \ depends on the largest eigenvalue (in absolute value). If this eigenvalue is \ greater than one, we expect that some part of the population will grow \ without bound. If it is smaller than one we expect that both populations \ will die out in time. If it is equal to one, we expect that the populations \ will stabilize. So let's make a graph of the size of the eigenvalues (in \ absolute value) of the matrix ", StyleBox["a", FontSlant->"Italic"], " as a function of the parameter ", StyleBox["k", FontSlant->"Italic"], ":" }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["s[k_] := Max[Abs[Eigenvalues[a]]]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Plot[s[k],{k,0,2}]", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ "Now narrow the graph a little and see if you can find a value of ", StyleBox["k", FontSlant->"Italic"], " for which the largest eigenvalue is about 1 and test this value on the \ population x below." }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["x = {100,1000}", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["k = .5", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a.a.a.a.x", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a.a.a.a.a.a.a.a.x", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a.a.a.a.a.a.a.a.a.a.a.a.x", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.a.x", "Input", ImageRegion->{{-0, 1}, {0, 1}}], Cell["Conclusions ???", "Text", ImageRegion->{{-0, 1}, {0, 1}}] }, Closed]], Cell[CellGroupData[{ Cell["Exercises", "Subsection", ImageRegion->{{-0, 1}, {0, 1}}], Cell["\<\ Find the eigenvalues and eigenvectors for the following matrices. \ Which are diagonalizable?\ \>", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["\<\ \:f2e63 5 7\:f2f6 1) A = \:f2e70 2 1\:f2f7 \:f2e81 4 6\:f2f8\ \>", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell["\<\ \:f2e61 0 3 5 7\:f2f6 \:f2e72 -1 7 0 8\:f2f7 (Hint: \ use NSolve[] to approximate the eigenvalues 2) B = \:f2f7-5 2 4 0 -1\:f2e7 and find \ each eigenspace separately.) \:f2e7 3 2 0 -6 3 \:f2e7 \:f2e8 1 0 2 -1 2\:f2f8\ \>", "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ " ", "\:f2e6", "20 5 0", "\:f2f6", StyleBox["\n ", FontSlant->"Italic"], "3) C = ", "\:f2e7", " -8 32 1 ", "\:f2e7", " \n ", "\:f2e8", " 1 1 23", "\:f2f8" }], "Text", ImageRegion->{{-0, 1}, {0, 1}}], Cell[TextData[{ " 4) For each matrix, find the characteristic polynomial P(x), and \ evaluate it at that matrix. \t(i.e., if the characteristic polynomial for \ the matrix D is x", StyleBox["2 ", FontVariations->{"CompatibilityType"->"Superscript"}], "+ 2 x - 1, evaluate \n D", StyleBox["2", FontVariations->{"CompatibilityType"->"Superscript"}], " + 2 D - I", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], ".)" }], "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. 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