DERIVE for Windows version 5.05 DfW file saved on 30 Sep 2002
ShowAction(A, H, K):=PROG(K := H`, RETURN (A·K)`)
length(v):=PROG(RETURN v™1^2 + v™2^2)
A:=[1436231/1480615, 631263/2861042; 2421259/3749017, 151919/211704]
B:=[COS(¹/3), - SIN(¹/3); SIN(¹/3), COS(¹/3)]
B1:=1/2·[1, -3; 1, 1]
C:=1/2·[1, -3; 1, 1]
D:=1/4·[5, -3; 3, -1]
H:=[-2, -2; -2, 2; -3, 1; 0, 4; 0, 4; 3, 1; 2, 2; 2, -2; 2, -2; -2, -2]
hCross:=APPROX(- 6096774193548387/2000000000000000)
vCross:=APPROX(45483870967741931/10000000000000000)
1ÿÿCTextObjçÿ`{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\froman Times New Roman;}{\f2\froman\fcharset0 Times New Roman;}{\f3\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\qc\cf1\b\f0\fs24 The Action of a Linear Operator
\par and an Introduction to Eigenvalues & Eigenvectors
\par \pard\b0
\par \cf0\f1 NOTE: \cf2 While working with this example we recommend that you open a 2D\f2 -\f1 Plot window. In the 2D\f2 -\f1 Plot window press the F11 key. In the \f2 Display Options window \f1 click on the Points tab. Set Connect to "Yes" and \f2 Size\f1 to "Small". \f2 Make sure Options > Simplify before Plotting is checked. \f1 Return to the Algebra window and choose the option Window\f2 \f1 >Tile Vertically. After doing this you will be able to work through the lesson and see the graphical illustrations side by side.\b
\par \cf0
\par Background - What Does Multiplica\f2 t\f1 ion By a Matrix Do?\b0
\par
\par In order to give a graphical illustration of the effect of matrix multiplication we will use a technique similar to that found in \ul Introduction to Linear Algebra - Second Edition\ulnone by Gilbert Strang (Wellesley and Cambridge Press, Wellesley, MA, 1998). We will work in two dimensional space and show the results of multiplying a figure by 2 \i X \i0 2 matrices.
\par
\par Our basic figure will be a simple stick house\f2 , \f1 defined by the matrix H. \cf3 Highlight this expression and then \f2 plot \f1 it in the 2D\f2 -\f1 Plot window.\cf0 \cf1\f3
\par }
ÿÿCExpnObj8"°Userð¿HH:=[[-2,-2],[-2,2],[-3,1],[0,4],[0,4],[3,1],[2,2],[2,-2],[2,-2],[-2,-2]]€ç8ÿ{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 This is a rather simple figure, but it will serve our purpose quite well. We begin by generating a random matrix (2 \i X\i0 2). \cf2\f1 Simplify\f0 the following two expressions and save the resulting matrix as a variable called, \i A.\i0 \cf1\f2
\par }
€8D€PUserð¿ RANDOM(0)€8\ø€Userð¿-[[RANDOM(1),RANDOM(1)],[RANDOM(1),RANDOM(1)]]€Œç7ÿ{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 The first expression is used to intialize the state of Derive's random number generator. The second expression generates a 2 \i X\i0 2 matrix. To show the action of we multiply the transpose of \i H\i0 on the left by \i A\i0 . This will move the ends of the lines making up the outline of the house.
\par
\par \ul Exercise\ulnone
\par 1)\cf2 Why do we need to take the Transpose of \i H\i0 ?
\par
\par \cf1 We have automated the process of taking the transpose of \i H\i0 and multiplying on the left by \i A \i0 by writing a small Derive function called \i ShowAction \i0 and saved its definition with this file. (If you wish to see the definition of the function, \f1 issue \f0 the \ul D\ulnone eclare\f1 > \f0 Function Definition\f1 command\f0 . Use the \f1 drop \f0 down menu \f1 beside the first text box to select \f0 "ShowAction(A,H,K)". The definition will appear in the second text box.) \cf2 Highlight the following expression and plot it in the 2D\f1 -p\f0 lot window.\cf1 \f2
\par }
€8C¸OUserð¿ShowAction(A,H)€[ç§ÿò{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 We can still recognize the basic shape of the house, but it is squashed and the bottom corners are no longer right angles.
\par
\par Not all matrices produce the same types of actions. We will first define a matrix that preserves all of the properties of the house except its orientation to the window. \cf2 Highlight expression #6. In the 2D\f1 -p\f0 lot window delete the last plot and then plot the result of multiplying \i H\i0 by \i B \i0 (ShowAction(B, H)).\cf1\f2
\par }
€8³ Userð¿1B:=[[COS(pi/3),-SIN(pi/3)],[SIN(pi/3),COS(pi/3)]]€8¸Userð¿ShowAction(B,H)€+çÃÿF{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fcharset2 DfW5 Printer;}{\f2\froman Times New Roman;}{\f3\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 Note that the house is exactly the same as the original except that it is rotated \cf2 (through what angle?)\cf1 . The matrix, \i B\i0 , has a special form using sines and cosines. Any matrix of this form is called a \ul rotation matrix\ulnone .
\par
\par \ul Exercise\ulnone
\par 2) \cf2 Show that a matrix having the form of B, replace \f1\'b9/\f2 3 by \f1\'a9\f2 , will have the effect of rotating a point (\i x\i0 ,\i y\i0 ) through the angle, \f1\'a9\f2 . \cf1
\par
\par Somewhat in the spirit of play, but also to illustrate the idea of a rotation and what happens when we raise this matrix to higher powers and look at the effect. \cf2 Highlight each of the following expressions individually and plot the result of multiplying \i H\i0 by higher powers of \i B\i0 . \cf1\f3
\par }
€8ÏÀçUserð¿ShowAction(B^2,H)€8óÀUserð¿ShowAction(B^3,H)€8À/Userð¿ ShowAction(B^4,H)€8;ÀSUserð¿
ShowAction(B^5,H)€8_ÀwUserð¿ShowAction(B^6,H)€ƒçõÿÏ{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fcharset2 DfW5 Printer;}{\f2\fmodern\fprq1\fcharset0 Times New Roman;}{\f3\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 Each successive power of \i B\i0 further rotates the house by an additional angle of \f1\'b9\f0 /3. Of course, since 6*\f1\'b9\f0 /3 = 2\f1\'b9\f0 , after 6 rotations the house is back in its original position and the cycle begins again.
\par
\par We will look at the action of one more matrix before returning to the first matrix, \i A\i0 . Notice from its form that this matrix, \i C\i0 , is not a rotation matrix. \cf2\f2 Delete all plots in \f0 the 2-D Plot window by pressing \f2 C\f0 trl-D while in that window, highlight expression #1 and redraw the house, and then highlight expression #13 and observe the action of \i C\i0 on \i H\i0 . \cf1\f3
\par }
€8È%Userð¿C:=1/2*[[1,-3],[1,1]]€81¸=Userð¿
ShowAction(C,H)€Iç‚ÿŽ{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 A look at the graph confirms our observation that this is not a rotation matrix. The angles where the lines join in the original house are not preserved. But yet, something interesting happens with this matrix. \cf2 Highlight ea\f1 c\f0 h of the following expressions one at a time and observe the action by \f1 deleting\f0 the previous plot and drawing the next one.\cf1\f2
\par }
€8ŽÀ¦Userð¿ShowAction(C^2,H)€8²ÀÊUserð¿ShowAction(C^3,H)€8ÖÀîUserð¿ShowAction(C^4,H)€8úÀUserð¿ShowAction(C^5,H)€8À6Userð¿ShowAction(C^6,H)€Bçhÿy{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 Even though \i C\i0 is not a rotation, it has some behaviors that are similar. Most obvious is that the sixth power of \i C\i0 places the house back in its original position and with its original shape. For comparison, let's look at the powers of \i B\i0 and \i C\i0 . \cf2\f1 Simplify\f0 each of the following expressions and compare the results.\f2
\par }
€8tÐŒUserð¿VECTOR(B^i,i,1,6)€8˜Ð°Userð¿VECTOR(C^i,i,1,6)€¼çAÿ˜{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 The results show the first six powers of \i B\i0 and \i C. \i0 In each case the sixth power is the identity matrix. This agrees with our graphical observation. Namely, that after six applications of the matrices the house is redrawn.
\par
\par \b An Introduction to Eigenvalues and Eigenvectors\b0
\par
\par What about the first matrix, \i A\i0 ? Does that return the house to its origional position and shape? Let's take a look.\i \cf2\i0\f1 Delete all plots from\f0 the 2D\f1 -p\f0 lot window and redraw the house. Highlight the following expressions and plot the results. This time do not erase the previous plots.\cf1\f2
\par }
€8MÀeUserð¿ShowAction(A^2,H)€8qÀ‰Userð¿ShowAction(A^3,H)€8•ÀUserð¿ShowAction(A^4,H)€8¹ÀÑUserð¿ShowAction(A^5,H)€Ýç ÿÁ{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 This is different! The action of each successive power seems to squash the house even more than the last one. Furthermore, the successive actions seem to be pulling the house in the same direction. There is no chance that the house will be restored by continually applying powers of \i A\i0 . \cf2 To see this, clear the plot window\f1 of plots\f0 , redraw the house, and plot the result of the following expression. \cf1\f2
\par }
€8" È: Userð¿ShowAction(A^25,H)€F ç¥ ÿ{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 What appears to be happening is that the action of higher powers of \i A\i0 appears to be collapsing the house to a straight line. Carrying this idea even further, we can say that the action of higher and higher powers of \i A\i0 collapses the \i entire plane \i0 to the straight line that appears on the graph.
\par
\par \ul Exercise\ulnone
\par 3) \cf2 Help to illustrate this claim by choosing any point, (\i x\i0 ,\i y\i0 ), in the 2D\f1 -p\f0 lot window and substitute it in the following expression.\cf1\f2
\par }
€8± ˜Õ Userð¿(A^25*[[x],[y]])`€á çÅ
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\viewkind4\uc1\pard\cf1\f0\fs24 Divide the result by the largest component of the resulting vector. (This results in a vector that should lie inside of the plot window and will lie on the same line as your result.) \f1 Before p\f0 lot\f1 ting\f0 the point \f1 use the 2D-plot window's Options > Display command to \f0 chang\f1 e Point\f0 Size \f1 to \f0 "Large". Does this point lie on the same line as the "collapsed" house?
\par
\par \cf2 We suspect that your answer to the last question was "yes" or at least "apparently".
\par
\par What happens to a point on the line under the action of \i A\i0 ? Since it already lies on the line and the action of \i A\i0 appears to be "pulling" the all of the points on the plane towards that line, it makes sense that a point on the line will remain on the line.
\par
\par \ul Exercise\ulnone
\par \cf1 4) Check this assertion by multiplying the result of exercise 3) by \i A\i0 and plotting the new point. Use expression #27 to perform the multiplication replacing \i x\i0 and \i y \i0 with the coordinates of the point from exercise 3. Note the coordinates of this result.
\par }
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\viewkind4\uc1\pard\cf1\f0\fs24 5) Prove that since matrix multiplication is a linear operation, the following statement is true. If (\i x\i0 , \i y\i0 ) lies on the line under consideration, then \i A\i0 (\i x\i0 , \i y\i0 ) = \f1\'bf\f0 (\i x\i0 , \i y\i0 ). HINT: Since every point on a line passing through the origin is of the form (\i x\i0 , \i\f1\'a9\f0 x\i0 ) for some \f1\'a9\f0 , then (\i x\i0 , \i y\i0 ) = (\i x\i0 , \i\f1\'a9\f0 x\i0 ) = \i x\i0 (\i 1\i0 , \i\f1\'a9\i0\f0 ).
\par 6)\cf2 \cf1 Find \f1\'bf\f0 for your matrix \i A\i0 .
\par
\par \cf2\b Illustrating Some Basic Facts About Eigenvalues\cf1\b0
\par
\par \cf2 The value, \f1\'bf\f0 , that you found in exercise 6) is called an \i eigenvalue\i0 or \i characteristic value\i0 of the matrix, \i A\i0 . The vectors lying along the line are called \i eigenvectors\i0 or \i characteristic vectors\i0 for the eigenvalue (the preferred term), \f1\'bf\f0 . Later in your Linear Algebra course you will learn the following facts. \cf1 Use your matrix, \i A\i0 , and eigenvalue, \f1\'bf\f0 , to verify that the following statements are true. \cf2
\par
\par I. The determinant of \i A\i0 - \f1\'bf\i\f0 I\i0 is 0.\f2
\par }
€8ÞðUserð¿DET(A-lambda*[[1,0],[0,1]])=0€ç!ÿ){\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 II. If \i v\i0 is a variable, the following expression is a polynomial in \i v\i0 .\f1
\par }
€8-ÐQUserð¿DET(A-v*[[1,0],[0,1]])€]çpÿo{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fcharset2 DfW5 Printer;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
{\colortbl ;\red0\green0\blue0;}
\viewkind4\uc1\pard\cf1\f0\fs24 III. The eigenvalue, \f1\'bf\f0 , is a root of the polynomial in expression #29. It is the largest of the two roots.\f2
\par }
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\viewkind4\uc1\pard\cf1\f0\fs24 The next fact will take some work to verify. We give a small hint - \cf2 choose a vector (\i x, y\i0 ) in a direction perpendicular to the line on your display and show that \i A\i0 (\i x, y\i0 ) = \f1\'b5\f0 (\i x, y\i0 ) where \f1\'b5\f0 is the other root.\cf1
\par
\par IV. The second root of the polynomial in expression #29 is also an eigenvalue for \i A\i0 .
\par
\par There is a way to display this last result geometrically, but it takes a deeper understanding of Linear Algebra and a little more work. We will suffice to end our discussion here.
\par
\par \b Eigenvalues in the Complex Plane\b0
\par
\par Oh yes, what about matrices, \i B\i0 and \i C\i0 ? They did not collapse the space to a line. To get a hint of the idea, \cf2 find the roots of the following polynomials. \cf1\f2
\par }
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Userð¿ DET(C-v*[[1,0],[0,1]])€é
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\viewkind4\uc1\pard\cf1\b\f0\fs24 Summary\b0
\par
\par In this activity we have looked at the action of different matrices. The matrix \i A\i0 was an example of a matrix which had real, but unequal eigenvalues. \i B\i0 had a special form that was typical of a rotation matrix. Its eigenvalues were complex valued and also roots of unity. \i C \i0 was an example of a more general matrix having complex valued eigenvalues. The only case that we did not consider was the one where the eigenvalues were equal (either real or complex). That is a story for another time.
\par
\par \fs20 Carl Leinbach: leinbach@gettysburg.edu\f1\fs24
\par }
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