how to use the EigenExploreDrag.mw worksheet


 Open this Maple worksheet EigenExploreDrag.mw.
 [EigenExploreDrag-EPC.mw, web version outside Maple, since Maple 2020 is not executing this drag feature properly, nor is Maple 2021!]

Execute the worksheet with the !!! icon on the toolbar. Then edit the matrix A at the end of the worksheet to use the interactive 2d plot showing the input vector x in red, and the output vector A x in blue. Click on the plot and left click mouse drag the red point around in a circle about the origin to explore for which directions the two position vectors line up, thus identifying an eigenvector direction and by counting squares in the grid, determine the multiple (eigenvalue) of the input vector which is the output vector.

If you try this matrix: a11 = 0, a12 = 1, a21 = -1, a22 = 0 , as you drag the red circle at the tip of the red arrow countercockwise around a circle about the origin with your mouse checking all directions for the input vector (red vector x), the output blue vector A x always rotates ahead of it by 90 degrees and has the same length. There are no special directions where the two arrows line up. [When using with simple matrices with integer eigenvectors and eigenvalues, if the red arrow is on the integer grid along special directions, the blue arrow will also have its tip on the integer grid and one can easily compare their lengths, i.e., compare the numbers of squares along the hypotenuse of the triangle locating the arrow.]

The rest are just possible ideas to play around with.

First change a12 to 1 from -1, corresponding to the matrix:  a11 = 0, a12 = 1, a21 = 1, a22 = 0  of the handout example.
Then drag the red circle around the origin in a circle noting the two lines through the origin where the blue arrow lines up with the red arrow. Along y = x, in either direction from the origin, they not only line up with the same direction, they also have the same length: conclude A x = 1 x, so vectors along this line have the eigenvalue 1. Try to put the red arrow tip on the integer grid to read off the output vector integer components. [Note that <1,1> or <-1,-1> are the smallest integer component vectors along this line, the first being the simplest one, a representative "eigenvector" among the whole family of such "eigenvectors".]  Along y = -x, in either direction from the origin, they not only line up with the opposite directions, they also have the same length: conclude A x = -1 x . Along this line the vectors  have the eigenvalue -1. All of the nonzero vectors along each line are called eigenvectors corresponding to the eigenvalue factor we determined.  [Note that <1,-1> or <-1,1> are the smallest integer component vectors along this line, either one could be chosen as the "simplest" representative "eigenvector" among the whole family of such "eigenvectors".]

Next consider the MIT video example matrix: a11 = 2, a12 = 0, a21 = 1, a22 = 3 .
Slowly rotate the red arrow around a circle of radius about 2. You see the arrows line up along the y axis. [Note that <0,1> or <0,-1> is the smallest integer component vector along this line, a representative "eigenvector" among the whole family of such "eigenvectors".] When the tip of the red arrow is at 2 units, the blue tip is at 6 units along the axis, in the positive or negative direction, a multiple of 3 times the input red vector. Or if the red tip is at 1 unit, the blue tip is at 3 units, always the same ratio 3 to 1 along the y axis, so the eigenvalue of all these vectors along the y axis is 3. The other special direction when the red tip is along y = - x. [Note that <-1,1> or <1,-1> is the smallest integer component vector along this line, a representative "eigenvector" among the whole family of such "eigenvectors".]

Next consider the MIT video example matrix: a11 = -1, a12 = 1, a21 = 2, a22 = 3.
Put the red arrow tip on the integer grid points along the directions where the arrows line up. Go slowly to find them. Read off the slopes of the two lines and the scalar multiple factors (eigenvalues) between input and output vectors for each of them. Better yet, find the smallest integer component vector along each line as a representative eigenvector for that line. Write them down together with their eigenvalues.

Next consider the matrix E&P3: 6.1.1: a11 = 4, a12 = -2, a21 = 1, a22 = 1 .
Here you have to be careful, both of the two lines are at slightly different positive slopes in the first quadrant. Put the red arrow tip on the integer grid points along the directions where the arrows line up. Go slowly to find them. Read off the slopes of the lines and the scalar multiple factors (eigenvalues) between input and output vectors for each of them. Better yet, find the smallest integer component vector along each line as a representative eigenvector for that line. Write them down together with their eigenvalues.

Repeat for the matrix E&P3: 6.1.2: a11 = 5, a12 = -6, a21 = 3, a22 = -4 . (eigendirections same as previous case, but different eigenvalues, opposite sign)

NO: Repeat for the matrix E&P3: 6.1.3: a11 = 8, a12 = -6, a21 = 3, a22 = -1 .  (eigendirections same as previous case, but different eigenvalues, same sign)

Repeat for the matrix E&P3: 6.1.7: a11 = 10, a12 = -8, a21 = 6, a22 = -4 .  (eigendirections close in direction, same sign eigenvalues)

NO: Repeat for the matrix E&P3: 6.1.9: a11 = 8, a12 = -10, a21 = 2, a22 = -1 .  (eigendirections close in direction, same sign eigenvalues)

In each of these examples, there are two lines of eigenvectors, each with its own eigenvalue.

[origin of this worksheet from a Duke University Java applet]