If your Java worked, this is what you would see
(screen shot).
[explanation of Duke website contents, including
what the grey square is in the applet window].
Here is a Maple worksheet
that does a similar job instead (see
exploreeigendrag.htm)
Click on each entry to get a popup window to change the default matrix entries of the 2x2 matrix A. Hit the enter key or click on OK to enter it.
The original default matrix is: a11 = 0, a12 = 1, a21 = -1, a22 = 0 .
For the default matrix 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 vectors, 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)
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: 6.1.9: 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.
[You may need to copy the applet website URL and include it in your Java security exception list to make this work in Internet Explorer and Firefox, but unfortunately Google Chrome has abandoned supporting Java so you probably won't be able to use that brower: check please and report back to bob. Mac users, what about Safari? For PCs under the Start Menu, Programs, Java, Configure Java look at the Security tab and copy and paste the URL for the applet into the Exception Site list using the Add button.]