Initializations

$TextStyle={FontFamily->"Times",FontSize->16};
Off[General::"spell1"]
Off[InverseFunction::"ifun"]
Off[Solve::"ifun"]

evecplot[p_]:=Show[Graphics[{Red,Dashing[{1,0}],Thickness[0.01],Arrow[{{0,0},evec[p]}]}],Graphics[{Blue,Dashing[{1,0}],Thickness[0.005],Arrow[{{0,0},A.evec[p]}]}],Axes->True,AxesLabel->{"(Subscript[x, 1])","(Subscript[x, 2])"},AspectRatio->1,DisplayFunction->Identity]

Eigensystem Basics

The n x n matrix A will take an n-vector x and map it into an n-vector y,

A x = y

For some vectors x, this mapping will produce a vector y which is parallel to the vector x, but scaled by some amount λ,

A x = λ x

The vectors for which this is true are called eigenvectors, and each eigenvector has an associated scaling parameter λ, which is the eigenvalue. We can rewrite our system of equations as follows

A x = λ x
A x - λ x = 0
(A  - λ I)x = 0

For this system to have a nonzero solution x, we require the determinant of the matrix A-λ I to be zero. We can solve the following equation for (λ^((k))), k=1,2,3,...,n.

det(A  - (λ^((k))) I) = 0

Once the values of  (λ^((k))) are known, we can solve the equations

(A  - (λ^((k))) I)(ξ^((k))) = 0

for the eigenvectors (ξ^((k))).

Example 7.3.15

A={{5,-1},{3,1}}
MatrixForm[A]
Eigenvalues[A]
Eigenvectors[A]

Our answer is correct! Yes! (pumping fist)

Here is the geometric interpretation of the eigensystem. evecplot is defined in the Initializations.

evec[1]={1,3}
evec[2]={1,1}

Show[{evecplot[1],evecplot[2]},DisplayFunction->$DisplayFunction,PlotRange->{All, All},ImageSize->{400,Automatic},AspectRatio->1]

Note that the red vectors are the eigenvectors ξ, and the blue vectors are the vector A ξ, which are in the same direction as the original eigenvector, but scaled by an amount given by the associated eigenvalue, λ (2 and 4, in this case). Pretty cool, huh?

Example 7.3.18

A={{1,I},{-I,1}}
MatrixForm[A]
Eigenvalues[A]
Eigenvectors[A]

Our answer is correct! Remember, eigenvectors are determined up to a constant multiplier. We really can't do a graphical interpretation of this like we did before, since the numbers are complex-valued.

Example 7.3.21

We can use Mathematica to help with some algebra along the way:

A={{1,0,0},{2,1,-2},{3,2,1}}
MatrixForm[A]
char=Det[A-λ*IdentityMatrix[3]]

Solve[char==0,λ]

Let's check our answer:

Eigenvalues[A]
Eigenvectors[A]