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Example: computing eigenvectors and eigenvalues
Make sure that you know the meaning and calculation of eigenvalues and eigenvectors for a matrix.
See the linear systems review note!
Here is a tiny example.
\[\begin{split}
A =
\begin{bmatrix}
0 & 1 \\
-2 & -3
\end{bmatrix}.
\end{split}\]
Solve \(\det(\lambda I - A)=0\):
\[\begin{split}
\det
\begin{bmatrix}
\lambda & -1 \\
2 & \lambda + 3
\end{bmatrix}
= \lambda (\lambda+3) + 2 = \lambda^2 + 3\lambda + 2
= (\lambda+2)(\lambda+1)=0.
\end{split}\]
Eigenvalues: \(-2\) and \(-1\).
For each eigenvalue, solve \(A v = \lambda v\).
For \(\lambda=-2\):
\[\begin{split}
\begin{bmatrix}
0 & 1 \\
-2 & -3
\end{bmatrix}
\begin{bmatrix}
v_1\\v_2
\end{bmatrix}
=
-2
\begin{bmatrix}
v_1\\v_2
\end{bmatrix}
\Rightarrow
v_2 = -2 v_1.
\end{split}\]
Choose \(v=\begin{bmatrix}-1 \\ 2\end{bmatrix}\)
For \(\lambda=-1\):
\[\begin{split}
\begin{bmatrix}
0 & 1 \\
-2 & -3
\end{bmatrix}
\begin{bmatrix}
v_1\\v_2
\end{bmatrix}
=
-1
\begin{bmatrix}
v_1\\v_2
\end{bmatrix}
\Rightarrow
v_2 = -v_1.
\end{split}\]
Choose \(v=\begin{bmatrix}-1 \\ 1\end{bmatrix}\)
Eigenvalues and eigenvectors:
\[\begin{split}
\begin{array}{c|c}
\text{Eigenvalue} & \text{Eigenvector} \\
\hline
-2 & \begin{bmatrix}-1\\2\end{bmatrix} \\
-1 & \begin{bmatrix}-1\\1\end{bmatrix}
\end{array}
\end{split}\]
Question: Can we always diagonalize the \(A\) matrix?
Answer: We may not be able to diagonalize fully, but we can get very close.
Example
\[\begin{split}
A =
\begin{bmatrix}
-5 & -4 & -2 & -1 \\
0 & -1 & 1 & 1 \\
1 & 1 & -3 & 0 \\
-1 & -1 & 1 & -2
\end{bmatrix},
\qquad
\text{eigenvalues: } -4,\ -4,\ -1,\ -2.
\end{split}\]
Eigenvectors:
For \(\lambda = -4\):
\[\begin{split}
\begin{bmatrix}
1 \\ 0 \\ -1 \\ 1
\end{bmatrix},
\qquad
\begin{bmatrix}
-1 \\ 0 \\ 1 \\ -1
\end{bmatrix}.
\end{split}\]
For \(\lambda = -1\):
\[\begin{split}
\begin{bmatrix}
-1 \\ 1 \\ 0 \\ 0
\end{bmatrix}.
\end{split}\]
For \(\lambda = -2\):
\[\begin{split}
\begin{bmatrix}
1 \\ -1 \\ 0 \\ -1
\end{bmatrix}.
\end{split}\]
Note that the eigenvectors corresponding to the same eigenvalue (\(-4\)) are not be linearly independent. This means that the associated Jordan block of that eigenvalue is of order 2. Thus the Jordan canonical form is:
\[\begin{split}
J =
\begin{bmatrix}
-4 & 1 & 0 & 0 \\
0 & -4 & 0 & 0 \\
0 & 0 & -2 & 0 \\
0 & 0 & 0 & -1
\end{bmatrix}.
\end{split}\]
(Note how the upper-left block of this matrix is of order 2.)