- #1
Kamekui
- 14
- 0
Homework Statement
Find the Jordan form and use it to find the matrix exponential.
Homework Equations
The Attempt at a Solution
Let A =\begin{bmatrix}
-3 &-1 &-1 &-1 \\
-1 & -3 & 1 & 1 \\
1 & -1 & -5 & -1 \\
1 & -1 & -1 & -5
\end{bmatrix}
det(A-λI)=λ4+16λ3+96λ2+256λ+256
(λ+4)4
→ λ=-4 (Multiplicity 4)
A+4I=\begin{bmatrix}
1 &-1 &-1 &-1 \\
-1 & 1 & 1 & 1 \\
1 & -1 & -1 & -1 \\
1 & -1 & -1 & -1
\end{bmatrix}
=\begin{bmatrix}
1 &-1 &-1 &-1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
v1 \\
v2 \\
v3 \\
v4
\end{bmatrix} =
v1\begin{bmatrix}
1 \\
1 \\
0 \\
0
\end{bmatrix}+v2\begin{bmatrix}
1 \\
0 \\
1 \\
0
\end{bmatrix} + v3\begin{bmatrix}
1 \\
0 \\
0 \\
1
\end{bmatrix}
Therefore, the Jordan Normal Form must look like:
J=\begin{bmatrix}
-4 &0 &0 &0 \\
0 & -4 & 0 & 0 \\
0 & 0 & -4 & 1 \\
0 & 0 & 0 & -4
\end{bmatrix}
Consider now that:
JE1=-4E1
JE2=-4E2
JE3=-4E3
JE4=E3-4E4
But this is equivalent to:
AE1=-4E1
AE2=-4E2
AE3=-4E3
AE4=E3-4E4
AE4=E3-4E4→ (A+4I)E4=E3
Now,
ker[(A+4I)2]=0
(A+4I)2E4=E3
0*E4=0
Therefore, E4 can be anything that is Linearly Independent of E1,E2, and E3.
Let E4=\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix}
(A+4I)E4=E3
\begin{bmatrix}
1 &-1 &-1 &-1 \\
-1 & 1 & 1 & 1 \\
1 & -1 & -1 & -1 \\
1 & -1 & -1 & -1
\end{bmatrix}*\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix}
=\begin{bmatrix}
-2 \\
2 \\
-2 \\
-2
\end{bmatrix}
I'm lost after this point