# Jordan Normal Form

1. Feb 29, 2012

### Kamekui

1. The problem statement, all variables and given/known data

Find the Jordan form and use it to find the matrix exponential.

2. Relevant equations

3. 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

2. Mar 1, 2012

### vela

Staff Emeritus
$$\begin{bmatrix} 1 &-1 &-1 &-1 \\ -1 & 1 & 1 & 1 \\ 1 & -1 & -1 & -1 \\ 1 & -1 & -1 & -1 \end{bmatrix} \ne \begin{bmatrix} 1 &-1 &-1 &-1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ I know what you mean, but don't use an equal sign there. The matrices aren't equal.

This isn't correct. $(A+4I)^2 = 0$, so the kernel is all of $\mathbb{R}^4$.

This isn't correct either. It's true that $(A+4I)^2\vec{x}=0$ is satisfied by any $\vec{x}$, but it's not true that the vector will also satisfy $(A+4I)\vec{x} = \vec{E}_3$.

You found the kernel of A+4I consists of vectors of the form
$$v1\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + v2\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} + v3\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$ What you want to do is find a particular linear combination $\vec{v}$ so that $(A+4I)\vec{x} = \vec{v}$ will have a solution. The vector $\vec{v}$ corresponds to what you called E3, and the solution $\vec{x}$, to E4.

3. Mar 2, 2012

### Kamekui

Ok, I went back and tried this:

Let
E4=
\begin{bmatrix}
0\\
1\\
1\\
0
\end{bmatrix}

Then,

(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}
0 \\
1 \\
1 \\
0
\end{bmatrix}
=\begin{bmatrix}
-2\\
2 \\
-2 \\
-2
\end{bmatrix}

Form a matrix T such that T-1AT=J

So, T=\begin{bmatrix}
1 & 1 & -2 & 1\\
1 & 0 & 2 & 1\\
0 & 1 & -2 & 1\\
0 & 0 & -2 & 1
\end{bmatrix}

and T-1=\begin{bmatrix}
1 & 0 & -1 & 0\\
0 & 0 & 1 & -1\\
-1/4 & 1/4 & 1/4 & -1/4\\
-1/2 & 1/2 & 1/2 & 1/2
\end{bmatrix}

T-1AT=\begin{bmatrix}
-4 & 0 & 0 & 0\\
0 & -4 & & 0\\
0 & 0 & -4 & 1\\
0 & 0 & 0 & -4
\end{bmatrix}
=J

If this is correct, computing the matrix exponential is easy, if its correct...

4. Mar 2, 2012

### vela

Staff Emeritus
Did you mean
$$T = \begin{bmatrix} 1 & 1 & -2 & 1\\ 1 & 0 & 2 & 1\\ 0 & 1 & -2 & 1\\ 0 & 0 & -2 & 1 \end{bmatrix}$$ or
$$T= \begin{bmatrix} 1 & 1 & -2 & 0\\ 1 & 0 & 2 & 1\\ 0 & 1 & -2 & 1\\ 0 & 0 & -2 & 0 \end{bmatrix}?$$