Jordan Form & Matrix Exponential: Homework Statement and Solution

[Kamekui]

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

 

[vela]

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}

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

Now,

ker[(A+4I)²]=0

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

(A+4I)²E₄=E₃
0*E₄=0

Therefore, E₄ can be anything that is Linearly Independent of E₁,E₂, and E₃.

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.

 

[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...

 

[vela]

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
$$T= \begin{bmatrix}
1 & 1 & -2 & 0\\
1 & 0 & 2 & 1\\
0 & 1 & -2 & 1\\
0 & 0 & -2 & 0
\end{bmatrix}?$$