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Let $$A=\begin{bmatrix}0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -4 & 0\end{bmatrix}$$

I want to determine $e^{At}$. I tried it using two methods.

First via the Jordan form:

The characteristic polynomial of $A$ is given by $$\det(I\xi-A)=\begin{vmatrix} \xi & -1 & 0 & 0 \\ 1 & \xi & 0 & 0 \\ 0 & 0 & \xi & -1 \\ 0 & 0 & 4 & \xi \end{vmatrix}=1\cdot\begin{vmatrix} 1 & 0 & 0 \\ 0 & \xi & -1 \\ 0 & 4 & \xi \end{vmatrix}+\xi\cdot\begin{vmatrix} \xi & 0 & 0 \\ 0 & \xi & -1 \\ 0 & 4 & \xi \end{vmatrix}$$

$$\begin{vmatrix} 1 & 0 & 0 \\ 0 & \xi & -1 \\ 0 & 4 & \xi \end{vmatrix}=1\cdot\begin{vmatrix} \xi & -1 \\ 4 & \xi \end{vmatrix}=\xi^{2}+4\qquad\text{and}\qquad \begin{vmatrix} \xi & 0 & 0 \\ 0 & \xi & -1 \\ 0 & 4 & \xi \end{vmatrix}=\xi\cdot\begin{vmatrix} \xi & -1 \\ 4 & \xi \end{vmatrix}=\xi(\xi^{2}+4)$$

$$\implies\det(I\xi-A)=\xi^{2}+4+\xi(\xi^{3}+4\xi)=\xi^{4}+5\xi^{2}+4=(\xi^{2}+4)(\xi^{2}+1)$$ Then the eigenvalues are given by $\lambda_{1,2}=\pm i$ and $\lambda_{3,4}=\pm 2i$. Now, computing the corresponding eigenvectors:

For $\lambda=i$, we have $$\begin{cases} -ix+y=0 \\ -x-iy=0 \\ iz+w=0 \\ -4z-iw=0 \end{cases}\implies\begin{cases} x=-y \\ 4z=-w \end{cases}$$ Then setting $x=a$ and $z=b$, for $a,b\in\mathbb{C}$, we obtain:

$$v_{1}=\begin{pmatrix} a \\ -a \\ b \\ -4b \end{pmatrix}=a\begin{pmatrix} 1 \\ -1 \\ 0 \\ 0 \end{pmatrix}+b\begin{pmatrix} 0 \\ 0 \\ 1 \\ -4 \end{pmatrix}$$

Similarly, for $\lambda=-i$, we compute, for complex $c,d$: $$v_{2}=c\begin{pmatrix} -1 \\ 1 \\ 0 \\ 0 \end{pmatrix}+d\begin{pmatrix} 0 \\ 0 \\ -1 \\ 4 \end{pmatrix}$$

For $\lambda=2i$, we get: $$v_{3}=j\begin{pmatrix} 1 \\ -1 \\ 0 \\ 0 \end{pmatrix}+k\begin{pmatrix} 0 \\ 0 \\ 1 \\ -2 \end{pmatrix}$$

And finally, for $\lambda=-2i$: $$v_{4}=m\begin{pmatrix} -1 \\ 1 \\ 0 \\ 0 \end{pmatrix}+n\begin{pmatrix} 0 \\ 0 \\ -1 \\ 2 \end{pmatrix}$$

Define $S=\begin{bmatrix} v_{1} & v_{2} & v_{3} & v_{4} \end{bmatrix}$. Then $$S=\begin{bmatrix} a & -c & j & -m \\ -a & c & -j & m \\ b & -d & k & -n \\ -4b & 4d & -2k & 2n \end{bmatrix}$$

But I have to calculate $S^{-1}$ via this method (which seems a bit too complicated), since $e^{At}=Se^{S^{-1}ASt}S^{-1}$.

Then I tried to compute $e^{At}$ using the theory of autonomous behaviours:

So I said that every strong solution of $\frac{d}{dt}x=Ax$ is of the form $$x(t)=B_{10}e^{-it}+B_{20}e^{it}+B_{30}e^{-2it}+B_{40}e^{2it}$$

The vectors $B_{ij}$ should satisfy the relations $$(-i\cdot I-A)B_{10}=0$$ $$(i\cdot I-A)B_{20}=0$$ $$(-2i\cdot I-A)B_{30}=0$$ $$(2i\cdot I-A)B_{40}=0$$

Solving these equations yield $B_{10}=v_{1}$, $B_{20}=v_{2}$, $B_{30}=v_{3}$, $B_{40}=v_{4}$, as calculated previously.

Hence every solution of $x$ can be written as $$x(t)=a\begin{bmatrix}-1 \\ 1 \\ 0 \\ 0\end{bmatrix}e^{-it}+b\begin{bmatrix}0 \\ 0 \\ -1 \\ 4\end{bmatrix}e^{-it}+c\begin{bmatrix}-1 \\ 1 \\ 0 \\ 0 \end{bmatrix}e^{it}+d\begin{bmatrix} 0 \\ 0 \\ 1 \\ 4 \end{bmatrix}e^{it} +j\begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}e^{-2it}+k\begin{bmatrix} 0 \\ 0 \\ -1 \\ 2 \end{bmatrix}e^{-2it}+m\begin{bmatrix} 1 \\ -1 \\ 0 \\ 0 \end{bmatrix}e^{2it}+n\begin{bmatrix}0 \\ 0 \\ 1 \\ -2 \end{bmatrix}e^{2it}$$

From this we can obtain four linearly independent solutions:

$$x_{1}(t)=\begin{bmatrix}-e^{it} \\ e^{it} \\ 0 \\ 0\end{bmatrix}\qquad x_{2}(t)=\begin{bmatrix}e^{2it} \\ -e^{2it} \\ 0 \\ 0\end{bmatrix}\qquad x_{3}(t)=\begin{bmatrix}0 \\ 0 \\ e^{it} \\ 4e^{it}\end{bmatrix} \qquad x_{4}(t)=\begin{bmatrix} 0 \\ 0 \\ e^{2it} \\ -2e^{2it}\end{bmatrix}$$

The matrix $X$ is defined as $X=\begin{bmatrix}x_{1} & x_{2} & x_{3} & x_{4}\end{bmatrix}$

Then $e^{At}=\Phi(t):=X(t)X^{-1}(0)=\text{something really complicated, so I think I went wrong somewhere}$

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  • $\begingroup$ A general remark: $A$ is block-diagonal, so any $f(A)$ is easiest to build block-wise. It is going to keep the same block structure as $A$. $\endgroup$ – A.Γ. Oct 18 '15 at 18:04
  • $\begingroup$ I think your Jordan normal form is much more complicated than it needs to be. With MATLAB I get $S=\begin{bmatrix} 0 & 0 & i & -i \\ 0 & 0 & 1 & 1 \\ i/2 & -i/2 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{bmatrix}$ with $J$ being the diagonal matrix with entries $(-2i,2i,-i,i)$. It would probably be convenient to reorder the eigenvalues as $(-i,i,-2i,2i)$ or similar, so that $S$ has the corresponding block structure to $A$, as A.G. suggested. $\endgroup$ – Ian Oct 18 '15 at 18:07
  • $\begingroup$ One mistake I see now: eigenvector for $\lambda=i$, $4z=-w$ is wrong, it does not solve the system. Must be $z=w=0$. BTW $x=-y$ is wrong too. $\endgroup$ – A.Γ. Oct 18 '15 at 18:21
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Hint The matrix $A$ is block diagonal, namely, $$A = A_1 \oplus A_2,$$ where $$A_1 := \pmatrix{0&1\\-1&0}, \qquad A_2 := \pmatrix{0&1\\-4&0},$$ and it's straightforward to show for square matrices $B_i$ (and in particular our matrices $A_i t$) that $$\exp(B_1 \oplus B_2) = \exp B_1 \oplus \exp B_2 .$$

A computational aside: To compute $\exp (A_1 t)$, we can proceed by finding the Jordan normal form and diagonalizing as in the question statement, but we can also use the characterization that $\exp (A_1 t)$ is the unique $1$-parameter family $C(t)$ of matrices that satisfies the homogeneous, linear, first-order IVP $$\frac{d}{dt} C(t) = A_1 C(t), \qquad C(0) = I .$$ In components $c_{ij}(t)$, this is just $$\left\{ \begin{array}{rclcrcl} c_{11}'(t) &=& \phantom{-}c_{21}(t), &\quad& c_{11}(0) &=& 1\\ c_{12}'(t) &=& \phantom{-}c_{22}(t), &\quad& c_{12}(0) &=& 0\\ c_{21}'(t) &=& -c_{11}(t), &\quad& c_{21}(0) &=& 0\\ c_{22}'(t) &=& -c_{12}(t), &\quad& c_{22}(0) &=& 1\\ \end{array} \right. .$$ Combining the first and third equations gives the familiar ODE $c_{11}''(t) = - c_{11}(t)$, which has general solution $c_1 \cos t + c_2 \sin t$, and substituting the initial conditions gives $c_{11}(t) = \cos t$. Proceeding similarly with the other entries gives that $$\exp A_1 t = \pmatrix{\cos t & \sin t\\ -\sin t & \cos t} .$$

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One problem for sure is how you treat complex linear systems. Look at your solution: $$ \begin{cases} -ix+y=0 \\ -x-iy=0 \\ iz+w=0 \\ -4z-iw=0 \end{cases}\implies\begin{cases} x=-y \\ 4z=-w \end{cases} $$ If I take $x=-y=1$, for example, I will get the first equation $$ -i-1=0. $$ It is clearly false. The solution is $x=-iy$, $w=z=0$, so an eigenvector is, for example, $$ \left[\matrix{-i\\1\\0\\0}\right]. $$

P.S. Note that all eigenvalues are distinct, so you cannot get more than one linear independent eigenvector for each.

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