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I'm trying to learn the process to discover $e^A$

For example, if $A$ is diagonalizable is easy:

$$A =\begin{pmatrix} 5 & -6 \\ 3 & -4 \\ \end{pmatrix}$$

Then we have the canonical form $$J_A =\begin{pmatrix} 2 & 0 \\ 0 & -1 \\ \end{pmatrix}$$

because the auto-values are $2$ and $-1$.

Am I right? so I continue

$$e^A =P\begin{pmatrix} e^2 & 0 \\ 0 & e^{-1} \\ \end{pmatrix}P^{-1}$$

Where $$P=\begin{pmatrix} 2 & 1 \\ 1 & 1 \\ \end{pmatrix}$$

Because the auto-vectors are (2,1) and (1,1).

If the auto-values the things become more complicated:

For example:

$$B =\begin{pmatrix} 0 & 1 \\ -2 & -2 \\ \end{pmatrix}$$

The auto-values are $-1+i$ and $-1-i$, then the canonical form is:

$$J_B =\begin{pmatrix} -1 & -1 \\ 1 & -1 \\ \end{pmatrix}$$

I don't know how to discover $e^B$ in the complex case. How do I have to proceed in this case?

I would like to know also if there are some pdfs or books with examples or problems with solutions about this subject.

I really need help

Thanks a lot.

EDIT

I found another example of a matrix whose some auto-values are complexes:

$$ C=\begin{pmatrix} 1 & 0 & -2 \\ -5 & 6 & 11 \\ 5 & -5 & -10 \\ \end{pmatrix} $$

Its canonical form

$$ J_C=\begin{pmatrix} 1 & 0 & 0 \\ 0 & -2 & 1 \\ 0 & -1 & -2 \\ \end{pmatrix} $$

Why $J_A$ is the matrix $A$ in the canonical form? the author doesn't use complexes numbers, why? How do I find $e^A$ in this case? in the same way?

EDIT 2

The book I'm using says that the complex Jordan block related to the auto-value $a+bi$ is

$$ \begin{pmatrix} a & b \\ -b & a \\ \end{pmatrix} $$

The book also says that it's using the real Jordan canonical form in contrast with the complex Jordan canonical form (see answer below).

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1  
The existence of some complex-valued eigenvalues does not alter the method of calculation of the exponential. –  sos440 Jun 15 '13 at 3:19
    
To compute exponential, it is good if you can diagonalize. However, it is not the only way. You can also try to make your matrix conjugate to an upper triangular matrix, which you can achieve in most cases. The exponential of an upper triangular matrix is not hard to compute, by using the power series. –  Secret Math Jun 15 '13 at 3:22
    
Alternatively, if you don't care about the matrix exponential itself and you really just want solutions to $x' = Ax$ then it's easy to rip solutions from $e^{tA}$ via the identity $e^{tA} = e^{\lambda t}(I +t(A-\lambda I)+\frac{t^2}{2!}(A-\lambda I)^2 + \cdots )$ by multiplying by eigenvectors and generalized eigenvectors which truncate the series. In the event of a complex vector you get a pair of real solutions one the real part the other the imaginary part. It's all nice an neat. –  James S. Cook Jun 15 '13 at 3:59
    
@user42912: You are very welcome! Look at this example: math.stackexchange.com/questions/207397/…. Regards –  Amzoti Jun 15 '13 at 7:15
    
The $\pmatrix{a&-b\\ b&a}$ in your second edit is called a real Jordan form. "Complex Jordan form" is what we usually called a Jordan form, which is a triangular bidiagonal matrix. It is a canonical form of a real or complex matrix. When the field is $\mathbb{R}$, a real matrix may or may not possess a Jordan form, depending on whether all its characteristic polynomial splits over $\mathbb{R}$ (i.e. whether its complex eigenvalues all lie on the real line). A real Jordan form is a real, block-bidiagonal matrix that is a canonical form of a real matrix. It always exists. –  user1551 Jun 15 '13 at 8:51

1 Answer 1

up vote 4 down vote accepted

You might want to have a look at Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later

Other references can be found in The matrix exponential: Any good books? We have:

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

We want to find the characteristic polynomial and eigenvalues by solving

$$|A -\lambda I| = 0 \rightarrow \lambda^2+2 \lambda+2 = (\lambda+(1-i)) (\lambda+(1+i))= 0 \rightarrow \lambda_1 = -1+i, ~~\lambda_2 = -1-i $$

If we try and find eigenvectors, we setup and solve: $[A - \lambda I]v_i = 0$.

So, we have:

$$[A - \lambda_1 I]v_1 = 0 = \begin{bmatrix}1-i & 1\\-2 &-1-i\end{bmatrix}v_1 = 0$$

This leads to a RREF of:

$$\begin{bmatrix}1 & \dfrac{1}{2}(1 +1)\\0 & 0\end{bmatrix}v_1 = 0 \rightarrow v_1 = \left(1, -\dfrac{1}{2}(1 + i)\right)$$

Doing the same process for the second eigenvalue yields: $v_2 = \left(1, -\dfrac{1}{2} + \dfrac{i}{2}\right)$

From all of this information, we can write the matrix exponential using the Jordan Normal Form.

We have the diagonal form:

$$A = P \cdot J\cdot P^{-1} = \begin{bmatrix}-1/2+i/2 & -1/2-i/2 \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}-1-i & 0 \\ 0 & -1+i\end{bmatrix} \cdot \begin{bmatrix}-i & 1/2-i/2 \\ i & 1/2+i/2 \end{bmatrix}$$

So, now we can take advantage of the diagonal form as:

$\displaystyle e^A = P \cdot e^J \cdot P^{-1} = \begin{bmatrix}-1/2+i/2 & -1/2-i/2 \\ 1 & 1\end{bmatrix} \cdot e^{\begin{bmatrix}-1-i & 0 \\ 0 & -1+i\end{bmatrix}} \cdot \begin{bmatrix}-i & 1/2-i/2 \\ i & 1/2+i/2 \end{bmatrix} = \begin{bmatrix}-1/2+i/2 & -1/2-i/2 \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}e^{-1-i} & 0 \\ 0 & e^{-1+i}\end{bmatrix} \cdot \begin{bmatrix}-i & 1/2-i/2 \\ i & 1/2+i/2 \end{bmatrix} = \begin{bmatrix} (1/2+i/2) e^{-1-i}+(1/2-i/2) e^{-1+i} & (1/2) i e^{-1-i}-(1/2) i e^{-1+i} \\-i e^{-1-i}+i e^{-1+i} & (1/2-i/2) e^{-1-i}+(1/2+i/2) e^{-1+i} \end{bmatrix}$

Other methods are shown in that paper I referenced above.

If we compare this for the real case, the process is the same and we end up with:

$\displaystyle e^A = P \cdot e^J \cdot P^{-1} = \begin{bmatrix}1 & 2\\1 & 1\end{bmatrix} \cdot e^{\begin{bmatrix}-1 & 0\\0 & 2\end{bmatrix}} \cdot \begin{bmatrix}-1 & 2\\1 & -1\end{bmatrix}= \begin{bmatrix}1 & 2\\1 & 1\end{bmatrix} \cdot {\begin{bmatrix}e^{-1} & 0\\0 & e^2\end{bmatrix}} \cdot \begin{bmatrix}-1 & 2\\1 & -1\end{bmatrix} = \begin{bmatrix}\dfrac{-1+2 e^3}{e} & -\dfrac{2 (-1+e^3)}{e}\\ \dfrac{-1+e^3}{e} & \dfrac{2-e^3}{e}\end{bmatrix}$

It is also worth noting that things can also get very ugly, for example, see: generalized eigenvector for 3x3 matrix with 1 eigenvalue, 2 eigenvectors

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In the real case, if the matrix is diagonalizable we put $e^a_1$ and $e^a_2$ ($a_1$ and $a_2$ are the auto-values) in the diagonal. If I understood well, the complex case is not so simple. –  user42912 Jun 15 '13 at 4:19
    
+1 Amzoti! Well done. –  amWhy Jun 15 '13 at 4:28
    
Yikes...I just clicked to see all the formatting it took! All that work! –  amWhy Jun 15 '13 at 4:32
    
thank you very much, I'm doing this right now –  user42912 Jun 15 '13 at 4:44
    
I did the real case and the complex case, but I think I have to do this in $\mathbb R$, please see the edit I made, thank you very much again. –  user42912 Jun 15 '13 at 6:23

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