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As far as Im aware the formula for calculating the exponential of a jordan block can only involve one eigenvalue, so how would this work for the case of a matrix having distinct eigenvalues ?

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- Thread starter JamesGoh
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- #1

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As far as Im aware the formula for calculating the exponential of a jordan block can only involve one eigenvalue, so how would this work for the case of a matrix having distinct eigenvalues ?

- #2

quasar987

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[tex]e^J=\sum_{k=0}^{\infty}\frac{J^k}{k!}=\sum_{k=0}^{\infty}\frac{\mathrm{diag}(B(\lambda_i)^k)}{k!}= \mathrm{diag} \left( \sum_{k=0}^{\infty}\frac{B( \lambda_i)^k}{k!}\right)=\mathrm{diag}(e^{B( \lambda_i)})[/tex]

More generally,

- #3

HallsofIvy

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[tex]\begin{bmatrix}1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2\end{bmatrix}[/tex]

is a Jordan matrix for a matrix having eigenvalues 1 and 2 (each with geometric multiplicity 2).

You can think of it as made of the two "elementary" Jordan matrices

[tex]\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}[/tex]

and

[tex]\begin{bmatrix} 2 & 1 \\ 0 & 2\end{bmatrix}[/tex]

The exponential of the first is, of course,

[tex]\begin{bmatrix}e & e \\ 0 & e\end{bmatrix}[/tex]

and of the second

[tex]\begin{bmatrix}e^2 & e^2 \\ 0 & e^2\end{bmatrix}[/tex]

so it should be no surprise that the exponential of the original matrix is

[tex]\begin{bmatrix}e & e & 0 & 0 \\ 0 & e & 0 & 0 \\ 0 & 0 & e^2 & e^2 \\ 0 & 0 & 0 & e^2\end{bmatrix}[/tex]

- #4

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How did you get the top right terms of both the individual Jordan matrices to be exponents ?

- #5

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How did you get the top right terms of both the individual Jordan matrices to be exponents ?

Not at all, finding the exponential of a Jordan canonical form is easy.

Indeed, in the Jordan canonical form, we are dealing with matrices of the form D+N, where D is diagonal and where N is nilpotent. If we have that DN=ND, then we have that

[tex]e^{D+N}=e^De^N[/tex]

The exponential of D can be easily find since it is diagonal. The exponential of N is also easy. Indeed, we know that [itex]N^k=0[/itex] for a certain k, so

[tex]e^N=I+N+\frac{N^2}{2}+...+\frac{N^{k-1}}{(k-1)!}[/tex]

which is a finite sum!!

In the case [tex]A=\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}+\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}=D+N[/tex]

We see that

[tex]e^D=\begin{pmatrix} e & 0\\ 0 & e \end{pmatrix}[/tex]

Furthermore, we know that [itex]N^2=0[/itex], so

[tex]e^N=I+N=\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}[/tex]

We get that

[tex]e^A=e^De^N=\begin{pmatrix} e & 0\\ 0 & e \end{pmatrix}\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}=\begin{pmatrix} e & e\\ 0 & e \end{pmatrix}[/tex]

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