# Finding the matrix $J = \operatorname{diag}(J_1(\lambda_1),J_2(\lambda_2),J_3(\lambda_3))$

EDITED

When I do my homework (stability theory), I must use the knowledge to the matrix. But I don't remember it. :( How can we find the matrix $J$, $e^{tJ}$? Thanks!

I have an example $$A=\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix}$$

We have $\lambda_1=0$, $\lambda_1=1$, $\lambda_1=2$.

We suppose that $A=PJP^{-1}$, where $$J = \operatorname{diag}(J_1(\lambda_1),J_2(\lambda_2),J_3(\lambda_3))$$

• $J_n(\lambda)$ is a Jordan matrix of order $n$ with the eigenvalue $\lambda$. Does that help? – Vedran Šego Oct 18 '13 at 10:58
• I'm sorry Vedran Šego! But as I said I don't remember it :( . Can you help me solve an example? Plz. – kimtahe6 Oct 18 '13 at 11:02
• That's why I gave you a link. This is just a diagonal matrix with the diagonal elements $0,1,2$, because your Jordan blocks are all of order $1$. – Vedran Šego Oct 18 '13 at 11:04

This answer does not really address the OP's question. I just want to illustrate that sometimes, when the matrix has a nice structure, we can calculate its matrix exponential without calculating a Jordan form. Obviously we are not always so lucky. That's why a general method using Jordan form is worth learning.

Note that in this particular example, we have $A=I+X$ with $X^3=X$. Since $I$ commutes with $X$, \begin{align*} \exp(A) &=\exp(I)\exp(X)\\ &=(eI)\left(I+X+\frac{X^2}{2!}+\frac{X^3}{3!}+\frac{X^4}{4!}+\frac{X^5}{5!}+\ldots\right)\\ &=e\left(I+X+\frac{X^2}{2!}+\frac{X}{3!}+\frac{X^2}{4!}+\frac{X}{5!}+\ldots\right)\\ &=e\left(I + \sinh(1)X + (\cosh(1)-1)X^2\right)\\ &=e\pmatrix{\cosh(1)&0&\sinh(1)\\ \cosh(1)-1&1&\sinh(1)\\ \sinh(1)&0&\cosh(1)}\\ &=\frac12\pmatrix{e^2+1&0&e^2-1\\ (e-1)^2&2e&e^2-1\\ e^2-1&0&e^2+1}. \end{align*}

• Thanks Vedran Šego and user1551! I accepted your ANS. Haha, it's really simple, when I read en.wikipedia.org/wiki/Matrix_exponential#cite_note-5 . 1/ Since $$A=PJP^{-1} \implies J=P^{-1}AP$$ 2/ We can express $J$ as a sum: $$J=\lambda I + N$$. It's simple when we evaluate the matrix $e^{\lambda I}$. Next, we compute the matrix $$e^{N}=I+N+\dfrac{N^2}{2}+\ldots$$. Thank you very much. – kimtahe6 Oct 19 '13 at 6:00

Given the OP's confusion in the comments, here is the full answer.

$J_n(\lambda)$ is a Jordan block of order $n$, with the eigenvalue $\lambda$. Generally, this means the square matrix with $\lambda$ on the diagonal, $1$ on the superdiagonal, and zero everywhere else:

$$J_n(\lambda) = \begin{bmatrix} \lambda & 1 \\ & \lambda & 1 \\ & & \ddots & \ddots \\ & & & \lambda & 1 \\ & & & & \lambda \end{bmatrix}.$$

You get $\operatorname{diag}(J_{n_1}(\lambda_1), \dots, J_{n_k}(\lambda_k))$ by simply combining this blocks, each right down from the previous one, thus obtaining a block diagonal matrix with blocks $J_{n_1}(\lambda_1), \dots, J_{n_k}(\lambda_k)$ on its diagonal.

But, since all your blocks are of order $1$, you simply get

$$J_1(0) = \begin{bmatrix} 0 \end{bmatrix}, \quad J_1(1) = \begin{bmatrix} 1 \end{bmatrix}, \quad J_1(2) = \begin{bmatrix} 2 \end{bmatrix}.$$

Making a diagonal matrix with this blocks results in an ordinary diagonal matrix:

$$\operatorname{diag}(J_1(0), J_1(1), J_1(2)) = \operatorname{diag}(0, 1, 2) = \begin{bmatrix} 0 \\ & 1 \\ & & 2 \end{bmatrix}.$$

For further details, please read the link I've supplied in the comments. The decomposition $A = S J S^{-1}$ is called the Jordan decomposition, and you can read about it here.

• Thanks Vedran Šego. How to find the matrix $e^{tJ}$ if $\lambda_1 \ne \lambda_2 \ne \lambda_3$? – kimtahe6 Oct 18 '13 at 11:37
• Any function of a diagonal matrix can be done elementwise, so $e^{tJ} = \operatorname{diag}(e^{\lambda_1 t}, e^{\lambda_2 t}, e^{\lambda_3 t})$, as long as all your Jordan blocks are of order $1$. – Vedran Šego Oct 18 '13 at 12:28