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Find the matrix exponential $e^A$ for $$ A = \begin{bmatrix} 2 & 1 & 1\\ 0 & 2 & 1\\ 0 & 0 & 2\\ \end{bmatrix}.$$

I think we should use the proberty

If $AB = BA$ then $e^{A+B} = e^A e^B$.

We can use that

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

Both matrices obviously commute. But I dont know how to calculate the exponential of

$$\begin{bmatrix} 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{bmatrix}.$$

Could you help me?

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You should decompose your matrix like this $$\begin{equation} \begin{pmatrix} 2 & 1 & 1 \\ 0 & 2 & 1\\ 0 & 0 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{pmatrix} + \begin{pmatrix} 0 & 1 & 1\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix} \end{equation} $$ The left one commutes with the right one and the right one is nilpotent. So it is easy to compute.

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    $\begingroup$ It might be useful to note that this is always possible (in $\mathbb{C}$) thanks to the Jordan-Chevalley decomposition : wiki/Jordan–Chevalley_decomposition $\endgroup$ – Caffeine Oct 29 '19 at 10:05
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    $\begingroup$ @GabrieleCassese Good point! I recently find that mastering the calculating of matrices adeptly is important for studying Lie theory and representation theory. In fact, J-C decomposition is significant in Lie algebra theory. $\endgroup$ – XT Chen Oct 29 '19 at 10:25
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We may use Cayley Hamilton theorem to find the exponential of $$ M=\begin{bmatrix} 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{bmatrix}.$$

Note that the eigenvalues of $M$ are $1,1,1$

We write $$e^{tM} = \alpha (t)I+ \beta (t)M+ \gamma (t) M^2.$$

To find the coefficient functions we replace $M$ with $\lambda$ and solve the resulting system.

$$e^{t\lambda} = \alpha (t)I+ \beta (t)\lambda + \gamma (t) \lambda ^2.$$

Differentiating with respect to $\lambda $ we get $$te^{\lambda t}=\beta + 2\lambda \gamma$$ Differentiating again, we get $$t^2 e^{\lambda t }=2\gamma$$

With t=1 and $\lambda =1$ we get $$\alpha = e/2, \beta =0, \gamma = e/2$$

with these values we find $$e^M =e \begin{bmatrix} 1 & 1 & 3/2\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{bmatrix}. $$

Apply the same method with $\lambda =2$ we get $$e^N =e^2 \begin{bmatrix} 1 & 1 & 3/2\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{bmatrix}. $$

for the original matrix $$N=\begin{bmatrix} 2 & 1 & 1\\ 0 & 2 & 1\\ 0 & 0 & 2\\ \end{bmatrix}$$

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Edit: As pointed out by @XTChen there is a much easier way to do this but I detail evaluation of the requested exponential nonetheless.

Note that $$\pmatrix{1&1&1\\0&1&1\\0&0&1}^n=\pmatrix{1&n&n(n+1)/2\\0&1&n\\0&0&1}$$ \begin{align} \sum_{n=0}^\infty & \frac1{n!}\pmatrix{1&1&1\\0&1&1\\0&0&1}^n \\ &=\sum_{n=0}^\infty\frac1{n!}\pmatrix{1&n&n(n+1)/2\\0&1&n\\0&0&1}\\ &=\pmatrix{1&0&0\\0&1&0\\0&0&1}+\sum_{n=1}^\infty\pmatrix{1/n!&1/(n-1)!&(n+1)/(2(n-1)!)\\0&1/n!&1/(n-1)!\\0&0&1/n!}\\ &=\pmatrix{1&0&0\\0&1&0\\0&0&1}+\pmatrix{e-1&e&3e/2\\0&e-1&e\\0&0&e-1}\\ &=\pmatrix{e&e&3e/2\\0&e&e\\0&0&e}\\ &=\frac12e\pmatrix{2&2&3\\0&2&2\\0&0&2}\\ \end{align}

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