current community

your communities

more stack exchange communities

company blog
Stack Exchange Inbox Reputation and Badges
tour help
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
up vote 9 down vote favorite
4

Consider the $3 \times 3$ matrix
$$A = \begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & -4 \\ 0 & 0 & 1 \end{pmatrix}.$$

I am trying to find $e^{At}$.

The only tool I have to find the exponential of a matrix is to diagonalize it. $A$'s eigenvalue is 1. Therefore, $A$ is not diagonalizable.

How does one find the exponential of a non-diagonalizable matrix?

My attempt:

Write $\begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & -4 \\ 0 & 0 & 1 \end{pmatrix} = M + N$, with $M = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ and $N = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 0 & -4 \\ 0 & 0 & 0 \end{pmatrix}$.

We have $N^3 = 0$, and therefore $\forall x > 3$, $N^x = 0$. Thus:

$$\begin{aligned} e^{At} &= e^{(M+N)t} = e^{Mt} e^{Nt} \\ &= \begin{pmatrix} e^t & 0 & 0 \\ 0 & e^t & 0 \\ 0 & 0 & e^t \end{pmatrix} \left(I + \begin{pmatrix} 0 & t & 2t \\ 0 & 0 & -4t \\ 0 & 0 & 0 \end{pmatrix}+\begin{pmatrix} 0 & 0 & -2t^2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\right) \\ &= e^t \begin{pmatrix} 1 & t & 2t \\ 0 & 1 & -4t \\ 0 & 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix} e^t & te^t & 2t(1-t)e^t \\ 0 & e^t & -4te^t \\ 0 & 0 & e^t \end{pmatrix}. \end{aligned}$$

Is that the right answer?

share|cite|improve this question
1  
If $A$ is what you say it is then the only eigenvalue is $\lambda=1$. That matrix is not diagonalizable. – David C. Ullrich 19 hours ago
    
can't eigenvalues be complex? – aribaldi 19 hours ago
2  
If you know about the Jordan Canonical Form (aka the Jordan Normal Form) you can use that to exponentiate a non-diagonalizable matrix. – David C. Ullrich 18 hours ago
1  
You may too prove by induction that $A^n$ is $$\begin{pmatrix} 1&n&4n-2n^2\\ 0&1&-4n\\ 0&0&1\ \end{pmatrix}$$ – Raymond Manzoni 17 hours ago
2  
An alternative way is to change one of the zeros to $\epsilon$ to obtain a diagonalizable matrix and then take the limit of $\epsilon\to 0$ in the result. This method will always work because the diagonalizable matrices are dense in the set of all matrices. – Count Iblis 13 hours ago
up vote 13 down vote accepted

this is my first answer on this site so if anyone can help to improve the quality of this answer, thanks in advance.

That said, let us get to business.

  1. Compute the Jordan form of this matrix, you can do it by hand or check this link. (or both).
  2. Now, we have the following case: $$ A = S J S^{-1}.$$ You will find $S$ and $S^{-1}$ on the previous link. For the sake of simplicity, $J$ is what actually matters, $$ J = \begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{pmatrix} $$ because:
  3. $e^A = e^{SJS^{-1}} = e^J$ And the matrix $J$ can be written as: $J = \lambda I + N$, where $I$ is the identity matrix and $N$ a nilpotent matrix.
  4. So, $e^J = e^{\lambda I + N} = \mathbf{e^{\lambda} \cdot e^N}$ By simple inspection, we get that: $$ J = \lambda I + N = 1 \cdot \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix} + \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ \end{pmatrix} $$ where you can check that $\lambda =1$ and N is $$ \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ \end{pmatrix} $$
  5. So, $e^A = e \cdot e^N$ we just apply the definition $ e^N \equiv \sum^{\infty}_{k=0} \frac{1}{k!} N^k$. And, of course, it converges fast: $N^2 \neq 0$ but $N^3=0$.
  6. Finally: $$ e^A = e \cdot \left[ 1 \cdot I + 1 \cdot N^1 + \frac{1}{2} N^2 \right] $$ where $$ N^2 = \begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{pmatrix} $$ then: $$ \mathbf{ e^A = e \cdot \begin{pmatrix} 1 & 1 & 1/2\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{pmatrix} } $$ Last but not least, $$ e^{At} = e^{A \cdot t} = e^{\lambda \cdot t} \cdot e^{N \cdot t} = e^t \cdot \begin{pmatrix} 1 & t & 1/2 t^2\\ 0 & 1 & t\\ 0 & 0 & 1\\ \end{pmatrix} $$ You replace $N$ by $At$ in the exp definition and that's it.
share|cite|improve this answer
2  
Good job for a first post. Click the "edited" text above Byron's name to see how he edited it to make it better. – GEdgar 16 hours ago
    
@Cesar You have correctly calculated $e^J,$ which unfortunately is not the same as $e^A.$ – Byron Schmuland 10 hours ago
up vote 4 down vote

Hint:

Write your matrix $A$ as $I+N$ where $I$ is the identity matrix and $N$ is a nilpotent matrix. Then use the definition of $e^{At}$ as a power series, noting that $N^k=0$ for some $k$.

share|cite|improve this answer
4  
(+1) - but probably (maybe?) one should add to the hint : $e^{(I +N)t}= e^{It}e^{Nt}$ because the matrices $I$ and $N$ commute... – peter a g 18 hours ago
    
@peterag: sure, but this fact is embedded in the idea of "use the definition of $e^{At}$ as a power series." writing down the expression, it is obvious what needs to be done. – symplectomorphic 17 hours ago
1  
To make my previous (useful or not) comment more useful (?) for the OP: if $AB=BA$, then $e^{A+B}=e^Ae^B$. – peter a g 17 hours ago
up vote 3 down vote

If you know about the Jordan Canonical Form you can use that.

Another method, probably more elementary, was mentioned in a comment. The comment was deleted; I don't know why. Note that $A=I+N$, where $N^3=0$. It follows that $$A^k=I+kN+\frac{k(k-1)}{2}N^2.$$You can use that to calculate $e^{At}=\sum t^kA^k/k!$.

Edit: Oh, that comment was converted to an answer. I'll leave this here anyway, being more detailed (at least regarding one approach).

share|cite|improve this answer
    
I added the comment as an answer. That's why. :) – symplectomorphic 17 hours ago
    
If you mean that $N^3$ is the zero matrix I think you are mistaken. $N^2 = 0$ – aribaldi 17 hours ago
2  
@aribaldi Huh? First, if $N^2=0$ as you claim that would show that I'm not mistaken, since $N^2=0$ implies $N^3=0$. Second, no, $N^2\ne0$. – David C. Ullrich 17 hours ago
    
If $I+N = A \implies N = \begin{pmatrix} 0 & 1 & 2\\ 0 & 0 & -4\\ 0 & 0 & 0\\ \end{pmatrix}$ right? And $N^2 = 0$ – aribaldi 17 hours ago
    
@DavidC.Ullrich No it isn't, you were right the first time. There is a single entry "-4" in the matrix $N^2$. – Byron Schmuland 16 hours ago
up vote 0 down vote

Hint: Find the Jordan matrix, by which the exponential can always be found.

share|cite|improve this answer
up vote 0 down vote

It's actually possible (though somewhat tedious) to calculate $\exp(tA)$ using the series definition. The definition states that, $$\exp(tA) = \sum_{n=0}^\infty \frac{t^n}{n!}A^n.$$

Now, since $\exp(tA)$ is a matrix, it is enough to find the values of $\exp(tA)e_i$ for $i=1,2,3$, where the $e_i$ are the standard basis vectors.
For $e_1$, we have $$\exp(tA)e_1 = \sum_{n=0}^\infty \frac{t^n}{n!}A^ne_1 = \sum_{n=0}^\infty \frac{t^n}{n!} e_1 = \exp(t)e_1$$ so the first column of $\exp(tA)$ is $$\left[ \exp(t)\quad 0\quad 0\right]^T$$

For $e_2$, we have that $Ae_2 = e_1 + e_2$, so $$A^n e_2 = A^{n-1}e_1 + a^{n-1}e_2 = e_1 + A^{n-1}e_n = \cdots = ne_1 + e_2$$ and consequently, $$\exp(tA)e_2 = \sum_{n=0}^\infty \frac{t^n}{n!}A^ne_2 = \sum_{n=0}^\infty \frac{t^n}{n!} ne_1 + \sum_{n=0}^\infty \frac{t^n}{n!} ne_2 = t\sum_{n=1}^\infty \frac{t^n}{(n-1)!} e_1 + \exp(t)e_2 = t\exp(t)e_1 + \exp(t)e_2$$ and the second row of the matrix is $$\left[t\exp(t)\quad \exp(t)\quad 0\right]^T$$

Finally, $$A e_3 = 2e_1 + -4e_2 + e_3$$ so $$A^n e_3 = 2A^{n-1}e_1 - 4A^{n-1}e_2 + A^{n-1}e_3 = 2e_1 - 4((n-1)e_1 + e_2) + A^{n-1} e_3=\cdots$$ $$ \cdots = 2n e_1 - 4\left(\frac{n(n-1)}{2}e_1 + ne_2\right) + e_3 = -(2n^2+4n) e_1 - 4ne_2 + e_3.$$ Thus, $$\exp(tA)e_3 = \sum_{n=0}^\infty \frac{t^n}{n!}A^ne_1 = \sum_{n=0}^\infty \frac{t^n}{n!}\left(-(2n^2-4n) e_1 - 4ne_2 + e_3\right)$$ $$ = -2\sum_{n=0}^\infty \frac{t^n}{n!}n(n-2)e_1 - 4\sum_{n=0}^\infty \frac{t^n}{n!}ne_2 + \sum_{n=0}^\infty \frac{t^n}{n!} e_3$$ $$= -2t\sum_{n=1}^\infty \frac{t^n}{(n-1)!}(n-2)e_1 -4 t\exp(t)e_2 + \exp(t)e_3$$ $$= -2t\sum_{n=0}^\infty \frac{t^n}{n!}(n-1)e_1 -4 t\exp(t)e_2 + \exp(t)e_3 $$ $$ = -2t\sum_{n=0}^\infty \frac{t^n}{n!}ne_1 + 2t\sum_{n=0}^\infty \frac{t^n}{n!}e_3 -4 t\exp(t)e_2 + \exp(t)e_3$$ $$= -2t \sum_{n=1}^\infty \frac{t^n}{(n-1)!} e_1 + 2t\exp(t)e_1 - 4t\exp(t)e_2 + \exp(t)e_3$$ $$= -2t^2 \sum_{n=0}^\infty \frac{t^n}{n!} e_1 + 2t\exp(t)e_1 - 4t\exp(t)e_2 + \exp(t)e_3$$ $$= -2t(t-1)\exp(t)e_1 -4t\exp(t)e_2 + \exp(t)e_3$$ so the final column of $\exp(tA)$ is $$\left[-2t(t-1) \exp(t) \quad -4t\exp(t) \quad \exp(t)\right]^T$$ Thus, $$\exp(tA) = \exp(t)\begin{bmatrix} 1 & t & -2t(t-1)\\ 0 & 1 & -4t\\ 0 & 0 & 1\\ \end{bmatrix}$$

share|cite|improve this answer
up vote 0 down vote

Using Jordan Decomposition

Since Jordan Decomposition gives $$ \begin{bmatrix} 1&1&2\\ 0&1&-4\\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 1&0&0\\ 0&1&\frac12\\ 0&0&-\frac14 \end{bmatrix} \begin{bmatrix} 1&1&0\\ 0&1&1\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1&0&0\\ 0&1&\frac12\\ 0&0&-\frac14 \end{bmatrix}^{\large\,-1}\tag{1} $$ we have $$ \exp\left(\begin{bmatrix} 1&1&2\\ 0&1&-4\\ 0&0&1 \end{bmatrix}\right) = \begin{bmatrix} 1&0&0\\ 0&1&\frac12\\ 0&0&-\frac14 \end{bmatrix} \exp\left(\begin{bmatrix} 1&1&0\\ 0&1&1\\ 0&0&1 \end{bmatrix}\right) \begin{bmatrix} 1&0&0\\ 0&1&\frac12\\ 0&0&-\frac14 \end{bmatrix}^{\large\,-1}\tag{2} $$ Since $$ \exp\left(\begin{bmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{bmatrix}\right) =\begin{bmatrix} 1&1&\frac12\\ 0&1&1\\ 0&0&1 \end{bmatrix}\tag{3} $$ we get $$ \exp\left(\begin{bmatrix} 1&1&0\\ 0&1&1\\ 0&0&1 \end{bmatrix}\right) =e\begin{bmatrix} 1&1&\frac12\\ 0&1&1\\ 0&0&1 \end{bmatrix}\tag{4} $$ and $$ \begin{align} \exp\left(\begin{bmatrix} 1&1&2\\ 0&1&-4\\ 0&0&1 \end{bmatrix}\right) &= e\begin{bmatrix} 1&0&0\\ 0&1&\frac12\\ 0&0&-\frac14 \end{bmatrix} \begin{bmatrix} 1&1&\frac12\\ 0&1&1\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1&0&0\\ 0&1&\frac12\\ 0&0&-\frac14 \end{bmatrix}^{\large\,-1}\\[6pt] &=e\begin{bmatrix} 1&1&0\\ 0&1&-4\\ 0&0&1 \end{bmatrix}\tag{5} \end{align} $$

Using Power Series for $\boldsymbol{\exp}$

Since $$ \begin{bmatrix} 0&1&2\\ 0&0&-4\\ 0&0&0 \end{bmatrix}^{\large\,2} =\begin{bmatrix} 0&0&-4\\ 0&0&0\\ 0&0&0 \end{bmatrix}\tag{6} $$ and $$ \begin{bmatrix} 0&1&2\\ 0&0&-4\\ 0&0&0 \end{bmatrix}^{\large\,3} =\begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&0&0 \end{bmatrix}\tag{7} $$ we can use the power series for $\exp$ to get $$ \exp\left(\begin{bmatrix} 0&1&2\\ 0&0&-4\\ 0&0&0 \end{bmatrix}\right) =\begin{bmatrix} 1&1&0\\ 0&1&-4\\ 0&0&1 \end{bmatrix}\tag{8} $$ Therefore, $$ \exp\left(\begin{bmatrix} 1&1&2\\ 0&1&-4\\ 0&0&1 \end{bmatrix}\right) =e\begin{bmatrix} 1&1&0\\ 0&1&-4\\ 0&0&1 \end{bmatrix}\tag{9} $$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.