# Integrating a matrix

Is it possible to integrate a matrix?

I've been working through a problem and come up with $$\int_{t_0}^t \begin{pmatrix}\sin(s)\cdot\cos(\beta s)\\ \cos(s)\cdot\cos(\beta s)\end{pmatrix}ds$$

I'm integrating from $t_0$ to $t$

Can this be done or do we think I went wrong somewhere?

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Isn't your integrand function simply $\;\tan s\;$ ...? –  DonAntonio Jul 23 '13 at 19:17
Did you mean $$\int \begin{pmatrix} \sin s & \cos (\beta s)\\ \cos s & \cos (\beta s)\end{pmatrix}\, ds$$ –  Daniel Fischer Jul 23 '13 at 19:17
Ah, okay, doesn't matter much. Anyway, just integrate componentwise. –  Daniel Fischer Jul 23 '13 at 19:19
It's a 2x1 matrix. I'd be greftful if you could do the edit for me! –  Steve Jul 23 '13 at 19:19
After the editing I see a $\,2\times 2\;$ matrix... –  DonAntonio Jul 23 '13 at 19:22

Matrices form a vector space. Therefore, you can simply integrate them componentwise.

In detail. Let $A:t\mapsto A(t)$ be a function from a real interval $I$ to the space of $m\times n$ real matrices. Every entry $a_{ij}$ is a real function of a real variable. If all entries are integrable functions, then you can define the integral of the matrix as the matrix of the integrals:

$$\int A(t)\,dt := \left( \int a_{ij}(t)\,dt \right).$$

$$\int \begin{pmatrix}\sin(s)&\cos(\beta s)\\ \cos(s)&\cos(\beta s)\end{pmatrix}ds = \begin{pmatrix}\int\sin(s)\,ds &\int\cos(\beta s)\,ds\\ \int\cos(s)\,ds&\int\cos(\beta s)\,ds\end{pmatrix} = \begin{pmatrix}-\cos(s)&\frac{1}{\beta}\sin(\beta s)\\ \sin(s)&\frac{1}{\beta}\sin(\beta s)\end{pmatrix}.$$

There is a more sophisticated operation, in case the matrix in question belongs to a Lie algebra: ordered exponentiation. It is to integration as exponentiation is to multiplication, and permits to go from a Lie algebra element (intuitively, a differential transformation) to a group element (a whole transformation). In this case, you need a $n\times n$ matrix-valued function.

It is explained quite well here: http://en.wikipedia.org/wiki/Ordered_exponential.

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Matrices...*of the same order*...form a vector space. Why this makes that "therefore" appear there is beyond my comprehension, though. –  DonAntonio Jul 23 '13 at 19:32
@DonAntonio: Because addition in that vector space is componentwise (and so is scalar multiplication). –  Ilmari Karonen Jul 23 '13 at 19:46
@geodude. Are you saying that if our matrix belongs to a Lie algebra, then we can define the integral of that matrix using ordered exponentiation? Do you have any further resource links on this subject? I've tried looking online but find either very specialized resources or nothing at all (except for the one link you cite) –  pbs Oct 9 '13 at 20:02
@pbs: Yes, precisely. I have the same problem, there isn't much material to gain insight on that! I used such a concept in this answer, if that can help: math.stackexchange.com/questions/223024/… For something in rigor, look here: dml.cz/bitstream/handle/10338.dmlcz/401130/…. –  geodude Oct 10 '13 at 20:51
I found these related links useful: en.wikipedia.org/wiki/Product_integral and en.wikipedia.org/wiki/Multiplicative_calculus. Fascinating! –  pbs Oct 11 '13 at 19:26