Integrate Matrix: Can Matrices Have Antiderivatives?

[daudaudaudau]

Hi. Is it possible to calculate the antiderivative of
$$\frac{1}{Ax+B}$$

if A and B are matrices? If they were scalars the result would be $1/A \log(Ax+B)$.

 

[coelho]

Well... i may be talking complete nonsense here, but... why not? After one of my professors blew my mind by saying that it was possible to calculate exponentials, sines, etc, of matrixes, i believe anything is possible in math...

His calculations were like this:

$$e^x=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots$$

So, if M is a matrix and I is the identity matrix:

$$e^M=I+M+\frac{1}{2!}(M.M)+\frac{1}{3!}(M.M.M)+\cdots$$

 

[henry_m]

Nearly! As coelho says, a lot of things carry over from real and complex arithmetic and calculus to apply to matrices, BUT you have to be extremely careful.

One reason for care is that matrices do not necessarily commute with one another, so it is important what order you write things down in. For example, if we differentiate 1/f(x) for f a real or complex function of x, we get -f'(x)/(f(x)²), and the order of things is irrelevant. But if f is a matrix-valued function, only one order will do: f'(x)= -f(x)^-1 f'(x) f(x)^-1 (You can check this by differentiating f(x)f(x)^-1=I). Similar things apply for differentiating things like exponentials, for which you get one of my favourite formulae; very nontrivial and yet so easy to remember:

$$\frac{\mathrm{d}}{\mathrm{d}x}\exp(f(x)) = \int_0^1 \exp(t f(x)) f'(x) \exp((1-t) f(x)) \mathrm{d}t$$

The other problem you have to contend with is multivaluedness. log, as definied as an inverse to exp, is already multivalued for real and complex arguments but the problem gets worse for matrices. The wiki article on matrix logarirthm is pretty good so look there for details.

Now I've got these caveats out the way, I'll give you an answer. I only did a quick calculation where I assumed A^-1B exists and is diagonal, but with more care I'm sure it can be proven in more generality. I got an antiderivative as log(x+A^-1B) A^-1=A^-1log(x+BA^-1). This is certainly a form we recognise from ordinary real/complex numbers. Bear in mind that stuff like log(ab)=log(a)+log(b) doesn't necessarily carry through into the complex case.

 

[daudaudaudau]

That's very interesting. Thank you both.

 

[HallsofIvy]

Yes, it is possible to calculate $e^x$, $ln(x)$, $sin(x)$, etc. for x any object that you can add and multiply.

Just use the Taylor's series for the function.