Robert Dodier wrote:
> Hi Slava,
>
>> hello, may be i'm not first, but i find bug: example A - matrix,
>> A^0 = matrix([1,1],[1,1]) or A^(-1)*A = matrix([1,1],[1,1]), expected
>> =matrix([1,0],[0,-1]).
>
>
> You want . (i.e., dot) for noncommutative multiplication here
> instead of * which is commutative multiplication.
> Also you want ^^ for exponentiation induced by . .
>
> E.g.
> A : matrix ([4, a], [b, 17]);
> display2d : false;
> A ^^ -1; => matrix([-17/(a*b-68),a/(a*b-68)],[b/(a*b-68),-4/(a*b-68)])
> A . %; => matrix([a*b/(a*b-68)-68/(a*b-68),0],[0,a*b/(a*b-68)-68/(a*b-68)])
> ratsimp (%); => matrix([1,0],[0,1])
> A ^^ 0; => matrix([1,0],[0,1])
>
> In all cases % = most recent preceding result.
>
> Hope this helps,
> Robert Dodier
On a related note:
I don't know if this is accurate, but I was taught in high school by my
maths teacher, that any formula involving matricies can be solved by a
polynomial equation with order = rank of matrix. This was as long as you
used the eigenvalues to set the coefficients of the polynomial.
(What does this have to do with maxima? I'm getting there.)
For example:
A: matrix([1,2],[3,4])
My maths teacher was saying that:
A^^3 = a_0.I + a_1.A
If the values of a_0 and a_1 satisfy the following equations (where n
and m are the eigenvectors of A):
n^^3 = a_0 + a_1 * n
m^^3 = a_0 + a_1 * m
Okay, so firstly, was my maths teacher correct? :)
Secondly, will it work with exponentials? (I'm imagining that the
exponential of a matrix would be given by the taylor expansion, but
replacing the commutive multiply by the dot multiply.)
Thirdly, when I enter the following into maxima:
exp(A), it calculates the exponent of individual terms.
(This is how my question related to maxima)
So is there a way to get an exponential of a matrix that uses the dot
multiply? A variable or setting?
Thank you,
Joal Heagney