If-A-and-B-are-two-matrices-of-suitable-order-does-there-exist-definition-for-A-B-

Question Number 2546 by Rasheed Soomro last updated on 22/Nov/15

If A and B are two matrices of suitable order

does there exist definition for A^{B} ?

Commented by Yozzi last updated on 22/Nov/15

A^{B} = e^{(l n A) B} i f A a n d B a r e s q a u r e

a n d n o n - s i n g u l a r .

e^{(l n A) B} = 1 + (l n A) B + \frac{{((l n A) B)}^{2}}{2!} + \frac{{((l n A) B)}^{3}}{3!} + \dots

∴ A^{B} = \underset{r = 0}{\sum^{\infty}} \frac{1}{r!} {[(l n A) B]}^{r} (M a t r i x e x p o n e n t i a l)

Commented by Rasheed Soomro last updated on 22/Nov/15

TH a n K S f o r S h a r i n g K n o w l e d g e . O n e t h i n g m o r e

I w o u l d l i k e t o a s k \underline{h o w l n A i s d e f i n e d ?}

D o A a n d B n e e d t o b e o f s a m e o r d e r ?

Commented by Yozzi last updated on 22/Nov/15

I f A c a n b e d i a g o n a l i s e d,

t o f i n d l n A, f i n d t h e e i g e n v e c t o r

m a t r i x V o f t h e m a t r i x A . W i t h

A^{^{'}} a s a d i a g i n a l m a t r i x o f t h e

e i g e n v a l u e s o f A f o r i t s c o r r e s p o n d i n g

V w r i t e A^{^{'}} = V^{- 1} A V (*) w h e r e V^{- 1} i s t h e

i n v e r s e m a t r i x o f V .

S u p p o s e A^{^{'}} = [\begin{matrix} a 0 \\ 0 b \end{matrix}] (f o r e x a m p l e) .

T h e n {l n A}^{^{'}} = [\begin{matrix} l n a 0 \\ 0 l n b \end{matrix}] .

T a k i n g l n o n b o t h s i d e s o f (*)

{l n A}^{^{'}} = V^{- 1} (l n A) V \Rightarrow l n A = V ({l n A}^{^{'}}) V^{- 1} .

Commented by Rasheed Soomro last updated on 22/Nov/15

{THANK}^{S s s s}!