Given-a-matrix-A-M-n-n-k-N-define-A-n-if-A-and-k-1-I-n-if-A-n-and-k-0-A-k-1-A-if-A-n-and-k-1-Prove-that-A-k-A-r-A-k-r-

Question Number 605 by magmarsenpai last updated on 10/Feb/15

G i v e n a m a t r i x A \in M_{n \times n} \forall k \in N d e f i n e :

A = {\begin{cases} \emptyset_{n} i f A = \emptyset a n d k ⩾ 1 \\ I_{n} i f A \neq \emptyset_{n} a n d k \neq 0 \\ A^{k - 1} A i f A \neq \emptyset_{n} a n d k ⩾ 1 \end{cases}

P r o v e t h a t A^{k} A^{r} = A^{k + r} \forall k, r \in N .

Commented by prakash jain last updated on 09/Feb/15

Question : k, r \in R or k, r \in N ?

A = {\begin{cases} \emptyset_{n} i f A = \emptyset a n d k ⩾ 1 \\ I_{n} i f A \neq \emptyset_{n} a n d k \neq 0 check if k = 0 . \\ A^{k - 1} A i f A \neq \emptyset_{n} a n d k ⩾ 1 \end{cases}

The definition in line 2 contradicts with

definition in line 3 .

If k, r are real then A^{k} needs to be defined

for k < 0 . A^{0.5} = A^{- 0.5} A = undefined

If k, r are postive integers including 0 .

The result follows from

property of matrix muliplication .

A^{r} = A \cdot A \cdot A \dots (r times)

A^{k} = A \cdot A \cdot A \dots (k times)

Answered by magmarsenpai last updated on 09/Feb/15

i s t r u e f o r k, r \in N

Answered by magmarsenpai last updated on 09/Feb/15

I s o l v e d s o :

w e k n o w :

A^{k} = A A . . A (k t i m e s)

A^{r} = A A . . A (r t i m e s)

E i t h e r : A = {(a_{i j})}_{n \times n} \forall i, j \in {1, \dots, n}

A^{k} A^{r} = C = {(c_{i j})}_{n x n}

w e h a v e :

c_{i j} = \underset{j = 1}{\sum^{n}} [(a_{i j}) (a_{i j}) . . (a_{i j})] [(a_{i j}) (a_{i j}) . . (a_{i j})]

c_{i j} = \underset{j = 1}{\sum^{n}} {(a_{i j})}^{k} {(a_{i j})}^{r} = \underset{j = 1}{\sum^{n}} {(a_{i j})}^{k + r}

∴ A^{k} A^{r} = A^{k + r}, w i l l b e w e l l ?

Answered by prakash jain last updated on 10/Feb/15

Matrix multiplication is associative

A \cdot A \cdot A = (A \cdot A) \cdot A = A \cdot (A \cdot A)

A \cdot A \cdot A (k + r t i m e s) = (A \cdot A \cdot A k t i m e s) \cdot (A \cdot A \cdot A r t i m e s)

or

A^{k + r} = A^{k} \cdot A^{r}