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) ∀ k,r ∈ N.


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Question Number 605 by magmarsenpai last updated on 10/Feb/15
$Given a matrix A ∈ M_(n × n ) ∀ k ∈ N define: A= { ((∅_n if A=∅ and k≥1)),((I_n if A≠∅_n and k≠0)),((A^(k−1) A if A≠∅_n and k≥1)) :} Prove that A^k A^r =A^(k+r) ∀ k,r ∈ N.$
$G i v e n a m a t r i x A \in M_{n \times n} \forall k \in N d e f i n e :$ $A = {\begin{cases} \emptyset_{n} i f A = \emptyset a n d k ⩾ 1 \\ I_{n} i f A \neq \emptyset_{n} a n d k \neq 0 \\ A^{k - 1} A i f A \neq \emptyset_{n} a n d k ⩾ 1 \end{cases}$ $P r o v e t h a t A^{k} A^{r} = A^{k + r} \forall k, r \in N .$
Commented by prakash jain last updated on 09/Feb/15
$Question: k, r∈R or k,r∈N? A= { ((∅_n if A=∅ and k≥1)),((I_n if A≠∅_n and k≠0 check if k=0.)),((A^(k−1) A if A≠∅_n and k≥1)) :} The definition in line 2 contradicts with definition in line 3. If k , r are real then A^k needs to be defined for k<0. A^(0.5) =A^(−0.5) A=undefined If k, r are postive integers including 0. The result follows from property of matrix muliplication. A^r =A∙A∙A...(r times) A^k =A∙A∙A ...(k times)$
$Question : k, r \in R or k, r \in N ?$ $A = {\begin{cases} \emptyset_{n} i f A = \emptyset a n d k ⩾ 1 \\ I_{n} i f A \neq \emptyset_{n} a n d k \neq 0 check if k = 0 . \\ A^{k - 1} A i f A \neq \emptyset_{n} a n d k ⩾ 1 \end{cases}$ $The definition in line 2 contradicts with$ $definition in line 3 .$ $If k, r are real then A^{k} needs to be defined$ $for k < 0 . A^{0.5} = A^{- 0.5} A = undefined$ $If k, r are postive integers including 0 .$ $The result follows from$ $property of matrix muliplication .$ $A^{r} = A \cdot A \cdot A . . . (r times)$ $A^{k} = A \cdot A \cdot A . . . (k times)$
	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
	$I solved so : we know : A^k =AA..A (k times) A^r =AA..A (r times) Either: A=(a_(i j) )_(n ×n) ∀ i,j∈{1,...,n} A^k A^r =C=(c_(i j) )_(n x n) we have: c_(i j) =Σ_(j=1) ^n [(a_(i j) )(a_(i j) )..(a_(i j) )][(a_(i j) )(a_(i j) )..(a_(i j) )] c_(i j) =Σ_(j=1) ^n (a_(i j) )^k (a_(i j) )^r =Σ_(j=1) ^n (a_(i j) )^(k+r) ∴ A^k A^r =A^(k+r) , will be well ?$
	$I s o l v e d s o :$ $w e k n o w :$ $A^{k} = A A . . A (k t i m e s)$ $A^{r} = A A . . A (r t i m e s)$ $E i t h e r : A = {(a_{i j})}_{n \times n} \forall i, j \in {1, . . ., n}$ $A^{k} A^{r} = C = {(c_{i j})}_{n x n}$ $w e h a v e :$ $c_{i j} = \underset{j = 1}{\sum^{n}} [(a_{i j}) (a_{i j}) . . (a_{i j})] [(a_{i j}) (a_{i j}) . . (a_{i j})]$ $c_{i j} = \underset{j = 1}{\sum^{n}} {(a_{i j})}^{k} {(a_{i j})}^{r} = \underset{j = 1}{\sum^{n}} {(a_{i j})}^{k + r}$ $∴ A^{k} A^{r} = A^{k + r}, w i l l b e w e l l ?$
	Answered by prakash jain last updated on 10/Feb/15

	$Matrix multiplication is associative$ $A \cdot A \cdot A = (A \cdot A) \cdot A = A \cdot (A \cdot A)$ $A \cdot A \cdot A (k + r t i m e s) = (A \cdot A \cdot A k t i m e s) \cdot (A \cdot A \cdot A r t i m e s)$ $or$ $A^{k + r} = A^{k} \cdot A^{r}$
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