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$\newcommand{\Mat}{\operatorname{Mat}}$ This may seem like a trivial fact but how would you show that:

Suppose $T \in L(V)$ Then $\Mat(T^n) = \Mat(T) \cdots \Mat(T)$ where there are $n$ of them?

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Induction (.......) –  dineshdileep Jan 10 '13 at 16:44
    
Base case n = 2 doesn't seem very clear. –  mathnoob Jan 10 '13 at 16:49
    
How would you define applying the transformation $T^2$? as applying $T$ twice? –  dineshdileep Jan 10 '13 at 16:51
    
Do not have standard definition but idea is like if: $T(x,y) = (y,x)$ then $T^2(x,y) = T(y,x) = T(x,y)$ –  mathnoob Jan 10 '13 at 16:54
    
nvm resolved. pg 51 of Axler gave the definition $M(TS) = M(T)M(S)$ –  mathnoob Jan 10 '13 at 17:33

1 Answer 1

up vote 0 down vote accepted

Suppose $M=(m_{ij})$ is the matrix representation of $T$ under a certain basis $B=\{v_1,\ldots,v_n\}$. Then \begin{align*} T^2(v_j) &= T\left(T(v_j)\right)\\ &= T\left( \sum_{k=1}^n m_{kj}v_k \right)\\ &= \sum_{k=1}^n m_{kj} T(v_k)\\ &= \sum_{k=1}^n m_{kj} \sum_{i=1}^n m_{ik} v_i\\ &= \sum_{i=1}^n \left(\sum_{k=1}^n m_{ik} m_{kj}\right) v_i. \end{align*} Since $\sum_{k=1}^n m_{ik} m_{kj}$ is, by definition, the $(i,j)$-th entry of $M^2$, the result follows.

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