# Composing Linear Transformations

Hello and thank you in advance;

The problem:

"Let V be a vector space and T a linear operator $T:V\rightarrow V$, show that $$[T^m]_B =[T]_B^m$$ Where $B$ is a basis(any) of $V$ and $T^m=T\circ T \circ T\circ....\circ T$($m$ times)"

I`ve quite understood how does it work but the way I tried to prove ended up in tons (~m~ tons lol) of sums into matrices, was wondering if there is a cleaner and concise way of doying it. Ideas?

Suppose $\dim V=n$ and let $B=(v_1,\ldots, v_n)$. Then $$[T]_B = \begin{bmatrix}Tv_1& \cdots &Tv_n\end{bmatrix},$$ that is, the $i^{\mathrm{th}}$ column of $[T]_B$ is $Tv_i$. Therefore $$[T^m]_B = \begin{bmatrix}T^mv_1& \cdots &T^mv_n\end{bmatrix}.$$ We show by induction that $[T]_B^m = [T^m]_B$. For $m=1$, the claim is evident. Assume the claim for some positive integer $m$, then \begin{align*} [T^{m+1}]_B &= [T^mT]_B\\ &= [T^m]_B[T]_B\\ &= [T]^m_B[T]_B\\ &= [T]^{m+1}_B. \end{align*}
• By that notation I mean the $i^{\mathrm{th}}$ colum of the matrix is $Tv_i$. Jun 7, 2015 at 4:52
• Do you mean: if $Tv_i=\sum_j A_{ji} v_j$, then the matrix is given by $A$? Jun 7, 2015 at 5:19