Exercise 2.9.8

Question

Prove (2.12):

If $T = \begin{pmatrix}1 & 1 \0 & 1 \end{pmatrix}$, then 
$$T^s =  \begin{pmatrix}1 & s \0 & 1 \end{pmatrix},\qquad s \in \mathbb{Z}\qquad \qquad (2.12)$$

Richard Ganaye · Accepted Answer

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\begin{proof}
Consider the proposition 
$$\mathscr{P}(s) : T^s =  \begin{pmatrix}1 & s \0 & 1 \end{pmatrix}.$$
Then $\mathscr{P}(0)$ is true, because $T^0 =I_2 = \begin{pmatrix}1 & 0 \0 & 1 \end{pmatrix}$.

Assume that $\mathscr{P}(s)$ is true for some $s \geq 0$. Then
\begin{align*}
T^{s+1} &= T^s\cdot T = \begin{pmatrix}1 & s \0 & 1 \end{pmatrix}\begin{pmatrix}1 & 1 \0 & 1 \end{pmatrix}\
&=\begin{pmatrix}1 & s+1 \0 & 1 \end{pmatrix}.
\end{align*}
This proves that, for all $s\in \N$,  $\mathscr{P}(s) \Rightarrow \mathscr{P}(s+1)$.

We conclude by induction that
$$\forall s \in \N,\ T^s =  \begin{pmatrix}1 & s \0 & 1 \end{pmatrix}.$$
Moreover, if $s\leq 0$, then $s = -s', s'\geq 0$, and
$$ T^s = T^{-s'} = (T^{s'})^{-1} = \begin{pmatrix}1 & s' \0 & 1 \end{pmatrix}^{-1} =  \begin{pmatrix}1 & -s' \0 & 1 \end{pmatrix}^{-1} 
=  \begin{pmatrix}1 & s \0 & 1 \end{pmatrix}^{-1}.$$
Therefore
$$\forall s \in \Z,\ T^s =  \begin{pmatrix}1 & s \0 & 1 \end{pmatrix}.$$
\end{proof}

Exercise 2.9.8

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Exercise 2.9.8

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