For a linear map T ∈ $L (V )$ and a polynomial P(x) = $\sum^n_{k=0} a_k x^k$, we define the map P(T) = $\sum^n_{k=0} a_k T^k$, where T^k denotes the composition of T with itself k times.
Let T ∈ $L (R^2)$ be defined on the standard basis by $T e_1 = e_2$, $T e_2 = e_1$. Let A = {$P(T)$, $P \in R[x]$} Show that A = {$aI + bT$, a, b ∈ R}. What are the invertible elements in A?
My answer is as the following:
$\sum^n_{k=0} a_k T^k$ = $a_0 + a_1T + a_2T^2 + a_3T^3 + ...$
I noted that $T^2$ basically flips the basis so if we have $T(a_1e_1, a_2e_2)$ then $T^2$ = $T(a_1e_2, a_2e_1)$, and so there are two groups: odd exponentials (a2, a1) and even exponentials (a1, a2) of T.
But I got stuck here, how does this relate to the identity matrix, if i am not mistaken, I? it has to be a 1x2 identity matrix but what i think is needed here is (1, -1) matrix, not (1, 1) matrix, since we need to flip the even exponentials of T to equal to odd exponentials.
In addition, what does the invertible elements mean?
$T^2 e_1 = e_1$ and $T^2 e_2 = e_2$. Because a linear transformation is completely determined by what it does on a basis, you know $T^2 = I$.