So I have an operator $T$ in $L(l^p(\mathbb{Z},\mathbb{C}))$ that looks like this (sorry that the formating isn't great):
\begin{bmatrix} \ddots 0 & \ddots0 & \ddots0 & \dots & \dots & \ddots0 \\ \ddots1 & 0 & 0 & \dots & \dots & 0 \ddots \\ \ddots0 & 1 & 0 & \dots & \dots & 0 \ddots \\ \ddots0 & 0 & 1 & \dots & \dots & 0 \ddots \\ \vdots & \dots & \ddots & \ddots & 0 & 0 \ddots \\ \vdots & \dots & \dots & 0\ddots & 1\ddots & 0 \ddots \\ \end{bmatrix}
So basically its 0 everywhere except for diagonal line directly under the main diagonal, where its allways 1.
Since the index of $T$ is $indT = dimKer(T) - dimCoker(T)$ I was trying to figure out how to calculate the dimension of the Kernel and coKernel of $T$ but i couldn't find anything about it ($cokerT:=\{l^p(\mathbb{Z},\mathbb{C})/ran(T)\}$.
If this were a matrix in $\mathbb{C}$ like: \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} I could use a basis for $\mathbb{C}^3$ like <(1,0,0),(0,1,0),(0,0,1)> and get the dimension of the Kernel = 1 and the coKernel = 1, but since $T$ has infinite entries I don't really know what to do.
If you know how to calculate the index of $T$ or know a book where this is done that would be great. Thanks in advance
Your operator is known as the bilateral shift. It is a unitary; as such, it is invertible and so its index is zero.