Showing that a linear transformation $T$ is not invertible but $T+I$ is. 
Assume that the linear transformation $T: \mathbb{R}^k \rightarrow
 \mathbb{R}^k$ has the property that $T^n=0$ for some integer $n >0$.
   Show that $T$ is not invertible but that $T+I$ is invertible.

First, I have some questions regarding the notation and definitions in the question. What exactly does $T^n$ mean? I haven't seen this notation before and it is not explained in the question. Does it just mean the $n$-th composition of $T$ with itself? For example $T^2 = T \circ T$, $T^3 = T \circ T \circ T$, so $T^n = \underbrace{T \circ T \circ \cdots \circ T}_{n \text{ times}}$? Also what does $T+I$ mean? Is that just another linear transformation defined as $(T+I)(\mathbb{x}) = T(\mathbb{x}) + I(\mathbb{x})$ where $I(\mathbb{x})$ is the identity linear transformation on $\mathbb{R}^k$?
Now actually proving the question. I know that if $T$ was invertible then it is one-to-one and onto. So to show it is not invertible, we just need to show that it is not one-to-one or not onto (or both). Similarly, to show that $T+I$ is invertible, we must show it is one-to-one and onto. I am also given a hint that the inverse of $T+I$ is a polynomial function of $T$, but I have no idea of where that comes into play for the question. Any help would be appreciated.
 A: Suppose on the contrary that $T$ is invertible.
Let $x$ be a non-zero vector, $ x \neq 0$.
Since $T^n=0$, $T^n(x)=0$, that is $T(T^{n-1}(x))=0$, this implies that $T^{n-1}(x)=0$.
Similarly, we can show that $T^{n-2}(x)=0$ and so on and eventually $x=0$ which is a contradiction.
Hint for the second part:
$$(I+T)\sum_{i=0}^\infty (-T)^i=I$$
Note that $\sum_{i=0}^\infty (-T)^i$ is actually a polynomial function of $T$
A: To show that $T$ is not invertible, it suffices to note that $\det(T^n)=\det(T)^n$.
Rather than showing that $T$ is invertible by directly showing that it's bijective, it's easier to find a formula for $S=(T+I)^{-1}$ and confirm that $(T+I)S=I$
Hint: Factor the "polynomial" $T^n+I$, just as you would factor $x^n+1$.
A: Hint: If $x^n=0$, then $$(1+x)(1-x+x^2-x^3+\ldots+(-1)^{n-1}x^{n-1})=1$$
A: As $T$ is nilpotent, therefore $0$ is the only eigenvalue and hence, $T$ can't be invertible.
Now, to show $T+I$ is invertible, we will show that $$\sum_{i=0}^{n-1}(-1)^iT^i$$ is its inverse. To that end, consider 
$$(T+I)\sum_{i=0}^{n-1}(-1)^iT^i=\sum_{i=0}^{n-1}(-1)^iT^{i+1}+\sum_{i=0}^{n-1}(-1)^iT^i=I$$
Thus, $T+I$ is invertible.
