Given that $(I-A)^k=0$, is A invertible? If $(I-A)^k=0$ for some positive integer $k$. Can we claim that $A$ is invertible?
 A: Hint:  The binomial theorem for real numbers states that:
$$(x+y)^k = \sum\limits_{i=0}^k \binom{k}{i}x^iy^{k-i}$$
Can you come up with something similar involving $(I+A)^k$ for matrices?
(why doesn't the generic binomial theorem work for arbitrary matrices $(A+B)^k$?)
Can you use what you've discovered to write $AB=I$ for some suitable choice of $B$?
What would this imply about the existence of $A^{-1}$?
A: a matrix $M$ is called nilpotent if there is a $k$ for which $M^k=0$.
it is easily proved that if $M$ is nilpotent then $I-M$ has an inverse. just use the formal series:
$$
(I-M)^{-1} = 1 + M + M^2 + ...
$$
and note that this terminates after a finite number of terms.
Now, the question states that the matrix $M=I-A$ is nilpotent, hence $I-M$ has an inverse. but $I-M$ is just $A$
A: Here is a proof for $\mathbb{C}$:
If $\lambda$ is an eigenvalue of $A$, then $(1-\lambda)^k$ is an eigenvalue of $(I-A)^k$ and hence $(1-\lambda)^k = 0$. In particular, $\lambda \neq 0$, and
hence $A$ is invertible.
