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If $(I-A)^k=0$ for some positive integer $k$. Can we claim that $A$ is invertible?

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    Can you find a way to write $AB=I$ for some suitable $B$ (*which may or may not look incredibly complicated and have something to do with $A$ itself*)? – 2017-01-19
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    @JMoravitz nice hint! I think I have proved this! – 2017-01-19

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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 $A^{-1}$?

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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 $A$

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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.