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Let $A$ be a real square matrix satisfying $A^{17}=0$.

a) Show that the matrix $I-A$ is invertible.
b) If $B$ is an invertible matrix, is $B-A$ also invertible? Proof or counterexample.
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a) If a square matrix $A$ satisfies $A^{17}=0$, then it follows that $I-A$ is invertible.
To prove this, we will use the formula for the inverse of a matrix expressed as a sum:
$(I-A)^{-1}=\sum_{k=0}^{\infty} A^k$
Since $A^{17}=0$, it follows that all higher powers of $A$ vanish as well. So, the sum is finite and well-defined, and the inverse exists.
b) Not necessarily. If $B$ is invertible, it doesn't necessarily mean that $B-A$ is invertible.
For a counterexample, consider the matrix $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$. Both matrices are invertible, but $B-A=\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]$ is not invertible.
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