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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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