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Matrix $A$ is an $n\times n$ matrix, and $I$ is the $n\times n$ identity matrix. Suppose that $A^2=0$, prove that $I-A$ and $I+A$ are both invertible.

Any clues for solving this problem please?

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    $\begingroup$ Hint: compute $(I+A)(I-A)$. $\endgroup$ – Dark Jun 20 '17 at 9:56
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More generally, if the matrix is nilpotent (i.e., $A^n=0$ for some $n \in \mathbb{N})$, then $I-A$ and $I+A$ are both invertible. In fact, $$(I-A)(I + A + A^2 + \cdots + A^{n-1})=I-A^n=I, \\ \quad (I+A)(I - A + A^2 + \cdots + (-1)^{n-1} A^{n-1})=I+A^n=I.$$

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    $\begingroup$ I do not quite understand how you did the simplification. Could you please show me some more about this? $\endgroup$ – markable Jun 20 '17 at 10:10
  • $\begingroup$ Just do matrix multiplication. For the first equality you have $$(I-A)(I + A + A^2 + \cdots + A^{n-1}) = I + A + A^2 + \cdots + A^{n-1} - (A + A^2 + \cdots + A^{n})\\ = I-A^n,$$ whereas the second one is obtained by the first one by replacing $A$ with $-A$. $\endgroup$ – Francesco Polizzi Jun 20 '17 at 10:21
  • $\begingroup$ Can I use the distributive property to compute the (I+A)(I+A), (A+A)(A-A), (A-A)(A-A), or the (A-I)(A-I)? $\endgroup$ – markable Jun 20 '17 at 10:28
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    $\begingroup$ of course you can $\endgroup$ – Francesco Polizzi Jun 20 '17 at 10:47
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HINT: try to compute $(I-A)\cdot(I+A)\ldots$

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    $\begingroup$ Can I get some more explanations for computing (I-A)(I+A) please? I do not quite get it. $\endgroup$ – markable Jun 20 '17 at 10:03
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    $\begingroup$ @markable, note that $$(I-A)(I+A) = I(I+A) - A(I+A) = I^2 + A - A - A^2 = I - O = I.$$ Here, we use the right-distributive property. $\endgroup$ – Decaf-Math Jun 20 '17 at 10:06
  • $\begingroup$ @markable Note that the $(1,1)$ element of $I-A$ is $1-a_{11}$. the diagonal elements of $I \pm A$ are all of the form $1 \pm a_{ii}$. The product of the matrices $(I+A)(I-A)$ will have diagonal elements of the form $1-a_{ii}^{2}$ etc etc $\endgroup$ – Kevin Jun 20 '17 at 10:06
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The annihilating polynomial is $a(t)=t^2$ for which there are two choices of minimal polynomial viz. $m(t)=t$ or $t^2$. In both cases, $A$ has only $0$ eigenvalue.

So, the only eigenvalue of $I\pm A$ is $1$, which implies that both are invertible.

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  • $\begingroup$ Can I use the distributive property to compute the (I+A)(I+A), (A+A)(A-A), (A-A)(A-A), or the (A-I)(A-I)? $\endgroup$ – markable Jun 20 '17 at 10:31
  • $\begingroup$ yes......the distribution of multiplication over addition. $\endgroup$ – Mathlover Jun 20 '17 at 10:38

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