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$A$ is a generic square matrix and $I$ is the identity matrix.

I failed to prove that, but I managed to disprove it:

\begin{align*} A &= [-1] \\ A^2 &= ([-1])^2 = [1] = -A \\ (I + A)^2 &= ([1] + [-1])^2 = [0] \neq A \end{align*}

So... am I missing something? Or is the text wrong?

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The text is wrong. $(I + A)^2 = I + 2A + A^2$. Thus $A^2 = -A \iff (I+A)^2 = (I+A)$. –  Daniel Fischer Jul 22 '13 at 18:14
    
I've taken the liberty of changing the title, in case anybody seeing it in the future either (a) worries that it's wrong, or (b) assumes it is true (and incorporates it into, say, a poorly proofread textbook...). –  Sharkos Jul 22 '13 at 19:30

1 Answer 1

up vote 2 down vote accepted

$$(I+A)^2$$ $$I^2+2A+A^2$$ $$I+2A+A^2$$

Now use $A^2=-A$

$$I+2A-A=I+A$$

Text is wrong

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What I can't understand (and this is why I spent more than a hour on that problem) is that the text is obviously wrong. How could the author didn't notice it? –  user16538 Jul 22 '13 at 18:19
    
I have no idea. This happens quite a lot. The text is probably really long. 1 or 2 mistakes are bound to happen... –  yankeefan11 Jul 22 '13 at 18:26
    
Actually the text is a one-liner... by the way thanks. And thanks Daniel Fischer for showing the correct equivalence. –  user16538 Jul 22 '13 at 18:29
    
One-liners signify that the author probably didn't think about the problem at all, so this is either a typo or just a brief mental lapse which nobody bothered to check. –  Sharkos Jul 22 '13 at 19:28

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