My Math Forum $a^2 = -I$

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 May 15th, 2017, 06:51 PM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 $a^2 = -I$ Let $\displaystyle A \epsilon M_n(R)$. If $A^2 = -I$ (where $I$ is the Identity matrix), then A) $n$ is even. B) $A= +$ or $- I$ C) all eigenvalues of $A$ are in $R$. D) $A$ is a diagonal matrix. Please explain this with an example if possible. Thank you so much. Last edited by skipjack; May 19th, 2017 at 07:49 AM.
May 17th, 2017, 10:23 PM   #2
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Quote:
 Originally Posted by Lalitha183 Let $\displaystyle A \epsilon M_n(R)$. If $A^2 = -I$ (where $I$ is the Identity matrix), then A) $n$ is even. B) $A= +$ or $- I$ C) all eigenvalues of $A$ are in $R$. D) $A$ is a diagonal matrix. Please explain this with an example if possible. Thank you so much.
Thank you

Last edited by skipjack; May 19th, 2017 at 07:49 AM.

 May 18th, 2017, 08:08 AM #3 Member   Joined: Jan 2016 From: Athens, OH Posts: 93 Thanks: 48 The only true statement is A. Thanks from zylo
May 18th, 2017, 08:25 AM   #4
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Quote:
 Originally Posted by johng40 The only true statement is A.
Thank you for your clear explanation

 May 19th, 2017, 07:43 AM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 (A$\displaystyle ^{2}$=-I)A$\displaystyle ^{-1}$ $\displaystyle \rightarrow$ A$\displaystyle ^{-1}$=-A Last edited by skipjack; May 19th, 2017 at 07:52 AM.
May 19th, 2017, 09:11 AM   #6
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Quote:
 Originally Posted by zylo (A$\displaystyle ^{2}$=-I)A$\displaystyle ^{-1}$ $\displaystyle \rightarrow$ A$\displaystyle ^{-1}$=-A
The above is generally not true, but it can be true, as johng40 points out, and the question is IF A^{2}=-I then.....

My apologies to johng40 and Lalitha183.

Just out of curiosity, what did skipjack edit in above post?

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