There is no real $3\times3$ matrix $A$ for which $A^2=-I$.

Question

I'm asked to prove or disprove this but I'm stuck. I feel like this is actually true but i'm not sure as to how I can formulate a proof for this statement. Any leads?

score 7 · Accepted Answer · edited Nov 22 '16 at 12:07

7

If $AA = -I_3$, then

$\det(AA) = \det(-I_3) $

Does this help?

edited Nov 22 '16 at 12:07

Ben Grossmann

234,171
12
184
355

answered Nov 22 '16 at 12:00

RGS

9,883

This does help, thank you. Now i can make det(A)^2 = -1 which is not possible, correct? – ShaniVl Nov 22 '16 at 12:08
@ShaniVl exactly that – RGS Nov 22 '16 at 12:18

There is no real $3\times3$ matrix $A$ for which $A^2=-I$.

1 Answers1

Linked