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If it is given that $B^2=0$. Then can I directly take the determinant on both sides and prove that determinant of $B=0$?

Given:$B2=0$

Taking determinant both sides

$|B^2|=|B|^2$

$|B|^2=|0|$

$|B|=0$

This was correct.

But I now have a doubt when $A^2=I$, why can't $A=I$ the same way we did for $B=0$

If $A^2=I$

$|A|^2=|I|$

$|A|=1$

Why is this wrong?

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  • 1
    $\begingroup$ $|A|=\pm 1$. $|A|$ is not always positive. $\endgroup$
    – Kavi Rama Murthy
    Commented Jan 16, 2022 at 11:40
  • 2
    $\begingroup$ What are the solutions of $x^2=1$ ? $\endgroup$
    – Lelouch
    Commented Jan 16, 2022 at 11:43
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    $\begingroup$ $|A|=±1$ Is this the correct answer? $\endgroup$
    – ExtroMechanic
    Commented Jan 16, 2022 at 11:45
  • $\begingroup$ The determinant can of course also be negative, so $\det(A)=-1$ is possible as well (in fact for odd size of $A$ , $A=-I$ is a concrete solution with determinant $-1$). Hence $|A|=\pm 1$ is correct. For $0$, we have only one solution because $-0$ and $+0$ fall together. $\endgroup$
    – Peter
    Commented Jan 16, 2022 at 11:51
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    $\begingroup$ Perhaps you considered $|A|$ to be something like the absolute value, which is why I would prefer the notation "det" $\endgroup$
    – Peter
    Commented Jan 16, 2022 at 11:56

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