If $A^{4}=I$, does this imply $A$ is invertible?

Question

If I have an $n\times n$ matrix, and $A^{4}=I_n$, does this imply that $A^{-1}$ exists?

My reasoning is

$A^{4}=I$, so $(A^{4})^{-1}=I=(A^{-1})^{4}$.

Is this valid? Thanks for your time in answering what is probably a super simple question.

Not correct: You already assume that $A$ has an inverse. It is much simpler: You can give a $B$ with $AB=I$. — gammatester, Aug 29 '16 at 13:08
Your reasoning doesn't imply the existence of $A^{-1}$. In order to write an equation like $(A^{-1})^4=I$ you have to make sure $A^{-1}$ exists (see the answers below). — Janik, Aug 29 '16 at 13:11

score 12 · Answer 1 · edited Aug 29 '16 at 13:12

12

The equation $A^4=I$ says precisely that $A^{-1}=A^3$.

edited Aug 29 '16 at 13:12

Pedro

122,002

answered Aug 29 '16 at 13:07

Janik

1,603

3

I don't think "choose" is the right word since we can't really choose an inverse. It either exists and is unique or it doesn't exist at all. But +1 for the concise and correct answer. – Aug 29 '16 at 13:10
This is indeed true, but i couldn't think of a proper term instead of "choose". I'll edit my post in order to get that clear. – Janik Aug 29 '16 at 13:12
1

@Janik I've suggested an alternative wording. – Pedro Aug 29 '16 at 13:12
@PedroTamaroff Thank you, I was a few seconds too late to do it myself. – Janik Aug 29 '16 at 13:14
I knew this going in, but wanted to check the logic of the reasoning I provided above. Thank you for your answer! – Babe in the Woods Aug 29 '16 at 13:17

Robert Z · Answer 2 · 2016-08-29T15:01:45.743

7

Take the determinant of $A^4$: $$[\det(A)]^4=\det(A^4)=\det(I_n)=1$$ so $\det(A)\not=0$ and $A$ is invertible.

Aternative approach: let $M:=A^3$ then $A\cdot M=M\cdot A=A^4=I_n$ which implies by definition that $A$ is invertible and its inverse is $M$.

edited Aug 29 '16 at 15:01

answered Aug 29 '16 at 13:08

Robert Z

145,942

Thank you. This makes sense. – Babe in the Woods Aug 29 '16 at 13:17
5

You're using a not quite trivial theorem here - in fact the result is trivial from the definition... – David C. Ullrich Aug 29 '16 at 13:51

score 7 · Accepted Answer · answered Aug 29 '16 at 13:10

7

Your reasoning is not correct. You claim that $(A^4)^{-1} = (A^{-1})^4$, but you don't know that yet because you don't know that $A^{-1}$ even exists.

You need to prove that $A^{-1}$ exists. You prove that by proving that there exists some matrix $B$ such that $AB=BA=I$.

answered Aug 29 '16 at 13:10

5xum

123,496
6
128
204

Thanks! I had a suspicion this was the case. – Babe in the Woods Aug 29 '16 at 13:18

score 1 · Answer 4 · answered Aug 29 '16 at 13:17

1

If $\lambda$ is an eigenvalue of $A$, then $\lambda^n$ is an eigenvalue of $A^n$ for $n \geq 1$. This rules out the possibility of zero eigenvalues.

answered Aug 29 '16 at 13:17

Calculon

5,725

score 0 · Answer 5 · answered Sep 03 '16 at 09:26

0

A^4=I ➡ A^(-1)=A^3. We can write it but we can write it when inverse exists. We shall first know that the inverse is exist or not.

answered Sep 03 '16 at 09:26

Sufaid Saleel

3,789

If $A^{4}=I$, does this imply $A$ is invertible?

5 Answers5