# $A^3 = I$ ($A$ is real Symmetric matrix). Does it imply that $A = I$?

Question of our assignment

$A$ is a $3×3$ real symmetric matrix such that $A^3 = I$ (Identity matrix). Does it imply that $A = I$? If so, why? If not, give an example.

Any help will be appreciated.

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is it a real or complex matrix ? What do you know about reduction and spectral theorem ? –  Yann Hamdaoui Apr 25 at 8:48
it is real symmetric matrix. I dont know anything about reduction and spectral theorem –  Ashish Apr 25 at 8:51

Yes, because of the following:

• $A$ is diagonalizable with real eigenvalues, since it is real symmetric;

• further these eigenvalues solve $\lambda^3=1$ due to $A^3=I$; the only real solution is $\lambda=1$.

Therefore $A=PIP^{-1}=I$.

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He said in a comment he didn't know anything about spectral theorem... The fact that the matrix is 3x3 leads me to think he has to do it by hand and solve equations –  Yann Hamdaoui Apr 25 at 9:05
@YannHamdaoui That also would be doable, although extremely stupid exercise. In that case I would suggest to first notice that $\mathrm{det}\,A=1$ and then simplify the matrix $A^2-A^{-1}=A^2-(\mathrm{det}\,A)\,A^{-1}$, which should be zero. –  O.L. Apr 25 at 9:17
O.L. I understand what you wrote but I suggest explaining that $$A = PIP^{-1} \Rightarrow A^3 = PI^3P^{-1} = PI P^{-1} = A = I$$ –  Ant Apr 25 at 12:17

For any vector $x$, let $y=(A-I)x$. Since $A$ is symmetric and $A^3=I$, we have \begin{align*} \|Ay\|^2 + \|A^2y\|^2 + \|y\|^2 &=y^T\left[A^TA+(A^2)^TA^2+I\right]y\\ &=y^T\left(A^2+A^4+I\right)y\\ &=y^T(A^2+A+I)y\\ &=y^T(A^2+A+I)(A-I)x\\ &=y^T(A^3-I)x\\ &=0. \end{align*} Therefore $\|y\|$ must be zero, i.e. $y=0$ or $Ax=x$. Since $x$ is arbitrary, we conclude that $A=I$.

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$A$ satisfies $x^3-1\in\mathbb R[x]$ which can be factored into irreducible factors as $$x^3-1=(x-1)(x^2+x+1)\text{ over \mathbb R}$$

Since real symmetric matrices are diagonalizable over $\mathbb R$ the minimal polynomial $m_A$ of $A$ must be factored into linear factors over $\mathbb R.$ Also $$m_A|x^3-1\text{ and }m_A(A)=0$$ implies that we are left with the only possibility that $A=I.$

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