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

Question

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.

Answer 1

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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Edit. In general, suppose $A$ is real symmetric and $A^m=I$ for some odd positive integer $m$. For any vector $x$, let $y=(A-I)x$. Since $A$ is symmetric, we have $\|A^ry\|^2=(A^ry)^T(A^ry)=y^TA^{2r}y$ for every nonnegative integer $r$. It follows that \begin{align*} \|y\|^2 &\le\sum_{\substack{0\le k\le m-1\\ k\text{ is even}}}\|A^{k/2}y\|^2+\sum_{\substack{0\le k\le m-1\\ k\text{ is odd}}}\|A^{(k+m)/2}y\|^2\\ &=y^T\left(\sum_{\substack{0\le k\le m-1\\ k\text{ is even}}}A^k+\sum_{\substack{0\le k\le m-1\\ k\text{ is odd}}}A^{k+m}\right)y\\ &=y^T\left(\sum_{k=0}^{m-1}A^k\right)y\quad\text{(because $A^m=I$)}\\ &=y^T\left(\sum_{k=0}^{m-1}A^k\right)(A-I)x\\ &=y^T(A^m-I)x\\ &=0.\\ \end{align*} Therefore $\|y\|$ must be zero, i.e., $y=0$ or $Ax=x$. Since $x$ is arbitrary, we conclude that $A=I$.

Similarly, if $A$ is real symmetric and $A^m=I$ for some even positive integer $m$, we may show that $A^2=I$. Indeed, if we put $y=(A^2-I)x$, then \begin{align*} \|y\|^2 &\le\|A^{(m-2)/2}y\|^2+\|A^{(m-4)/2}y\|^2+\cdots+\|Ay\|^2+\|y\|^2\\ &=y^T\left(A^{m-2}+A^{m-4}+\cdots+A^2+I\right)y\\ &=y^T\left(A^{m-2}+A^{m-4}+\cdots+A^2+I\right)(A^2-I)x\\ &=y^T(A^m-I)x\\ &=0\\ \end{align*} and hence $y=(A^2-I)x$ must be zero for any vector $x$.

Answer 2

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

Answer 3

$A$ 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 $A=I.$