Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

3 Answers 3

up vote 5 down vote accepted

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

share|improve this answer
    
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
add comment

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 $A=I$.

share|improve this answer
add comment

$A$ 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 $A=I.$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.