3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ is it a real or complex matrix ? What do you know about reduction and spectral theorem ? $\endgroup$ – yago Apr 25 '14 at 8:48
  • $\begingroup$ it is real symmetric matrix. I dont know anything about reduction and spectral theorem $\endgroup$ – Ashish Apr 25 '14 at 8:51
6
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – yago Apr 25 '14 at 9:05
  • $\begingroup$ @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. $\endgroup$ – Start wearing purple Apr 25 '14 at 9:17
  • $\begingroup$ O.L. I understand what you wrote but I suggest explaining that $$A = PIP^{-1} \Rightarrow A^3 = PI^3P^{-1} = PI P^{-1} = A = I$$ $\endgroup$ – Ant Apr 25 '14 at 12:17
6
$\begingroup$

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 $A=I$.

$\endgroup$
1
$\begingroup$

$A$ 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 $A=I.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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