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Does a real $3\times 3$ matrix $A$ that satisfies conditions $\operatorname{tr}(A)=0$ and $A^2+A^T=I$ ($I$ is an identity matrix) exist?

Thank you for your help.

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  • $\begingroup$ Is $A$ real or complex valued? $\endgroup$ – jkn Apr 24 '13 at 19:20
  • $\begingroup$ Sorry, It should be real, I forgot to mention that. I have tried writing down the components of $A^T$ and $A^2$ and creating equations, and it really looks like a quadric, resp. quadratic form, but to find a solution for those 11 equations is a work for a slave, then I thought maybe it could be a matrix of a projection, but got nowhere. $\endgroup$ – user74200 Apr 24 '13 at 19:30
  • $\begingroup$ Agreed, manually solving the simultaneous equations is not what is intended here. $\endgroup$ – Michael Grant Apr 24 '13 at 19:31
  • $\begingroup$ There is in the 2x2 case... $\endgroup$ – Eleven-Eleven Apr 24 '13 at 19:32
  • $\begingroup$ In the 2x2 case, $a_{11}=\varphi, a_{12} = a_{21} = 0$ and $a_{22}=-{\varphi}$. $\endgroup$ – Eleven-Eleven Apr 24 '13 at 19:47
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[Many thanks to user1551 for this contribution.] First, let us show that the eigenvalues of $A$ must be real. The equation $A^2+A^T=I$ implies that $$A^T=I−A^2 \quad\text{and}\quad (A^T)^2+A=I.$$ Substituting the first equation into the second yields $$I−2A^2+A^4+A=I \quad\Longrightarrow\quad A^4−2A^2+A=A(A−I)(A^2+A−I)=0.$$ The eigenvalues must therefore satisfy $\lambda(\lambda−1)(\lambda^2+\lambda−1)=0$, which has the roots $\{0,1,(-1+\sqrt{5})/2,(-1-\sqrt{5})2\}$, all real. In fact, we will show that only the last two are possible.

Let $A=QUQ^T$ be the Schur decomposition of $A$, where $Q$ is unitary and $U$ is upper triangular. Because the eigenvalues of $A$ are real, both $Q$ and $U$ are real as well. Then $$A^2+A^T=I\quad\Longrightarrow QUQ^TQUQ^T+QU^TQ^T=I\quad\Longrightarrow\quad U^2+U^T=I.$$ $U^2$ is upper triangular, and $U^T$ is lower triangular. The only way it is possible to satisfy the equation is if $U$ is diagonal.

(Alternatively, $A$ commutes with $A^T$ because $A^T=I-A^2$. Therefore $A$ is a normal matrix. Since we have shown that all eigenvalues of $A$ are real, it follows that $A$ is orthogonally diagonalisable as $QUQ^T$ for some real orthogonal matrix $Q$ and real diagonal matrix $U$.)

If $U$ is diagonal, then $A$ was symmetric; and the diagonal elements of $U$ are the eigenvalues of $A$. Each must separately satisfy $\lambda^2+\lambda-1=0$. So $\lambda=\frac{-1\pm\sqrt{5}}{2}$. But there are three eigenvalues, so one of them is repeated. There is no way for the resulting sum to be zero.

Since the sum of the eigenvalues is equal to $\mathop{\textrm{Tr}}(A)$, this is a contradiction.

EDIT: I'm going to add that it does not matter that the matrix is $3\times 3$. There is no square matrix of any size that satisfies the conditions put forth. There is no way to choose $n>0$ values from the set $\{(-1+\sqrt{5})/2,(-1-\sqrt{5})/2\}$ that have a zero sum.

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    $\begingroup$ $x^2 - x - 1$ cannot possibly be the characteristic polynomial. It's quadratic. $\endgroup$ – EuYu Apr 24 '13 at 19:41
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    $\begingroup$ No need to delete the answer, you can push through rather easily. The eigenvalues must lie amongst the roots of the polynomial. It's easy to see that no combination of the roots will give $0$. $\endgroup$ – EuYu Apr 24 '13 at 19:48
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    $\begingroup$ How can you say $U^T U = I$? $U$ is unitary not orthogonal. $A^2 is Q U^2 Q^*$ but $A^T $ is $(Q^*)^T U^T Q$ $\endgroup$ – Arin Chaudhuri Apr 24 '13 at 21:06
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    $\begingroup$ @ACARCHAU The equation $A^2+A^T=I$ implies that (a) $A^T=I-A^2$ and (b) $(A^T)^2+A=I$. Substitute (a) into (b), we get $I-2A^2+A^4+A=I$, which simplifies to $A^4-2A^2+A=A(A-I)(A^2+A-I)=0$. The polynomial $x(x-1)(x^2+x-1)$ has only real roots. Therefore $A$ is orthogonally triangularisable. $\endgroup$ – user1551 Apr 24 '13 at 21:35
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    $\begingroup$ The Schur decomposition for real matrices might involve complex numbers, google for real schur decomposition. In this particular case it does have a real schur decomposition, but it needs @user1551's argument, it is not obvious. $\endgroup$ – Arin Chaudhuri Apr 24 '13 at 21:42
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I consider the case where $A$ is restricted to real matrices so $A^\top=A^*$.

Let $\lambda$ be any eigenvalue of $A$, and let $x$, $\|x\| > 0$ be a corresponding eigenvector.

Multiply on the left by $x^*$ and on the right by $x$ to get

$x^* (A^2 x) + (Ax)^*x = \|x\|^2$, i.e., $\lambda^2\|x\|^2 + \overline{\lambda}\|x\|^2 = \|x\|^2$.

As the right hand side is non-zero we must have $\lambda \neq 0$ and $ \lambda^2 + \overline{\lambda} = 1.$

Let $\lambda = a + ib$ with $a,b$ real. We get $$a^2 -b^2 + a = 1$$ and $$2ab - b = 0.$$ Note, $a \neq 1/2$ as then $a^2 + a - 1 = b^2$, a consequence of the first equation becomes impossible. SO the second equation above implies $b=0$ and all eigenvalues are real and are roots of $x^2 + x - 1 = 0$. The rest follows as above solution by Michael Grant, i.e., the trace being the sum of three numbers in $\{ \frac{-1 \pm \sqrt{5}}{2} \}$ cannot be $0$.

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