# Conditions on a $n\times n$ Hermitian matrix such that its extremal eigenvectors have equal magnitude entries

Is it possible to find (necessary and sufficient) conditions on a general $n\times n$ Hermitian matrix $A$, such that its extremal eigenvectors (the eigenvectors corresponding to the maximum and minimum eigenvalues of $A$) have (pairwise) equal magnitude entries?

As an example, consider the case $n=2$. A basis in the space of $2\times 2$ Hermitian matrices is given by the identity matrix $\mathbb I_2$ together with the three Pauli matrices $$\sigma_1 = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)\;,\qquad \sigma_2 = \left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right)\;,\qquad \sigma_3 = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)\;.\qquad$$ The eigenvectors of the general $2\times 2$ Hermitian matrix $A=\alpha \mathbb I_2 + \beta \sigma_1 + \gamma \sigma_2 + \delta \sigma_3$ ($\alpha,\beta,\gamma,\delta\in\mathbb R$) are $$\left(-\frac{\sqrt{\beta^2+\gamma^2+\delta^2}-\delta}{\beta+i\gamma},\,1\right)^T\;,\qquad \left(\frac{\sqrt{\beta^2+\gamma^2+\delta^2}+\delta}{\beta+i\gamma},\,1\right)^T\;.$$ Therefore, if $\delta=0$, then the extremal eigenvalues have the desired property.

As an other example, consider the case $n=3$. A basis in the space of $3\times 3$ Hermitian matrices is given by the identity matrix $\mathbb I_3$ together with the eight Gell-Mann matrices $\{\lambda_i\}_{i=1,\dots,8}$. For instance, consider the subspace spanned by $\lambda_4,\lambda_6,\lambda_7$, i.e. set $A=\alpha \lambda_4 + \beta \lambda_6 + \gamma \lambda_7$. The extremal eigenvectors of $A$ are $$\left(-\frac{\alpha }{\sqrt{\alpha ^2+\beta ^2+\gamma ^2}},-\frac{\beta -i \gamma }{\sqrt{\alpha ^2+\beta ^2+\gamma ^2}},\,1\right)^T\;,\qquad \left(\frac{\alpha }{\sqrt{\alpha ^2+\beta ^2+\gamma ^2}},\frac{\beta -i \gamma }{\sqrt{\alpha ^2+\beta ^2+\gamma ^2}},\,1\right)^T\;.$$ The desired property holds in this case. However, I don't know what is the maximal subspace on which it holds.

In conclusion, the question asks for general conditions on a $n\times n$ Hermitian matrix $A$, for $n>2$, such that the extremal eigenvectors of $A$ have equal magnitude entries.

• You can easily construct as many such matrices as you wish by just taking two orthogonal vectors $w_1$, $w_n$ with equal magnitude entries (not an issue in the complex field), complementing them to an orthonormal basis in any way you desire, choosing real numbers $\lambda_1<\lambda_j<\lambda n$ and putting $A=\sum_j\lambda_j w_j\otimes w_j$. Moreover, this is a full parametric description but it amounts to nothing more than a reformulation. Thus, when asking for conditions, you should always specify the exact terms in which you want the description. Otherwise the question is meaningless. – fedja Nov 21 '17 at 17:36
• I agree yours is a trivial rewriting of the property above. However, what I was asking is whether there exists an alternative (nontrivial, possibly simpler) characterization of the Hermitian matrices satisfying the property (i.e. in terms which do not involve directly computing the extremal eigenvectors and checking whether the property holds or otherwise). To be clear: for the case $n=2$, an acceptable answer is $A$ has the property iff it belongs to the span of $\mathbb I_2,\sigma_1,\sigma_2$. For general $n$, I wouldn't know if a similar characterization may be given. – jvn99 Nov 21 '17 at 21:48
• Since the image of a semialgebraic set under a polynomial mapping is a semialgebraic set, you can write a finite number of polynomial identities and inequalities that give you a full description directly in terms of the matrix entries. But again, what is the point of that? Well, perhaps somebody else will be able to understand what you want better than I :-) – fedja Nov 21 '17 at 22:58
• Also, if you think a bit, your "acceptable" answer is no different from my answer: just notice that the general parameterization is $w_1=(1,-u), w_2=(1,u)$ with $|u|=1$, so we get $a\begin{bmatrix} 1 & -u \\ -\bar u & 1\end{bmatrix}+b\begin{bmatrix} 1 & u \\ \bar u & 1\end{bmatrix}$ and your description follows almost at once without any fancy names. – fedja Nov 21 '17 at 23:11
• Agreed, but one could conjecture from the case $n=2$ that, for any $n$, decomposing $A$ on the basis of the generalized Gell-Mann matrices, it is always the case that the family of matrices $A$ with the property above coincides with the span of the non-diagonal generators. This looks to me as an alternative, simpler characterization (however, now I know it is untrue). – jvn99 Nov 22 '17 at 10:44