# Are the only local minima of $\angle(v, Av)$ the eigenvectors?

Let $A$ be an invertible $n \times n$ complex matrix. For $v \in \mathbb{CP}^{n-1}$, define $$d(v) = \frac{|\langle A \tilde{v}, \tilde{v} \rangle |^2}{ \langle A \tilde{v}, A \tilde{v} \rangle \langle \tilde{v}, \tilde{v} \rangle}$$ where $\langle \ , \ \rangle$ is the standard Hermitian inner product and $\tilde{v}$ is any lift of $v$ to $\mathbb{C}^n \setminus \{ 0 \}$

So $d(v) \leq 1$, with equality precisely if $v$ is an eigenvector (by Cauchy-Schwarz).

Are the eigenvectors the only local maxima of $d$?

Motivation: If this is true, than we can prove that complex matrices have complex eigenvectors by a proof analogous to the standard proof that real symmetric matrices have real eigenvectors.

• Your motivation would be more compelling if your comparison with the real symmetric case were more apt, i.e., if you were to assume that $A$ be Hermitian symmetric (in which case, the answer to your question is readily found by reducing to the diagonal case). The property that makes real symmetric matrices manageable is that they are self-adjoint with respect to the standard inner product; the analogous problem in the complex case would be to take $A$ to be self-adjoint with respect to the standard Hermitian inner product. Otherwise, the problems are quite different. – Robert Bryant Nov 18 '12 at 17:50

Thank you for this interesting question! (Long time I was expecting the opposite answer.)

True motivation. Let $V$ be a finite-dimensional $\mathbb C$-linear space equipped with a positive-definite hermitian form $\langle-,-\rangle$. Pick a $1$-dimensional $\mathbb C$-linear subspace $p\subset V$. The orthogonal decomposition $V=p\oplus p^\perp$ provides the natural identification and inclusion in $$\text{T}_p{\mathbb P}_{\mathbb C}V=\text{Lin}_{\mathbb C}(p,V/p)=\text{Lin}_{\mathbb C}(p,p^\perp)\subset\text{Lin}_{\mathbb C}(L,L).$$ The rule $\langle t_1,t_2\rangle:=\text{tr}(t_1\circ t_2^*)$ defines a positive-definite hermitian form on $\text{Lin}_{\mathbb C}(L,L)$ and thus induces a hermitian structure on ${\mathbb P}_{\mathbb C}V$ known as Fubini-Study. The Fubini-Study distance is well known to satisfy the inequalities $0\le\text{dist}(p_1,p_2)\le\frac\pi2$ and the identity $$\cos\text{dist}(p_1,p_2)=\frac{\langle p_1,p_2\rangle\langle p_2,p_1\rangle}{\langle p_1,p_1\rangle\langle p_2,p_2\rangle}=:\text{ta}(p_1,p_2)$$ (see, for instance, arXiv:0702714).

Question. Are the fixed points of $A$ the only local minima of $\text{dist}(Ap,p)$, where $A$ is an arbitrary holomorphic automorphism of ${\mathbb P}_{\mathbb C}V$ ?

Answer. Suppose that a local maximum $x=p$ of $\text{ta}(Ax,x)$ is not fixed by $A$. Taking suitable representatives $p\in V$ and $A\in\text{GL}_{\mathbb C}V$, we assume $\langle p,p\rangle=\langle Ap,Ap\rangle=1$ and $g:=\langle Ap,p\rangle\ge0$. If $g=0$, then $\langle Ax,x\rangle=0$ for all $x\in V$ sufficiently close to $p$, hence, for all $x\in V$. Taking an eigenvector $x$ of $A$, we arrive at a contradiction. So, $0<g<1$ (because the hermitian form on $V$ is positive-definite).

Since $\dim_{\mathbb C}p^\perp+\dim_{\mathbb C}({\mathbb C}p+{\mathbb C}Ap)>\dim_{\mathbb C}V$, there exists $0\ne w\in p^\perp$ such that $(1-gA)w\in{\mathbb C}p+{\mathbb C}Ap$. We will show that, for some $0\ne c\in{\mathbb C}$, an arbitrarily small deformation $p'=p+tcw$ of $p$, $t\in{\mathbb R}$, provides $\text{dist}(Ap',p')<\text{dist}(Ap,p)$. We assume $\langle w,w\rangle=1$ and write $w-gAw=ap+bAp$ for some $a,b\in{\mathbb C}$. It follows that $g\langle Aw,p\rangle+a+bg=0$.

For any $v\in p^\perp$ such that $\langle v,v\rangle=1$, we define $$w_1(t):=\big\langle A(p+tv),p+tv\big\rangle\big\langle p+tv,A(p+tv)\big\rangle,$$ $$w_2(t):=\big\langle A(p+tv),A(p+tv)\big\rangle(1+t^2)$$ so that $$\text{ta}\big(A(p+tv),p+tv\big)=\frac{w_1(t)}{w_2(t)}=:\varphi(t).$$ The fact that $x=p$ is a local maximum of $\text{ta}(Ax,x)$ implies $\varphi'(0)=0$ and $\varphi''(0)\le0$.

Taking derivatives with respect to $t$, we obtain $$w'_1(t)=2\text{Re}\Big(\big(\langle Av,p+tv\rangle+\langle Ap+tAv,v\rangle\big)\langle p+tv,Ap+tAv\rangle\Big),$$ $$w'_2(t)=2\text{Re}\langle Av,Ap+tAv\rangle(1+t^2)+2\langle Ap+tAv,Ap+tAv\rangle t.$$ Therefore, $$w_1(0)=g^2,\qquad w'_1(0)=2g\text{Re}\big(\langle Av,p\rangle+\langle Ap,v\rangle\big),$$ $$w''_1(0)=4g\text{Re}\langle Av,v\rangle+2\big|\langle Av,p\rangle+\langle Ap,v\rangle\big|^2,$$ $$w_2(0)=1,\qquad w'_2(0)=2\text{Re}\langle Av,Ap\rangle,\qquad w''_2(0)=2\langle Av,Av\rangle+2.$$ The condition $\varphi'(0)=0$ is equivalent to $w'_1(0)w_2(0)-w_1(0)w'_2(0)=0$, i.e., to $2g\text{Re}\big(\langle Av,p\rangle+\langle v,Ap\rangle-g\langle Av,Ap\rangle\big)=0$. Since it holds for any $v\in p^\perp$, we obtain $\langle Av,p\rangle+\langle v,Ap\rangle-g\langle Av,Ap\rangle=0$. In particular, $\langle Aw,p\rangle+ag+b=0$, hence, $g\langle Aw,p\rangle+ag^2+bg=0$. It follows from $g\langle Aw,p\rangle+a+bg=0$ that $a=0$. So, $w-gAw=bAp$ and $\langle Aw,p\rangle=-b$.

The equality $w'_1(0)w_2(0)-w_1(0)w'_2(0)=0$ implies $\varphi''(0)=\frac{w''_1(0)w_2(0)-w_1(0)w''_2(0)}{w_2^2(0)}$. As $$w''_1(0)w_2(0)-w_1(0)w''_2(0)=$$ $$=4g\text{Re}\langle Av,v\rangle+2\big|\langle Av,p\rangle+\langle Ap,v\rangle\big|^2-2g^2\langle Av,Av\rangle-2g^2,$$ we obtain $2g\text{Re}\langle Av,v\rangle+\big|\langle Av,p\rangle+\langle Ap,v\rangle\big|^2\le g^2\langle Av,Av\rangle+g^2$, i.e., $$2\text{Re}\big(\langle Av,p\rangle\langle v,Ap\rangle\big)+\big|\langle Av,p\rangle\big|^2+\big|\langle Ap,v\rangle\big|^2+1-g^2\le\langle v-gAv,v-gAv\rangle.$$ Replacing $v$ by $uv$ with a suitable unitary $u\in{\mathbb C}$, $|u|=1$, we obtain $\text{Re}\big(\langle Av,p\rangle\langle v,Ap\rangle\big)\ge0$ and conclude that $$\big|\langle Av,p\rangle\big|^2+1-g^2\le\langle v-gAv,v-gAv\rangle.$$ In particular, for $v=w$, we obtain $|b|^2+1-g^2\le|b|^2$. A contradiction.

So, the answer is yes.