Let $H$ be a symmetric matrix. Show that if $u$ and $v$ are eigenvectors of $H$ corresponding to distinct eigenvalues, then $u$ and $v$ are $H$-conjugate.
Need some help, what and how should I approach this? I think $H$ should be considered a Hessian.
Thank you in advance!
On
I do not exactly what you want. Some results: $u^THv=u^Tv=u^TH^THv=0$.
Proof. There is an orthogonal matrix $U$ s.t. $U^THU=D=diag(\lambda_i)$ where the $(\lambda_i)$ are real numbers. Up to mult. by some non-zero complex, there are $i,j$ s.t. $u=C_i,v=C_j$ where $C_i,C_j$ are columns of $U$ associated to $\lambda_i\not= \lambda_j$ (that is $Hu=\lambda_i u,Hv=\lambda_j v$). Then $u^THv=d_{i,j}=0$.
Note that $u,v$ are orthogonal vectors, that is $u^Tv=0$. Moreover $u^T(H^TH)v=0$; indeed $u^T(H^TH)v=(Hu)^T(Hv)=\lambda_i\mu_j(u^Tv)=0$.
Suppose that $u, v$ are eigenvectors of $H$ with eigenvalues $\lambda_u, \lambda_v$, respectively, which are not equal: $\lambda_u \neq \lambda_v$. Since $H$ is symmetric, we know that eigenvectors corresponding to distinct eigenvalues are orthogonal: $\langle u, v \rangle = 0$. (If you don't know this or want a proof, say so!)
Now let's check if $u,v$ are $H$-conjugate, which means that $\langle u, Hv \rangle = 0$: $$ \langle u, Hv \rangle = \langle u, \lambda_v v \rangle = \bar \lambda_v \langle u, v \rangle = \bar \lambda_v \cdot 0 = 0. $$ There you have it!