# Orthogonality with respect to the standard inner product on $\mathbb{C}^{n\times1}$

If $x$ and $y$ are eigenvectors of a Hermitian matrix corresponding to distinct eigenvalues, then $x$ and $y$ are orthogonal with respect to the standard inner product on $\mathbb{C}^{n\times1}$.

What I know that $x$ and $y$ are hermitian if $x = x^*$ and $y = y^*$ and that the standard inner product of $\mathbb{C}^{n\times1}$ is $y^*x$. But I don't know how to relate the orthogonality with respect to the standard inner product $\mathbb{C}^{n\times1}$.

• math.stackexchange.com/questions/900540/… – Tsemo Aristide May 14 '18 at 14:41
• Orthogonality means that $y^\star x=0$. Now use the fact, that $x$ and $y$ are eigenvectors of a Hermitian matrix. Hint: Try multiplying the product in two different ways with the matrix and show that the result should be the same. Then use the fact, that the eigenvalues are different. – ctst May 14 '18 at 14:41
• You might want to name that Hermitian matrix and use that instead of $x=x^*, \ y=y^*$. – Berci May 14 '18 at 14:44
• the product of x and y? – Migz May 14 '18 at 14:45
• hi @Berci what do you mean name that Hermitian matrix? – Migz May 14 '18 at 14:46

Here $x$ and $y$ are not "hermitian". That applies to square matrices, not to vectors. The situation you have is that there is a hermitian matrix $H$ with $$Hx=\lambda x,\ \ Hy=\mu y,$$ with $\lambda\ne\mu$.
Now, as $\lambda,\mu$ are both real (from $H$ hermitian), $$\lambda y^*x=y^*(\lambda x)=y^*Hx=(Hy)^*x=(\mu y)^*x={\mu}y^*x.$$ From $\lambda\ne\mu$, the above equality can only happen if $y^*x=0$.