If $x$ and $y\in\Bbb{R}^{n}$ are eigenvectors for $\lambda\neq\mu$, respectively, show $x^{T}\cdotp y = 0$

Question

For $x^{T}\cdotp y = 0$, I understand that I can either look at it through matrix multiplication $x^{T}y^{T} = 0$ as you can't do that multiplication. I'm very sure this isn't the right way of looking at it but am unsure how else to think about proving this.

Answer 1

The statement in the question is unclear, speaking of eigenvalues and eigenvectors without mentioning a matrix. For some reason the OP is unwilling to fix the question.

A reader not already familiar with the correct version of the result in question might get the impression we want to prove this:

False Fact If $A$ is a square matrix, $Ax=\lambda x$, $Ay=\mu y$ and $\lambda\ne \mu$ then $x^Ty=0$.

Of course that's false: Consider $A=\begin{bmatrix}1&1\\0&2\end{bmatrix}$, $x=(1,0)^T$, $y=(1,1)^T$.

The correct version is this:

True Fact If $A$ is a symmetric square matrix, $Ax=\lambda x$, $Ay=\mu y$ and $\lambda\ne \mu$ then $x^Ty=0$.

This is easy. First, the whole point to symmetric matricies:

Lemma. If $A$ is symmetric then $x^TAy=y^TAx$.

Proof: Since a $1\times 1$ matrix is its own transpose and $(AB)^T=B^TA^T$ we see that $$x^TAy=(x^TAy)^T=y^TA^Tx=y^TAx.$$

Having established that the True Fact is easy:

$$0=x^TAy-y^TAx=\mu x^Ty-\lambda y^Tx=(\mu-\lambda)x^Ty,$$which implies that $x^Ty=0$, since $\mu-\lambda\ne0$.