How Eigenvectors of $A^T$ are perpendicular to eigen vectors of A? Assuming $A$ to be a square $n\times n$ matrix with $n$ distinct eigen values , then $A$ has $n$ eigen vectors $u_1,u_2,\dotsc,u_n$. Now $A^T$ has the same eigen values as $A$ and eigen vectors $v_1,v_2,\dotsc,v_n$.
Now what I observed is that for $A$, eigen value $e_1$ the eigen vector $v_1$ is perpendicular to all eigen vectors of $A^T$ which correspond to eigen value $e_i \neq e_1$ . 
Can some one tell if I am correct and if yes can prove it ?
 A: If $\def\vec#1{{\bf#1}}A\vec v=\lambda\vec v$ and $A\vec w=\mu\vec w$ with $\lambda\ne\mu$ then
$$\eqalign{(\lambda-\mu)\vec v\cdot\vec w
  &=(\lambda-\mu)\vec v^T\vec w\cr
  &=(\lambda\vec v^T)\vec w-\vec v^T(\mu\vec w)\cr
  &=(A\vec v)^T\vec w-\vec v^T(A^T\vec w)\cr
  &=\vec v^TA^T\vec w-\vec v^TA^T\vec w\cr
  &=0\ ,\cr}$$
and $\lambda-\mu\ne0$ so $\vec v\cdot\vec w=0$.
A: You are correct. To prove this, consider the following eigenvalue equation for $u$:
$$ A u = \lambda u $$
Take the transpose of this: 
$$ u^T A^T  = \lambda u^T $$
Now, multiply this expression on the left with $v$:
$$ u^T a^T v = \lambda u^T v $$
But if $v$ is an eigenvector of $A^T$ with eigenvalue $\mu \neq \lambda$, we have
$$ A^Tv = \mu v $$
so 
$$ u^T A^T v = \mu u^T v. $$
Therefore, 
$$\lambda u^T v = \mu u^T v $$
which is only true (since $\lambda \neq \mu$) if $u^T v = 0$. Since this is just the dot product, the vector $u$ must be orthogonal to the vector $v$. Thus, for an eigenvector $u_i$ with eigenvalue $\lambda_i$ of $A$, any eigenvector of $A^T$ with a different eigenvalue must be orthogonal to $u_i$.
