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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 ?

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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 $u_i$.

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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 $\vec v\cdot\vec w=0$.

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