# If a square matrix has distinct eigen values, then are the eigen vectors orthogonal?

Here the square matrix $A$ need not be symmetric, and $a_{ij}\in\mathbb{C}$.

I know that the eigen vectors are orthonormal for a symmetric matrix with distinct eigen values. But is this true for a general square matrix? So far, i only find that such eigen vectors are linearly independent.

Are there any counter-examples to the proposition in the title?

In particular, $A$ is a $2\times2$ matrix. $\lambda=\{\lambda_1, \lambda_2\}$ ($\lambda_1 \ne \lambda_2$) and $x=\{x_1, x_2\}$ are its eigen values and eigen vectors respectively. I am thinking of why $A-\lambda_1I$ has columns dependent on $x_2$. (Problem 11 from Gilbert Strang, Introduction to Linear Algebra)

With this "philosophical" answer in mind, it is not hard to cook up an example: if $P$ is an invertible $2 \times 2$ matrix, the matrix $$P \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} P^{-1}$$ has two distinct eigenvalues (1 and 2) but its eigenvectors are (up to a multiplicative factor) $P e_1$ and $P e_2$. If you choose $P$ correctly, these vectors have no reason to be orthogonal. For example, they aren't if $P = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$.
The matrix $$\begin{bmatrix} -1 & 1 \\ -2 & 2 \end{bmatrix}$$ provides an example.