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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)

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The answer must be no.

The reason is that having different eigenvalues is a purely linear property (it doesn't depend on the inner product you're interested in), whereas orthogonality strongly depends on it.

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 $P = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$.

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  • $\begingroup$ thanks for the explanation! i didn't think of solving the problem backwards :s $\endgroup$ – garyF Sep 8 '15 at 18:57
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No, eigenvectors relative to distinct eigenvalues need not be orthogonal.

The matrix $$ \begin{bmatrix} -1 & 1 \\ -2 & 2 \end{bmatrix} $$ provides an example.

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