Show non-symmetric matrix has non-orthogonal eigenvectors

Question

I'm struggling with a problem from Boas's Mathematical Methods in the Physical Sciences. The question is, for a 2x2 matrix M s.t. M is real, not symmetric, with eigenvalues real and not equal, show that the eigenvectors of M are not orthogonal.

I've tried to manipulate MC = CD in Einstein notation (C, a matrix that diagonalizes M and D, the matrix of eigenvalues) with little luck. I've also tried manipulation of arbitrary elements of M*C and C*D to get an expression for the inner product of the eigenvectors, but I'm not finding any relationships that force the dot product to be non-zero when the off-diagonal elements of M aren't equal.

Any suggestions for a better approach would be appreciated. I've already turned in the assignment, but this question's still bugging me!

Answer 1

By contradiction if there's two orthogonal eigenvectors then there's a change matrix $C$ which's an orthogonal matrix i.e. $C^{-1}=C^t$ and then $$M=CDC^t$$ is a symmetric matrix. Contradiction.

Answer 2

If $M$ admits orthogonal eigenvectors, we may in addition assume that the eigenvectors has norm $1$. Let $A$ be the matrix whose columns are these two eigenvectors. Then $A$ is orthogonal, that is, $A^{-1}=A^T$. Also, since the vectors in $A$ are eigenvectors of $M$, $MA=AD$, where $D$ is the diagonal matrix whose entries along the diagonal are the eigenvalues of $M$. This means that $$ M=ADA^{-1}=ADA^{T}. $$ Using this representation, it is clear that $M=M^T$, that is, $M$ is symmetric, against the assumption.

Answer 3

Let $V$ be the matrix whose columns are the eigenvectors of $M$.

If $M$ has a complete set of linearly independent eigenvectors (which is not always true for non-symmetric real matrices, but is one of the assumptions of your problem) then any vector $w$ can be decomposed into a linear combination $\alpha_1 v_1 + \alpha_2 v_2$ of the two eigenvectors. Finding the alphas amounts to solving the linear system $$V\alpha = w.$$

Then $$Mw = \alpha_1Mv_1 + \alpha_2 Mv_2 = \alpha_1 \lambda_1 v_1 + \alpha_2 \lambda_2 v_2 = VD\alpha$$ where $D = \left[\begin{array}{cc}\lambda_1 & 0\\0 &\lambda_2\end{array}\right]$. Putting it all together, $$Mw = VDV^{-1}w$$ for every vector $w$, so $M = VDV^{-1}$. Now if the eigenvectors are orthogonal and unit length (the latter can be assumed without loss of generality), $V^{-1} = V^T$ and $M=VDV^T$. The right-hand side is symmetric, so the left-hand side must be as well, a contradiction.