Multiplying a matrix with its eigenvectors stretches or contracts the vector without changing its "direction". Is this true for complex eigenvalues? I tried to prove this as follows -
Suppose A is a square matrix with a complex eigenvalue $\lambda$ and its corresponding eigenvector x.
Then, by definition
Ax = $\lambda$x
Angle between x and the transformed vector $\lambda$x is given by
cos$\theta$ = $\dfrac{<\textbf{x},\lambda\textbf{x}>}{\parallel \textbf{x}\parallel\parallel\lambda\textbf{x}\parallel}$
$\Rightarrow$  cos$\theta$ = $\dfrac{\overline{\lambda}<\textbf{x},\textbf{x}>}{|\lambda|\parallel \textbf{x}\parallel\parallel\textbf{x}\parallel}$
$\Rightarrow$  cos$\theta$ = $\dfrac{\overline{\lambda}<\textbf{x},\textbf{x}>}{|\lambda|\parallel\textbf{x}\parallel^2}$
Now, $\parallel\textbf{x}\parallel^2 = <\textbf{x},\textbf{x}>$ 
So I get,
cos$\theta$ = $\dfrac{\overline{\lambda}}{|\lambda|}$
where $\parallel\textbf{x}\parallel$ represents the norm of vector x and $<\textbf{x},\textbf{y}>$ is the inner product of two vectors x and y
The value of cos$\theta$ should have come out to be 1 or -1 but this is not the case here. Also, how is the cosine of the angle coming out to be a complex number? Should I take the absolute value of the inner product ?
 A: This question is similar to the question about roots of polynomials. Suppose you have a quadratic equation $\, ax^2 + bx + c = 0 \,$ with real coefficients. If there are two real roots, then the polynomial factors into two linear factors. What if there are two conjugate complex roots? The two factors are not real but complex. We were expecting real roots, but then the actual roots are complex.
A similar situation arises with a two dimensional real vector space with a linear map given by a $\, 2\times 2\,$ matrix. The eigenvector equation leads to solving a quadratic. If the two roots are real then we can find the eigenvectors that go along with those eigenvalues. If the roots are two conjugate complex roots, then what? Can we multiply vectors in a real vector space by a complex scalar?
Maybe. For example, if the map is rotation by an angle then the eigenvalue is a unit complex number corresponding to that angle. Given a standard orthonormal basis for the space, define the scalar multiplication by $\,i\,$ to take the first basis vector to the second and that second vector to the negative of the first. Now we have a one dimensional complex vector space.
If you are asking about a complex vector space, then according to the Wikipedia article inner product space, your formula should be modified to read
 $\, \cos\theta=\dfrac{|\langle\textbf{x},\lambda\textbf{x}\rangle|}{\parallel \textbf{x}\!\parallel\,\parallel\!\lambda\textbf{x}\parallel}.\,$ The final result is now $\, \cos\theta = |\overline\lambda| / |\lambda| = 1\,$ and the angle $\, \theta \,$ is $0.\,$ This means that $\, \lambda\textbf{x} \,$ is in the same "direction" as $\, \textbf{x} \,$ in a complex vector space. What this means is that the usual ideas we have about "direction" and angles in real vector spaces must be greatly modified when applied to complex vector spaces.
