A complex eigenvector is calculated from a $2 \times 2$ square matrix A. Expressed as a sum of real and imaginary parts

$\lambda_{A} = \begin{bmatrix}a+c&0\\0&a-c\end{bmatrix} + \begin{bmatrix}ib&0\\0&ib\end{bmatrix}$

If the eigenvectors are real (i.e. $b=0$), I know the transformation has the effect of scaling. What is the transformation associated with the "imaginary matrix"? I'm under the impression a rotation is involved, but it is not clear to me as to why or how to interpret is visually.


I have two $2 \times 2$ square matrices (A and B) for which I obtain their eigenvalues ($\lambda_{A_{1,2}}$ and $\lambda_{B_{1,2}}$) and eigenvectors ($\textbf{v}_{A_{1,2}}$ and $\textbf{v}_{B_{1,2}}$). It is my objective to understand qualitatively how the two transformations differ given knowledge of their eigenvalues and eigenvectors.

Both matrices have the same set of eigenvectors ($\textbf{v}_{A_{1,2}} = \textbf{v}_{B_{1,2}}$), but different corresponding eigenvalues ($\lambda_{A_{1,2}} \neq \lambda_{B_{1,2}}$). In my particular case, the eigenvalues are determined to be $\lambda_{A_{1,2}} \in\mathbb C$ and $\lambda_{B_{1,2}} \in \mathbb R$. Both the eigenvectors are determined to be $\textbf{v}_{A_{1,2}},\textbf{v}_{B_{1,2}} \in \mathbb R^2$.

If I consider $\operatorname{Re}(\lambda_{A_{1,2}})$, I can make an "apples to apples" comparison with $\lambda_{B_{1,2}}$, in which the real part corresponds to isotropic scaling. For the sake of discussion, lets assume $\operatorname{Re}(\lambda_{A_{1,2}}) = \lambda_{B_{1,2}}$, as it is my intention to better understand the effects of the imaginary component.

Since $\lambda_{A_{1,2}} \in\mathbb C$, there is a additional transformation corresponding to $\operatorname{Im}(\lambda_{A_{1,2}})$. I believe imaginary eigenvalues correspond to a rotation but I'm having a hard time envisioning this when the eigenvectors lie in $\mathbb R^2$. If it is a rotation, what is the axis of rotation?

I'm mostly interested in qualitative/visual solutions here so I can get a feel for what is going on. Also,since I'm talking about linear transforms here, is there any value in thinking of it in a different way? In other words, impose the complex nature of the eigenvalue onto the eigenvector then compare eigenvectors in a visual manor? I don't know, I'm just trying to get a grasp on the imaginary effects.

EDIT: I've provided a very specific case to go along with my question


$A = B = \begin{bmatrix}z&-c\\-c&z\end{bmatrix}$

in which $c \in \mathbb R$. Regardless of how $z$ is defined, the eigenvectors (for both matrices) are given by

$\textbf{v}_{A,B} = \begin{bmatrix}-1&1\\1&1\end{bmatrix}$

with components $(x_1,x_2) | x_i \in \mathbb R$. It is my understanding that these vectors are therefore said to lie in $\mathbb R^2$. (I could be wrong with this lose understanding?)

The eigenvalues on the other hand, depend on where $z = a+ib | a,b \in \mathbb R$ lies in the complex plane. For the A-matrix, I say that $b\neq0$, hence $\lambda_{A_{1,2}} \in\mathbb C$ and for the B-matrix, I say that $b=0$, hence $\lambda_{B_{1,2}} \in\mathbb R$. The corresponding eigenvalues are defined by the diagonals in the following

$\lambda_{A} = \begin{bmatrix}a+c+ib&0\\0&a-c+ib\end{bmatrix} \quad \lambda_{B} = \begin{bmatrix}a+c&0\\0&a-c\end{bmatrix}$

I should note that the eigenvalues for $\lambda_{A}$ are NOT complex conjugate pairs ($ib > 0$ for both). I'm pretty sure I calculated those right. Maybe for a complex matrix transformation, this is OK?

I believe this abstract example helps clarify my question. When comparing the transformation matrices, A and B, I'm trying to understand how to visualize the transformation effects from $\operatorname{Im}(\lambda_{A_{1,2}}) = ib$. Where does a resultant vector end up following the transformation ($\mathbb R^2$ or $\mathbb C$)? Also, please see my questions in the original posting above.

  • $\begingroup$ Are you working in $\mathbb R^2$ or $\mathbb C^2$? The interpretation of complex eigenvalues and eigenvectors is quite different. $\endgroup$ – amd Aug 15 '18 at 1:15
  • $\begingroup$ For the sake of building on what you've already kindly provided, lets stick to $\mathbb R^2$. If you could include connections to my example in your solution presentation, it might help me see things more clearly. If you find places to slip in some comments when working with $\mathbb C^2$, great. I think it can only help, but I also don't expect a full discussion on that. Maybe a "see X for further information" would suffice. $\endgroup$ – ThatsRightJack Aug 15 '18 at 2:43

You didn’t say what vector space you’re working in, but from context I’m going to guess it’s $\mathbb R^2$.

A rotation occurs within some plane. The notion of an axis of rotation is unique to three-dimensional spaces, in which the orthogonal complement of a plane is one-dimensional. There’s only one plane available in $\mathbb R^2$ for rotation, namely $\mathbb R^2$ itself.

A consequence of the fundamental theorem of algebra is that complex eigenvalues of real matrices come in conjugate pairs. To understand the action of a linear transformation $L:\mathbb R^2\to\mathbb R^2$ that has complex eigenvalues, it’s helpful to look first at the action of matrices of the form $$\begin{bmatrix}a&-b\\b&a\end{bmatrix},$$ which have eigenvalues $a\pm ib$. When $a$ and $b$ are not both zero, we can pull out a factor of $r = |a+ib|=\sqrt{a^2+b^2}$ to write this matrix as $$r\begin{bmatrix}\cos\theta&-\sin\theta \\ \sin\theta&\cos\theta\end{bmatrix},$$ i.e., the composition of a rotation through an angle of $\theta = \arg(a+ib)$ and uniform scaling by a factor of $r$. This is precisely the geometric interpretation of multiplication by the complex number $a+ib$. In fact, the set of matrices of this form—a subset of the so-called conformal matrices—are isomorphic to the complex numbers.

Moving to the more general case, every $2\times2$ matrix with complex eigenvalues is similar to a matrix of the above form. To visualize the action of such a matrix, we can use a slightly different take on the action of a conformal matrix with positive determinant. Imagine a family of circles all centered at the origin. We can view the rotation component of the transformation as movement along a circle from this family and the dilation component as shifting to a different circle. Applying a change of basis is equivalent to transforming this family of circles into a family of homothetic ellipses. The rotation part of the transformation now moves along one of these ellipses instead of along a circular arc, while the dilation moves to another ellipse.

This all applies in higher-dimensional spaces, too. A complex conjugate eigenvalue pair of a real matrix indicates that the transformation performs one of these scaled “elliptical rotations” in the plane spanned by the associated eigenvectors. If $L \mathbf v = \lambda \mathbf v$, then $L \overline{\mathbf v} = \overline\lambda \overline{\mathbf v}$, so the eigenvectors of complex conjugate eigenvalues also come in conjugate pairs. The real and imaginary parts of a complex vector are (complex) linear combinations of the vector with its conjugate, so also lie in the span of the conjugate vectors, therefore you can identify the plane of rotation by extracting real and imaginary parts of one of the complex eigenvectors.

  • $\begingroup$ Your answer was very helpful, so thank you for your response. However, there were some specifics with my question that may conflict with your answer (or I just didn't clearly make the connection). I made some edits and added an example. Would you mind reviewing those and update your answer? $\endgroup$ – ThatsRightJack Aug 15 '18 at 1:04
  • $\begingroup$ Just wanted to add this Q/A as an additional reference. I found it quite useful and it seems to further support your answer: math.stackexchange.com/questions/241097/… $\endgroup$ – ThatsRightJack Aug 15 '18 at 2:23

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