6.6 Repeated and Complex Eigenvalues

1Reading¶

Material related to this page, as well as additional exercises, can be found in ALA 8.2.

2Learning Objectives¶

By the end of this page, you should know:

how to define repeated and complex eigenvalues,
that every $n \times n$ matrix has at least 1 and at most $n$ distinct eigenvalues.

3Repeated Eigenvalues¶

Recall this example from a few pages ago where we say an example of a $3\times 3$ matrix with a double eigenvalue:

\begin{align*} A = \bm 2&-1&-1\\0&3&1\\0&1&3 \em \end{align*}

(1)

with

\begin{align*} \lambda_1 = 2,&\quad \vv{v_1} = \bm 1\\0\\0 \em, \vv{\hat{v_1}} = \bm 0\\1\\-1 \em\\ \lambda_2 = 4,& \quad \vv{v_2} = \bm -1\\1\\1 \em. \end{align*}

(2)

In this case, even though $A$ only has 2 distinct eigenvalues, it still has three linearly independent eigenvectors: as we’ll see later, this is important as it will allow us to use the eigenvectors of $A$ to define a basis for $\mathbb{R}^3$ .

This doesn’t always happen though. Next, we’ll see a simple example of a $2\times 2$ matrix with only one eigenvector!

Example 1 (A

2\times 2

matrix with only one eigenvector)

Let $A = \bm 2&1\\0&2\em$ . Then $A$ has a double eigenvalue of $\lambda_1 = 2$ , because $\det(A - \lambda I) = (2 - \lambda)^2 = 0 \iff \lambda = 2$ .

The associated eigenvector equation $(A - 2I) \vv v = \vv 0$ then becomes:

\begin{align*} \bm 0&1\\0&0\em\bm v_1\\v_2 \em = \bm 0\\0\em \iff v_2 = 0, v_1 \text{ free} \end{align*}

(3)

i.e., $\vv v = a\bm 1\\0\em$ is an eigenvector, and we set $\vv{v_1} = \bm 1\\0\em$ .

Thus, even though $\lambda = 2$ is a double eigenvalue, it only admits a one dimensional eigenspace. The list of eigenvalues/vectors is in a sense incomplete.

For the next few lectuers, we will avoid such degenerate examples, but we will read how to handle them when we study a motivating application of linear dynamical systems a few pages down the line.

4Complex Eigenvalues¶

So far, all of the examples we’ve considered have had real eigenvalues. In general, however, complex eigenvalues (and eigenvectors) are also important. We will assume you already have previous experience working with complex numbers, particularly finding complex roots of polynomial equations, but a quick refresher can be found here:

https://math.mit.edu/~stoopn/18.031/complexnumbers.pdf

Example 2 (A

2\times 2

matrix with an imaginary eigenvalue)

Let $A = \bm 0&-1\\1&0\em$ corresponding to a $90\degree$ rotation.

Its eigenvalues are determined by solutions to

\begin{align*} \det(A - \lambda I) = \det \bm -\lambda & -1 \\ 1 & -\lambda \em = \lambda^2 + 1 = 0, \end{align*}

(4)

i.e., $\lambda^2 = -1$ . There is no $\lambda \in \mathbb R$ satisfying this equation, but as you know, this means we have to expand our candidate solutions too include complex numbers. In this case, $\lambda_1 = i$ and $\lambda_2 = -i$ , for $i = \sqrt{-1}$ the imaginary number, are two solutions.

The corresponding eigenvector equations, which now will include complex numbers, becomes:

For $\lambda_1 = i$ :

\begin{align*} &(A - iI) \vv v = \bm -i&-1\\1&-i \em \bm v_1\\v_2 \em = \bm 0\\0 \em \\ &\iff \begin{array}{c} -iv_1 - v_2 = 0\\ v_1 - iv_2 = 0 \end{array} \\ &\iff v_2 = -iv_1 \end{align*}

(5)

so that $\vv v = a\bm 1\\-i \em$ for any $a \in \mathbb{C}$ , and we set $\vv{v_1} = \bm 1\\-i \em$ as the eigenvector associated with $\lambda_1 = i$ .

For $\lambda_2 = -i$ : following the same procedure, we see that $\vv{v_2} = \bm 1 \\ i \em$ is the corresponding eigenvector.

Thus, we have

\begin{align*} \lambda_1 = i, &\quad\vv{v_1} = \bm 1\\-i\em\\ \lambda_2 = -i, &\quad\vv{v_2} = \bm 1\\i\em\\ \end{align*}

(6)

A couple of observations about the example above:

$\lambda_1$ and $\lambda_2$ are complex conjugates, i.e., $\lambda_{1} = \overline{\lambda_2}$ , as are $\vv{v_1}$ and $\vv{v_2}$ . This is a general fact about real matrices:

The eigenvalues for our example, which defines a pure rotation in $\mathbb{R}^2$ , are purely imaginary. This isn’t a coincidence! When we discuss symmetric and skew symmetric matrices later in this class, this will be further explained, but for now, you should start associating imaginary components of eigenvalues with rotations.

5Basic Properties of Eigenvalues¶

We want to belabor the derivation of the following properties of eigenvalues: they mostly follow from properties of the determinant and the Funamental Theorem of Algebra (which states that any degree $n$ polynomial equation with complex coefficients has exactly $n$ complex roots, counting multiplicities). The TLDR is that for any $A \in \mathbb{R}^{n\times n}$ , its characteristic polynomial is a degree $n$ polynomial with real coefficients and can thus be factored as:

\begin{align*} \det (A - \lambda I) = (-1)^n (\lambda - \lambda_1)(\lambda - \lambda_2) \dots (\lambda - \lambda_n) \end{align*}

(7)

where the complex numbers $\lambda_1, ..., \lambda_n$ , some of which may be repeated, are the eigenvalues of $A$ . Therefore we immediately conclude that: