7.2 Repeated Eigenvalues, Jordan Forms, and Linear Dynamical Systems

Lecture notes

1Reading¶

Material related to this page, as well as additional exercises, can be found in ALA 8.6, 10.1 and 10.3.

2Learning Objectives¶

By the end of this page, you should know:

• examples of matrices with repeated Eigenvalues
• what are Jordan Forms
• algebraic and geometric multiplicity of Eigenvalues
• how to solve linear dynamical systems with repeated Eigenvalues

3Repeated Eigenvalues¶

Let’s revisit the matrix $A = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$ we saw in the previous lecture. This matrix has an eigenvalue $\lambda = 2$ of algebraic multiplicity 2: ($\det(A-\lambda I) = (\lambda-2)^2 = 0$ $\Leftrightarrow \lambda_1 = \lambda_2 = 2$) but geometric multiplicity 1, i.e., only one linearly independent eigenvector

exists. How can we solve $\dot{\mathbf{x}} = A\mathbf{x}$ in this case? Taking the approach that we’ve seen so far, we would write a candidate solution as

But this won’t work! What if $\mathbf{x}(0) = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$? There is no $c_1 \in \mathbb{R}$ such that $\mathbf{x}(0) = \begin{bmatrix} c_1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$. Does this mean no solution to $\dot{\mathbf{x}} = A\mathbf{x}$ exists? This would be deeply unsettling! The issue here is that we are “missing” an eigenvector. To remedy this, we’ll introduce the idea of a generalized eigenvector. We will only consider 2×2 matrices, in which case a generalized eigenvector $\mathbf{v}_2$ for an eigenvalue λ with eigenvector $\mathbf{v}_1$ is given by the solution to the linear system:

For our example, we compute $\mathbf{v}_2$ by solving:

Now, how can we construct a solution using $\mathbf{v}_2$? If we try the strategy we used for eigenvalue/vector pairs, things do not quite work out:

If $\mathbf{x}_2(t) = e^{2t}\mathbf{v}_2$ then $\dot{\mathbf{x}}_2(t) = 2e^{2t}\mathbf{v}_2 = 2\mathbf{x}_2$ but $A\mathbf{x}_2(t) = A(e^{2t}\mathbf{v}_2) = e^{2t}(2\mathbf{v}_2 + \mathbf{v}_1) = 2\mathbf{x}_2 + \mathbf{v}_1e^{2t}$,

where we used the fact that the generalized eigenvector $\mathbf{v}_2$ satisfies

$A\mathbf{v}_2 = \lambda\mathbf{v}_2 + \mathbf{v}_1$,

which is obtained by rearranging (3). So we’ll have to try something else. Let’s see if

does better. This guess is made because we need to find a way to have $e^{2t}\mathbf{v}_1$ appear in $\dot{\mathbf{x}}$.

First we compute

This looks promising! Now let’s check

Success! We therefore can write solutions to our initial value problem as linear combinations of

Let’s check if we can find $c_1$ and $c_2$ so that $\mathbf{x}(0) = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$:

and $\mathbf{x}(t) = \begin{bmatrix} te^{2t} \\ e^{2t} \end{bmatrix}$ is the solution to our initial value problem.

42×2 Jordan Blocks¶

In the complete matrix setting, we saw that we could diagonalize the matrix $A$ using a similarity transformation defined by the eigenvectors of $A$, i.e., for $V = [\mathbf{v}_1 \, \mathbf{v}_2 \, \ldots \, \mathbf{v}_n]$, we have that

We saw that this was very useful when solving systems of linear equations.

In the case of incomplete matrices, similarity transformations defined in terms of generalized eigenvectors and Jordan blocks play an analogous role.

For example, consider the matrix $A = \begin{bmatrix} 1 & 1 \\ -1 & 3 \end{bmatrix}$. This matrix has a repeated eigenvalue at $\lambda = 2$, and one eigenvector ${\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}}$. We therefore compute the generalized eigenvector by solving $(A-\lambda I)\mathbf{v}_2 = \mathbf{v}_1$:

One solution is $\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$. We construct our similarity transformation as before, and set $V = \bm \mathbf{v}_1 & \mathbf{v}_2 \em = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$, and compute $V^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}$

Let’s see what happens if we compute $V^{-1}AV$. In the complete case, this would give us a diagonal matrix. In this case, we get

which we’ll recognize as our previous example! It turns out that all $2 \times 2$ matrices with $\lambda=2$ having algebraic multiplicity 2 and geometric multiplicity 1 are similar to the Jordan Block

and this similarity transformation is defined by $V = \bm \mathbf{v}_1 & \mathbf{v}_2 \em$ composed of the eigenvector $\mathbf{v}_1$ and generalized eigenvector $\mathbf{v}_2$ of the original matrix.

We can generalize this idea to any $2 \times 2$ matrix with only one eigenvector:

Using this theorem, we can conclude, much in the same way we did for diagonalizable $A$, that if $A = VJ_\lambda V^{-1}$, then

is a general solution to $\dot{\mathbf{x}} = A\mathbf{x}$.