two dimensional linear differential equation with $1$ eigenvector

I have the following linear differential equation:

\begin{equation} x' = \begin{pmatrix}3&-4\\1&-1\end{pmatrix}x \end{equation} The corresponding characteristic equation is: \begin{equation} \lambda^2-2\lambda+1 = (\lambda-1)(\lambda-1) \implies \lambda_1=\lambda_2=1 \end{equation} A corresponding eigenvector is $\begin{pmatrix} 2\\1\end{pmatrix}$.

By plotting the solutions to the equation I know that it is a degenerate node, i.e there is only one eigenvector for the matrix (I believe the terminology is that the eigenspace has dimension 10.

1. How do I determine this without plotting the differential equation?
2. How do I know that the eigenspace only has dimension 1?
• en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem – Amzoti Jul 10 '13 at 2:41
• You might want to learn the terms algebraic and geometric multiplicity for eigenvalues. The RREF of $[A- \lambda I]v_i = 0$ can help you determine rank info. Also, you need a generalized eigenvector for the system above. I can do all the steps if you'd like to show it manually. Regards – Amzoti Jul 10 '13 at 2:57
• I know how to solve the system but thanks! What I want to know is if I am remembering my linear algebra correctly: $\dim(row space) = \dim(column space) = dim (A) = rank(A)$? – CodeKingPlusPlus Jul 10 '13 at 3:19
• Then see if you can get your head around algebraic and geometric multiplicity, generalized eigenvectors and rank nullity (Rank–nullity theorem) and you'll be fine. – Amzoti Jul 10 '13 at 3:21
• $(A-\lambda I)x = 0$. Can't I just then row-reduce $A=\begin{pmatrix}-4&-4\\-2&2\end{pmatrix}$ to $\begin{pmatrix}-4&4\\0&0\end{pmatrix}$. There is one pivot, and so the $rank(A)=1 \implies$ dimension of the null space is $1$? – CodeKingPlusPlus Jul 10 '13 at 3:28

Terminology

When we are solving for eigenvalues of a system, and an eigenvalue is repeated, then one worries as to whether there exist enough linearly independent eigenvectors.

If we have an eigenvalue with multiplicity $n$ which has less than $n$ linearly independent eigenvectors, we will not have enough solutions. With this in mind, we deﬁne the algebraic multiplicity of an eigenvalue to be the number of times it is a root of the characteristic equation. We deﬁne the geometric multiplicity of an eigenvalue to be the number of linearly independent eigenvectors for the eigenvalue.

Mathematically, we can state that the algebraic multiplicity of an eigenvalue $\lambda$ is, by definition, the largest integer $k$ such that $(x−\lambda)^k$ divides the characteristic polynomial. The geometric multiplicity of $\lambda$ is the dimension of its eigenspace, that is, it is the dimension of $\{X \in \mathbb{C}^{n×1} : AX=λX\}$, where $n$ is the dimension of the matrix.

The null space of the matrix is called the eigenspace associated with the eigenvalue $\lambda$.

When the geometric multiplicity of an eigenvalue is less than the algebraic multiplicity, we say the matrix is defective. In the case of defective matrices, we must search for additional solutions using generalized eigenvectors.

Analyze System

In this system, we have:

$$x' = \begin{pmatrix}3&-4\\1&-1\end{pmatrix}x$$

Aside: The matrix here is rank $= 2$. This means that, $\dim(A) = \text{rank}(A) + \text{nullity}(A) = 2 = 2 + 0$, in other words, $A$ is an invertible matrix.

The corresponding characteristic equation for $A$ is:

$$\lambda^2-2\lambda+1 = (\lambda-1)(\lambda-1) \implies \lambda_1=\lambda_2=1$$

The algebraic multiplicity for the eigenvalue $\lambda_1 = 1$ is $2$. Lets find the geometric multiplicity.

To find an eigenvector, we set up and solve $[A- \lambda I]v_i = 0$, so we have:

$$[A - 1 I]v_1 = \begin{pmatrix}2 & -4\\ 1 & -2\end{pmatrix}v_1 = 0.$$

The RREF of this matrix is:

$$\begin{pmatrix}1 & -2\\ 0 & 0\end{pmatrix}v_1 = 0.$$

This gives us an eigenvector $v_1 = (2, 1)$.

Observations

• The rank of the RREF matrix is $1$ and we know, from the rank-nullity theorem, that the $\dim = 2 = \text{rank} + \text{nullity} = 1 + \text{nullity} \rightarrow \text{nullity} = 1$.
• Note that sometimes this is more generally called $\dim (\text{image}~ T) + \dim (\ker ~T) = \dim ~V$
• We know that the nullity is the geometric multiplicity, which is $1$. This means we can only find one independent eigenvector, so we have what is called a deficient matrix.
• The eigenspace for $\lambda = 1$ is the nullspace of $[A - \lambda I] = [A- 1 I] = \text{Span}\left\{\begin{pmatrix}2\\ 1\end{pmatrix}\right\}$. Notice how this agrees with the eigenvector we found above (as it should)?
• From the eigenspace terminology above, we can write $E(1)= \left\{X \in \mathbb{C}^{2×1} : AX = 1X = \begin{pmatrix}2\\ 1\end{pmatrix}\right\}$
• Note that there is a nice way of getting everything using the factorization of the characteristic polynomial, but that is for another day.
• From all of this, we still need to find a second independent (generalized) eigenvector.

To find a second eigenvector, we try:

$$[A - 1I]v_2 = v_1 \rightarrow \begin{pmatrix}1 & -2\\ 0 & 0\end{pmatrix}v_2 = \begin{pmatrix}2\\1\end{pmatrix}$$

This leads to: $a = 1 + 2b \rightarrow \text{let}~~ b = 0, a = 1 \rightarrow v_2 = (1, 0)$.

So, we can write our general solution as:

$$x(t) = \begin{bmatrix}x_1(t)\\ x_2(t)\end{bmatrix} = e^t\left[ c_1 v_1 + c_2(v_1 t + v_2)\right] = e^t\left[ c_1 \begin{pmatrix}2\\1\end{pmatrix} + c_2\left(\begin{pmatrix}2\\1\end{pmatrix} t + \begin{pmatrix}1\\0\end{pmatrix}\right)\right] = e^t\left[ c_1 \begin{pmatrix}2\\1\end{pmatrix} + c_2\begin{pmatrix}2t + 1\\t\end{pmatrix}\right]$$

If we wanted to write the matrix exponential, we would have:

$$e^{At} = e^t\begin{pmatrix}2 t+1 & -4 t \\ t & 1-2 t \end{pmatrix}$$

We can also draw the phase portrait for the system. We have a critical point at $(x, y) = (0,0)$. From the eigenvalues, we have a positive, repeated real root $\lambda = 1 \rightarrow$ a degenerate node. The phase portrait is as follows. • Nice work! $+1$ – Namaste Jul 11 '13 at 0:30
• this one took a lot of work! – Namaste Jul 11 '13 at 0:35
• @amWhy: Yes, this one took a LONG time to write up! Thanks for the support! – Amzoti Jul 11 '13 at 0:35
• @Amzoti: Thanks for teaching me all nice points above. I really appreciate you. :+) – mrs Jul 11 '13 at 8:15
• @BabakS.: I also learn from you and others and really like the site for this reason. Hope all is well my friend! – Amzoti Jul 11 '13 at 13:09