[Math] Generalized Eigenvector for 4×4 matrix

linear algebraordinary differential equations

I'm working on Systems of Differential Equations and I'm looking to find the Generalized eigenvector for the following matrix:

$\left[\begin{array}{rrrr}
3 &-4 &1 &0 \\
4& 3 &0 &1 \\
0 &0 &3 &-4 \\
0\ \ &0\ \ &4\ \
&\>\>\>3
\end{array}\right]$

I've found the eigenvalues, r = 3+4i, r = 3-4i, and one of the corresponding Eigenvectors

E1 = $\left[\begin{array}{rrrr}
i \\
1 \\
0 \\
0\\
\end{array}\right]$

So I'm looking for one generalized eigenvector. For some reason this question in particular has been causing me some problems. Any help is greatly appreciated

Best Answer

Given:

$$A = \begin{pmatrix} 3 &-4 &1 &0 \\ 4& 3 &0 &1 \\ 0 &0 &3 &-4 \\ 0 & 0 & 4 & 3 \end{pmatrix}$$

You have correctly found the eigenvalues:

$$ \lambda_1 = 3+4i, \lambda_2 = 3-4i$$

Each eigenvalue has an algebraic multiplicity of two.

For $\lambda_1 = 3 + 4i$, the row reduced echelon form (RREF) of $[A-\lambda_1 I]v_1 = 0$, is:

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

For this eigenvalue, you have correctly found one of the corresponding eigenvectors

$$v_1 = \begin{pmatrix} i \\ 1 \\ 0 \\ 0\\ \end{pmatrix}$$

However, we only have a single eigenvector (geometric multplicity is one) and need to find a second generalized eigenvector. One approach is to use $[A - \lambda_1 I]v_2 = v_1$, yielding the augmented matrix:

$$ \left[\begin{array}{rrrr|r} 4i &-4 &1 &0 & i \\ 4& 4i &0 &1 &1 \\ 0 &0 & 4i &-4 &0\\ 0 & 0 & 4 & 4i&0 \end{array}\right] $$

The RREF of this matrix is:

$$ \left[\begin{array}{rrrr|r} 1 &-i & 0 & 0 & 0 \\ 0& 0 &1 &0 &i \\ 0 &0 & 0 &1 &1 \\ 0 & 0 & 0 & 0&0 \end{array}\right] $$

From this RREF we have:

$$d = 1, c = i, a = i b$$

Choose $b = 0 \implies a = 0$, yielding a second generalized eigenvector:

$$v_2 = \begin{pmatrix} 0 \\ 0 \\ i \\ 1 \end{pmatrix}$$

It is worth noting that the eigenvalues come in complex conjugate pairs and so do the eigenvectors. In other words, you can just write the other two eigenvectors for $\lambda_2$ from the work above by taking the complex conjugate of each eigenvector.