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I have $x' = A x$ system. The matrix $A$ is $3\times 3$. Repeated eigenvalue $\lambda=1$ of multiplicity $3$. There are two "normal" eigenvectors associated with this $\lambda$ (i.e. each of rank 1) say $v_1,v_2$, so defect is 1.

I need to find a third generalized eigenvector $v_3$ (which will be of rank 2).

All the methods in the textbook I tried fail for this one example to find $v_3$. (They do work for many other examples, but not for this one below). I tried

  1. Solving for $v_3$ from $ (A-\lambda I) v_3 = v_2$. Does not work, since inconsistent system
  2. Solving for $v_3$ from $ (A-\lambda I) v_3 = v_1$. Does not work, since inconsistent system
  3. Solving for $v_3$ from linear combination of $v_1,v_2$. Hence solving $(A-\lambda I) v_3 =\alpha v_1 + \beta v_2$ for $v_3$ where $\alpha,\beta$ are constants to be found. This also did not work to find $\alpha,\beta$. inconsistent system.

What other methods are there that guarantee finding third eigenvector for this case? I am only asking here about $\lambda$ of multiplicity $3$ and defect of $1$. i.e. have two normal eigenvectors and need to find missing third one (generalized one).

Here is the $A$ matrix which shows this problem. $$ A=\left[\begin{array}{ccc} 1&0&0\\ 1&3&1\\-2&-4&-1\end{array}\right] $$ The eigenvalue is $\lambda=1$ and the two normal vectors found are $$ v_1=\left[\begin{array}{c}-1\\0\\1\end{array}\right] $$ $$ v_2=\left[\begin{array}{c}-2\\1\\0\end{array}\right] $$

The above values are correct, checked on the computer.

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1 Answer 1

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The $1$-eigenspace is the kernel of the map $I-A$, i.e., the null-space of the matrix

$$\begin{bmatrix} 0 & 0 & 0 \\ -1 & -2 & -1 \\ 2 & 4 & 2 \end{bmatrix}.$$

If $(I-A)x=0$, then

$$\begin{bmatrix} 0 & 0 & 0 \\ -1 & -2 & -1 \\ 2 & 4 & 2 \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0\\ -x_1-2x_2-x_3\\ 2x_1+4x_2+2x_3 \end{bmatrix}=\begin{bmatrix} 0\\0\\0 \end{bmatrix}.$$

The only constraints are then $-x_1-2x_2-x_3=0$ and $2x_1+4x_2+2x_3=0$, which is really just the one equation $x_1+2x_2+x_3=0$. It helps to pick parameters $x_2=s$ and $x_3=t$ (these are the "free-variables"). Then the null-space consists of all $(x_1,x_2,x_3)$ satisfying \begin{align} x_1&=-2s-t\\ x_2&=s\\ x_3&=t \end{align}

The null-space is thus

$$\operatorname{null}(I-A)\operatorname{span}\left\{\begin{bmatrix}-2\\ 1\\ 0 \end{bmatrix},\begin{bmatrix}-1\\ 0\\ 1 \end{bmatrix} \right\}$$

giving the two eigenvectors also verified by computer.

To find generalized eigenvectors, we need to find the null-space of $(I-A)^2$, but it turns out $(I-A)^2=0$. Thus

$$\operatorname{null}\left((I-A)^2\right)=\operatorname{span}\left\{\begin{bmatrix}1\\ 0\\ 0 \end{bmatrix},\begin{bmatrix}0\\ 1\\ 0 \end{bmatrix},\begin{bmatrix}0\\ 0\\ 1 \end{bmatrix} \right\}.$$

Conveniently, any vector not in the span of the two ordinary eigenvectors will thus work. Since those two lie in the plane $x_1+2x_2+x_3=0$, we can take the normal to this plane, i.e., $$\begin{bmatrix} 1\\2\\1 \end{bmatrix}.$$

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