Finding a generalized eigenvector

Question

I'm asked to find a Jordan normal form and a Jordan basis for a matrix A such that $$A= \begin{bmatrix} 1 & 0 & 0 & 2 \\ 2 & -1 & 0 & 2\\ 2 & 0 & -1 & 2 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

I was able to find the eigenvalues $\lambda_{1}=1$ and $\lambda_{2}=-1$ just fine, and both have algebraic multiplicities equal to 2. For the first eigenvalue, I also got a single eigenvector, whereas for $\lambda_{2}$ I found two. The problem starts when I try to complete the Jordan basis. I assume I need to find a generalized eigenvector from $\lambda_{1}$, since it's geometric multiplicity is just 1. From other similar questions, I gathered that:

$$Ker(A-\lambda_{1}{I})^k$$

Is the generalized eigenspace which also contains the eigenvector I need to fill the basis. Then, it seems I'd just need to work with the following matrix: $$(A-1{I})= \begin{bmatrix} 0 & 0 & 0 & 2 \\ 2 & -2 & 0 & 2\\ 2 & 0 & -2 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

But the problem is that the matrix $(A-1I)$ isn't nilpotent, so I can't never seem to get to a point where $(A-1I)^k$ is the zero matrix, and thus I can't find a "final" matrix for the sequence $Ker(A-1I) \subset Ker(A-1I)^2 \subset \dots \subset Ker(A-1I)^k \subset \dots$

According to the method I've used to find a Jordan basis, I need to find generalized eigenvectors from $Ker(A-\lambda{I})^k$ that aren't eigenvectors from $Ker(A-\lambda{I})^{k-1}$, assuming $k$ is the index of nilpotence of the matrix. However, since the sequence is infinite in this case, I don't know how to find the missing eigenvector. Could someone please point out the flaw in my process?

Answer 1

If $A-\lambda I$ is nilpotent, that means some power is zero, that means that the generalised eigenspace is the whole space ${\Bbb F}^n$, for your example ${\Bbb C}^4$ (or ${\Bbb R}^4$). But in this case you already know that that is not true, because you have eigenvectors and generalised eigenvectors associated with two different eigenvalues.

In fact, the generalised eigenspace for $\lambda$ will always be $\ker(A-\lambda I)^a$, where $a$ is the algebraic multiplicity. Sometimes this will also be true for some exponent less than $a$, as in your example with $\lambda=-1$: the generalised eigenspace is $\ker(A+I)$, but it is also $\ker(A+I)^2$, which turns out to be the same (check it!!).

For $\lambda=1$ the generalised eigenspace is $\ker(A-I)^2$, and in this case it isn't equal to $\ker(A-I)$. The matrices $$(A-I)^2\ ,\quad (A-I)^3\ ,\quad (A-I)^4\ ,\ldots$$ may not all be the same, but their kernels will be the same.

So to sum up: the generalised eigenspace you want is $\ker(A-I)^2$.

Answer 2

so I think the quastion for the Eigenvalue $\lambda_1 = -1$ is clear and as you already said this is the Eigenvalue with geometric multiplicity 2. So you can pick up two vectors from the eigen space namely the standard basis vectors: \begin{align} \{e_2,e_3 \} \end{align} Now to the eigenvalue $\lambda_2 = 1$. We can find a eigenvector from the eigenspace, which is of dimension one, this is : \begin{align} b := e_1+e_2+e_3 \end{align} and choose any vector $v \in \Re^4$, which is linear independent to $\{e_1,e_2,b\}$ and obtain that this vector must be in the space: \begin{align} Ker(A-I)^2 \setminus Ker(A-I) \end{align} So your Jordan basis is now given by the construction by: \begin{align} \{ v,(A-I)v,e_2,e_3 \} \end{align} Note that the Matrix $A-I$ does not have to nilpotent on $\Re^4$, but only on the restriction to the generlized eigenspace to $\lambda_2 = 1$. You can check it by the choice of the eigenspace spanned by $\{e_2,e_3\}$. The restriction of $(A-I)^k$ is not the zero Mapping for all integers $k$. If you namely assume that there is an integer $k_0$ with: \begin{align} (A-I)^{k_0} = 0 \end{align} then $(A-I)^{k_0-1}$ is an eigenvector to the eigenvalue $\lambda_2 = 1$, which is a linear combination of $\{e_2,e_3\}$, which can not happen as the eigenvectors to different eigenvalues are linear independent.

Answer 3

Just solve the equations. $(A-\lambda_1I)\vec x=\vec0$ $$\begin{bmatrix}0&0&0&2\\2&-2&0&2\\2&0&-2&2\\0&0&0&0\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$$ Doing elimination $$\begin{align}\begin{bmatrix}0&0&0&2\\2&-2&0&2\\2&0&-2&2\\0&0&0&0\end{bmatrix}&\underrightarrow{R_1\leftrightarrow R_3}\begin{bmatrix}2&0&-2&2\\2&-2&0&2\\0&0&0&2\\0&0&0&0\end{bmatrix}\underrightarrow{R_1/2}\begin{bmatrix}1&0&-1&1\\2&-2&0&2\\0&0&0&2\\0&0&0&0\end{bmatrix}\\ &\underrightarrow{R_2-2R_1}\begin{bmatrix}1&0&-1&1\\0&-2&2&0\\0&0&0&2\\0&0&0&0\end{bmatrix}\underrightarrow{R_2/(-2)}\begin{bmatrix}1&0&-1&1\\0&1&-1&0\\0&0&0&2\\0&0&0&0\end{bmatrix}\\ &\underrightarrow{R_3/2}\begin{bmatrix}1&0&-1&1\\0&1&-1&0\\0&0&0&1\\0&0&0&0\end{bmatrix}\underrightarrow{R_1-R_3}\begin{bmatrix}1&0&-1&0\\0&1&-1&0\\0&0&0&1\\0&0&0&0\end{bmatrix}\end{align}$$ The solution is $\vec x=x_3\begin{bmatrix}1&1&1&0\end{bmatrix}^T$. We choose $x_3=\frac1{\sqrt3}$ and now we have to solve $(A-\lambda_1I)\vec y=\vec x$. Doing elimination $$\begin{align}\begin{bmatrix}0&0&0&2&\frac1{\sqrt3}\\2&-2&0&2&\frac1{\sqrt3}\\2&0&-2&2&\frac1{\sqrt3}\\0&0&0&0&0\end{bmatrix}&\underrightarrow{\text{same stuff}}\begin{bmatrix}1&0&-1&0&0\\0&1&-1&0&0\\0&0&0&1&\frac1{2\sqrt3}\\0&0&0&0&0\end{bmatrix}\end{align}$$ The solution is $\vec y=\begin{bmatrix}0&0&0&\frac1{2\sqrt3}\end{bmatrix}^T+\sqrt3y_3\vec x$. If we want $\vec y\cdot\vec x=0$ we choose $y_3=0$.