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I have this matrix: $$\begin{pmatrix}1& 0& 0& -3\\ 0& 1& 0& 3\\ 0&0& 1& -1\\0& 0& 0& 1 \end{pmatrix}$$

There is one eigenvalue $1$ with three eigenvectors $\left\{ \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}\right\}$.

How do I find a fourth (generalized) eigenvector? If I begin to solve $(I-A)x=v$, where $v$ is an eigenvector, I will still get the fourth coordinate equal zero.

eMathHelp
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3 Answers3

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The problem here is that your eigenspace to the eigenvalue $1$ is more than $1$-dimensional. Hence you cannot just choose a basis and try to form Jordan chains on these basis elements. Indeed, if you try to solve $(I - A)x = v$ for one of your basis vectors $v$, you get nothing. But this does not mean that $(I - A)x = v$ does not have a solution for any eigenvector $v$. Indeed, if you take the eigenvector $$v = \begin{pmatrix} 3 \\ -3 \\ 1 \\ 0\end{pmatrix},$$ then you get the solution $$x = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1\end{pmatrix}.$$ In this general case it is actually easier to compute the kernel of $(I - A)^2$, which in this case happens to be the zero matrix, and complete the basis in this way. This shows that you can in fact take any vector that does not vanish in the last component as your generalized eigenvector.

Klaus
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As the other answer states, $x = (x_1,x_2,x_3,x_4)$ will be a generalized eigenvector but not an eigenvector if and only if $x_4 \neq 0$.

However, when applying the usual procedure recommended for finding the Jordan form, it is convenient to be able to find a generalized eigenvector by solving the equation $(A - I)x = v$ for a suitable eigenvector $v$ so that the vectors $x,v$ can be used in a chain.

With that in mind, suppose that we don't know the generalized eigenvectors of this matrix. we are looking for an eigenvector $v$ such that the equation $$ (A - I)x = v $$ has a solution. Note that we have $$ A - I = \pmatrix{0& 0& 0& -3\\ 0& 0& 0& 3\\ 0&0& 0& -1\\0& 0& 0& 0}. $$ In order for $(A - I)x = v$ to have a solution, $v$ must be an element of the column-space of $A - I$. So, $v$ must be a (non-zero) multiple of $(-3,3,1,0)$. We can simply take $v = (-3,3,1,0)$ (and note that $v$ is indeed an eigenvector of $A-I$). We now solve the equation $(A - I)x = v$ in the usual fashion: $$ \left( \begin{array}{cccc|c} 0& 0& 0& -3&-3\\ 0& 0& 0& 3&3\\ 0&0& 0& -1&-1\\0& 0& 0& 0&0 \end{array} \right) \leadsto \left( \begin{array}{cccc|c} 0& 0& 0& 1&1\\ 0& 0& 0& 0&0\\ 0&0& 0& 0&0\\0& 0& 0& 0&0 \end{array} \right). $$ In other words, our system of equations is equivalent to $x_4 = 1$, so any vector $x$ with $x_4 = 1$ solves the equation $(A - I)x = v$.

With that having be done, we can extend $x,v$ to get the following Jordan basis of $A$: $$ \{v,x,e_1,e_2\} = \{(-3,3,-1,0),(0,0,0,1),(1,0,0,0),(0,1,0,0)\}. $$

As Klaus's answer notes, there are other strategies for handling eigenspaces of dimension greater than $1$.

Ben Grossmann
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    for what it's worth: with integer matrix elements and integer eigenvalues, I like to reverse the order, take $b_4$ any vector with $(A-I)^2 b_4 = 0$ but $(A-I)b_4 \neq 0.$ Then $b_3 = (A-I)b_4$ is a nonzero genuine eigenvector, and one may choose $b_2, b_1$ eigenvectors. Put those as columns of $B.$ – Will Jagy Sep 13 '22 at 15:51
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    @WillJagy I agree. That said, the more generally difficult approach is typically the one presented in introductory linear algebra/differential equations classes, so I figure this kind of answer might be useful for posterity's sake – Ben Grossmann Sep 13 '22 at 16:37
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Yes, for any $x$, the vector $v=(I-A)x$ has its fourth coordinate equal to $0$, hence $v$ belongs to your eigen-hyperplane, hence $x$ is a generalized eigenvector, of rank $2$ if its own fourth coordinate is not $0.$

Anne Bauval
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  • If its 4th coordinate equals 0, then it is linearly dependent from the eigenvectors. – eMathHelp Sep 13 '22 at 14:19
  • Because generalized eigenvector should be linearly independent. – eMathHelp Sep 13 '22 at 14:23
  • Your answer doesn't provide a generalized eigenvector. – eMathHelp Sep 13 '22 at 14:26
  • It does: my conclusion was to take an $x$ with fourth coordinate $\ne0.$ To be more precise, according to https://en.wikipedia.org/wiki/Generalized_eigenvector#Overview_and_definition and references thereof, such an $x$ will be a generalized eigenvector of rank 2, whereas all other nonzero $x$'s will be g.e. of rank $1$, i.e. ordinary eigenvectors. – Anne Bauval Sep 13 '22 at 14:30