Upper triangular matrix/Finding Jordan normal form/Generalized eigenspace/Method

We describe how to find to a linear trigonalizable mapping ${}\varphi \colon V\rightarrow V$ a basis, such that describing matrix with respect to this basis is in Jordan normal form. For this, we determine, for every eigenvalue ${}\lambda \in K$, the minimal exponent ${}s$ with

${}\operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}^{s}=\operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}^{s+1}\,.$

This kernel is the generalized eigenspace to ${}\lambda$. We set

${}V_{i}=\operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}^{i}\subseteq \operatorname {GeEig} _{\lambda }(\varphi )\,$

for ${}i=1,\ldots ,s$. This yields the chain

${}V_{1}=\operatorname {Eig} _{}{\left(\lambda \right)}\subseteq V_{2}\subset \cdots \subset V_{s-1}\subset V_{s}=\operatorname {GeEig} _{\lambda }(\varphi )\,.$

Now, we choose a vector ${}u$ from ${}V_{s}\setminus V_{s-1}$. The vectors

$u,(\varphi -\lambda \operatorname {Id} )(u),(\varphi -\lambda \operatorname {Id} )^{2}(u),\ldots ,(\varphi -\lambda \operatorname {Id} )^{s-1}(u)$

form a basis for a Jordan block. If this basis generates the generalized eigenspace, then we are done. Otherwise, we look in ${}V_{s}\setminus V_{s-1}$ for another vector which is linearly independent to ${}u$ and to ${}V_{s-1}$. Again, we add this vector and all its successive images. If ${}V_{s}\setminus V_{s-1}$ is exhausted, then we look whether ${}V_{s-1}\setminus V_{s-2}$ is already covered, and so on. If the generalized eigenspace to ${}\lambda$ is covered, then we continue with the next eigenvalue.

Under certain circumstances, we can also start with a basis of the eigenspace. If, for example, the eigenspace to ${}\lambda$ is one-dimensional, then we can choose an eigenvector ${}v$ for ${}\lambda$, and we can find successively preimages under ${}\varphi -\lambda \operatorname {Id} _{V}$ of the vectors, that is, we have to solve the equation

${}v={\left(\varphi -\lambda \operatorname {Id} _{V}\right)}(v')\,,$

then

${}v'={\left(\varphi -\lambda \operatorname {Id} _{V}\right)}(v^{\prime \prime })\,,$

etc.

If, for example, the eigenspace is ${}k$-dimensional and the generalized eigenspace is ${}(k+1)$-dimensional, then we only have to find a preimage for one eigenvector under ${}\varphi -\lambda \operatorname {Id} _{V}$.