4
$\begingroup$

I am currently reading two linear algebra books. One is Hoffman/Kunze's and the other one is Friedberg/Insel/Spence's. They define Jordan canonical form of linear operator in different ways. In Hoffman's book, Jordan forms are obtained by the primary decomposition and cyclic decomposition. In this case, 1's can appear some sub-diagonal entries. For example,

$M = \begin{array}{cc} 2 & 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 0 & 3 \\ \end{array}$

But in Friedberg's (And almost any other linear algebra books for engineers..), 1's can appear some super-diagonal entries. For example,

$M' = \begin{array}{cc} 2 & 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 3 \\ \end{array}$

Can someone explain why they define differently and relation of each other?

$\endgroup$
  • 3
    $\begingroup$ Since every matrix is similar to its transpose, then you get the same equivalence classes. Notice that $M$ and $M'$ are transpose. As to why: I'm not sure entirely. I think it's a matter of taste. $\endgroup$ – Philip Hoskins Jul 3 '16 at 9:06
1
$\begingroup$

Assume that $T \colon V \rightarrow V$ is a nilpotent operator and we can find a vector $v \in V$ for which the set $\{ v, Tv, T^2v, \dots, T^{n-1}v \}$ is a basis for $V$ so that the Jordan form of $T$ has one block. Thus, we have the following picture:

$$ v \mapsto Tv \mapsto \dots \mapsto T^{n-1}v \mapsto 0. $$

Such a basis is called a chain basis for $V$ (with respect to $T$). The matrix of $T$ with respect to the ordered basis $\mathcal{B} = (v, Tv, \dots, T^{n-1}v)$ is given by

$$ [T]_{\mathcal{B}} = \begin{pmatrix} 0 & 0 & \dots & \dots & 0 \\ 1 & 0 & \dots & \dots & 0 \\ 0 & 1 & 0 & \dots & \vdots \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \dots & 0 & 1 & 0 \end{pmatrix} $$

while the matrix of $T$ with respect to the same basis ordered differently as $\mathcal{B}' = (T^{n-1}v, T^{n-2}v, \dots, v)$ is given by

$$ [T]_{\mathcal{B}'} = \begin{pmatrix} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ 0 & 0 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 & 1 \\ 0 & \dots & \dots & 0 & 0 \end{pmatrix}. $$

Thus, the two variants of the Jordan form of $T$ in this case correspond to choosing two different orders for a chain basis of $V$ with respect to $T$. Choosing between the two different orders is largely a matter of convention. This obviously generalizes when the Jordan form of $T$ has more than one Jordan block.

$\endgroup$
1
$\begingroup$

the deference is that here a change base by a particular permutation on the vectors base and we get in the first base one lower Jordan matrix and in the second base an upper Jordan matrix, as in your case the first base is of the form $\{f(u),u,f(v),v,w\}$ where $u$ is a particular eigenvector associated to the eigenvalue $2$, and $v,w$ two eigenvectors associated to the eigenvalue $3$ ( withe $v$ convenable), and the second Jordan form is the matrix of $f$ relatively to the base $\{u,f(u),v,f(v),w\}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.