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In our lecture notes, there's the following example problem.

Find a Jordan normal form matrix that is similar to the following. $$A=\begin{bmatrix}2 & 0 & 0 & 0\\-1 & 1 & 0 & 0\\0 & -1 & 0 & -1\\1 & 1 & 1 & 2\end{bmatrix}$$

The solution begins by demonstrating that the matrix $XI-A$ is equivalent to a matrix of the form

$$B=\begin{bmatrix}-1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & -(X-1) & 0\\0 & 0 & 0 & (X-2)(X-1)^2\end{bmatrix}$$

From here, the notes immediately deduce that the Jordan normal form of $A$ is $$J=\begin{bmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 1 & 1 & 0\\0 & 0 & 0 & 2\end{bmatrix}.$$

I'm struggling to understand this final step - how do we get from $B$ to $J$? I get that $J$ consists of three elementary Jordan matrices, but how are these matrices actually found?

Rodrigo de Azevedo
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goblin GONE
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  • There must be a typo somewhere. If $A$ is equiavalent to $B$ (which is a diagonal matrix), then Jordan form of $A$ must be diagonal, too. In fact, in this case, $B\big|_{X=0}$ is the Jordan form of $A$. – TZakrevskiy Jun 15 '13 at 10:27
  • @TZakrevskiy, I've just triple-checked that everything is written down correctly, so if there's a typo, it's not my fault. – goblin GONE Jun 15 '13 at 10:29
  • Well, maybe they use some weird definition of "equivalent"?. – TZakrevskiy Jun 15 '13 at 10:47
  • @TZakrevskiy, two matrices $U$ and $V$ are equivalent, according to the notes, if there exist invertible matrices $G$ and $H$ such that $GUH=V.$ Note however that we're not claiming that $A$ and $B$ are equivalent, but rather that $XI-A$ and $B$ are equivalent. – goblin GONE Jun 15 '13 at 10:50
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    My bad, I confused equivalence of matrices and similarity of matrices. – TZakrevskiy Jun 15 '13 at 10:56
  • @TZakrevskiy, ahh no worries. – goblin GONE Jun 15 '13 at 10:57

3 Answers3

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The method used in the lecture notes is based $\def\KX{K[X]}$on the structure theorem for finitely generated $\KX$-modules (more generally f.g. modules over a PID) rather than on purely linear algebraic considerations. What the given computation shows is that the $\KX$-module, defined by having $X$ act on $K^4$ by the matrix$~A$, is isomorphic to $\KX/\langle1\rangle\oplus\KX/\langle1\rangle\oplus\KX/\langle X-1\rangle\oplus\KX/\langle(X-2)(X-1)^2\rangle$ (the first two summands are trivial and can be omitted). The final factor can be further decomposed as $\KX/\langle(X-2)(X-1)^2\rangle\cong\KX/\langle X-2\rangle\oplus\KX/\langle(X-1)^2\rangle$ because of a theorem for which I am still wondering if it has a name in English (in French it is the lemme des noyaux, it is related to the Chinese remainder theorem, but not the same as it is about modules rather than rings).

So our $\KX$-module is isomorphic to $\KX/\langle X-1\rangle\oplus\KX/\langle(X-1)^2\rangle\oplus\KX/\langle X-2\rangle$, where I have moved to the front the two factors that contain eigenvectors for the eigenvalue$~1$. The outermost summands are of dimension$~1$, and $X$ acts as a scalar $1$ respectively $2$ on each of them; in the middle factor of dimension$~2$, the action of $X$ has eigenvalue$~1$ but is not diagonalisable: it is a Jordan block of size$~2$. So this is how one gets two Jordan blocks of sizes $1,2$ for $\lambda=1$ and a single Jordan block of size$~1$ for $\lambda=2$.

I'll add a word about the initial computation leading to the diagonal form. The algorithm applied is that of computing the Smith normal form of the matrix $XI-A$ over $\KX$ (which is a PID), using row and column operations with scalars in $\KX$ to arrive at a diagonal form in which successive diagonal entries each divide the next one. I am not sure whether your lecture notes explain why one should start with $XI-A$, and why the result describes a cyclic decomposition of the $\KX$-module defined by$~A$, so here it is. Intermediate matrices describe a presentation of a $\KX$-module as quotient of the free module $\KX^n$, where $n$ is the number of rows, by the sub-module $N$ generated by the columns of the matrix. During the algorithm, column operations correspond to changing to different generators of $N$, while row operations correspond to choosing a different basis of $\KX^n$ than the standard one, and changing coordinates (of the generators of$~N$) to the new basis. At the end of the transformation one arrives at a situation in which all generators of$~N$ are (polynomial) multiples of the basis of $\KX^n$ used, whence the diagonal form, and this describes the module as a direct sum of cyclic sub-modules.

So why does $XI-A$ describe the $\KX$-module$~M$ defined on $K^n$ by the action of$~A$? Because the standard basis $(e_1,\ldots,e_n)$ of $K^n$ is certainly a set of generators for$~M$ (probably quite redundant) so we have a surjective morphism $f:\KX^n\to M$, and column$~j$ of $XI-A$ describes $Xe_j-\sum_i A_{i,j}e_i\in\KX^n$ which element, given that $X$ acts on $M$ as $A$, is in the kernel $N$ of $f$ by definition; it is not hard to see that these elements generate all of$~N$, so that they give a complete presentation of$~M$.

Whether this method is a very practical manner to find a Jordan normal form is questionable; in my experience performing the Smith normal form algorithm over $\KX$ is extremely tedious an error prone by hand. However, apart from the theoretical importance, it is certainly an algorithm, while I think methods of finding the Jordan normal form by linear algebra only tend to be hard to describe as a complete algorithm (in practice many shortcuts are possible, but a complete method covering all possibilities is long to describe). I've tried to describe a linear algebraic algorithm (as well as the one above) for the somewhat coarser Rational Canonical (or Frobenius Normal) Form in this answer.

Community
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Marc van Leeuwen
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  • Thanks for this. I just reread this answer four years later (how time flies!). Can you tell me if this is correct? Given $P \in \mathbb{C}[X]$, we get a corresponding collection of matrices $\Phi(P)$ by assigning to each root of $P$ an appropriate Jordan matrix, choosing an order on the roots, and then stacking them into a square matrix. We can also go the other way; given a complex matrix $A$, we can form the matrix $X-A$ and then compute the determinant; lets call the result $\Psi(A) := \mathrm{det}(X-A)$. The important observation is that $\mathrm{JNF}(A) = \Phi(\Psi(A))$, where... – goblin GONE Jul 20 '17 at 16:48
  • $\mathrm{JNF}(A)$ is the set of Jordan normal forms of $A$. Is this right? If so, I might write it up as an answer. – goblin GONE Jul 20 '17 at 16:48
  • No that cannot work. If I understand you correctly $\Psi(A)$ is just the characteristic polynomial of $A$; from this information alone the Jordan normal form cannot be deduced. Nor can it be in a non contrived way from any other polynomial whose roots are eigenvalues of $A$, simply because there are too many different possible Jordan normal forms. For instance there are 11 different Jordan normal forms of $6\times6$ nilpotent (all eigenvalues $0$) matrices. – Marc van Leeuwen Jul 20 '17 at 23:29
  • How does one find invertible matrix $T$ such that $A = TJT^{-1}$ using this method? Or does that have to be found by other means? – goblin GONE Jun 10 '18 at 10:26
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$A$ has eigenvalue $\lambda_1=2$ of multiplicity $1$ and $\lambda_2=1$ of multiplicty $3$. Apparently, the Jordan block corresponding to $2$ is a $1\times 1$ matrix and thus uniteresting for us.

Next, we study $\text{rank} (A-I)$. It's easy to check that the rank is equal to $2$, so we conclude that $A$ has $4-2=2$ linearly independent eigenvectors corresponding to eigenvalue $1$. It yields the structure of Jordan blocks: one is $1\times 1$ matrix and the other one is a $2\times 2$ (other possibilities are ruled out: three $1\times 1$ matrices would require 3 eigenvectors and one $3\times 3$ would require only one eigenvector). Thus, we conclude that the Jordan form is indeed $$J=\begin{bmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 1 & 1 & 0\\0 & 0 & 0 & 2\end{bmatrix}.$$

In more general case, the algorithm is a little bit more complex: you study eigenvalues, then eigenvectors, and then generalized eigenvectors.

Marc van Leeuwen
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TZakrevskiy
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  • So basically, since $\mathrm{rank}(A)-\mathrm{rank}(A-\lambda I) = 4-2=2$ for $\lambda=1,$ thus $J$ must have precisely two elementary Jordan elementary matrices corresponding to the eigenvalue $1$?? Is this a general principle? – goblin GONE Jun 15 '13 at 10:56
  • Yes, the number of eigenvectors corresponding to a given eigenvalue is equal to the number of Jordan cells corresponding to that eigenvalue. – TZakrevskiy Jun 15 '13 at 10:59
  • This is a correct computation, but the method is fundamentally different than the one used in the text cited. – Marc van Leeuwen Jun 15 '13 at 11:17
  • @MarcvanLeeuwen Indeed it is; and, as an additional benefit, it is intuitive and relies only on the idea behind Jordan decomposition. – TZakrevskiy Jun 15 '13 at 11:40
  • This method is certainly simpler, at least for hand computation, in most concrete cases (as I indicate in my answer). But giving a complete algorithm which in particular will in all cases provide a Jordan basis as well is surprisingly difficult. – Marc van Leeuwen Jun 15 '13 at 13:06
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Let

$${ A = {\begin{pmatrix} 2 &0 &0 &0 \\ -1 &1 &0 &0 \\ 0 &-1 &0 &-1 \\ 1 &1 &1 &2 \end{pmatrix}} . }$$

The goal is to find a Jordan normal form matrix similar to ${ A . }$ We can follow the procedure as in the proof here.

Note that

$${ \det(tI - A) = (t-2) (t-1) ^3 . }$$

Hence ${ 1, 2 }$ are the eigenvalues of ${ A . }$

Consider the eigenvalue ${ 1 . }$ Consider the powers

$${ (A - 1I) , (A - 1I) ^2, \ldots }$$

Note that the powers

$${ (A - 1I) = {\begin{pmatrix} 1 &0 &0 &0 \\ -1 &0 &0 &0 \\ 0 &-1 &-1 &-1 \\ 1 &1 &1 &1 \end{pmatrix}} , }$$

$${ {\begin{aligned} &\, (A - 1 I) ^2 \\ = &\, {\begin{pmatrix} 1 &0 &0 &0 \\ -1 &0 &0 &0 \\ 0 &-1 &-1 &-1 \\ 1 &1 &1 &1 \end{pmatrix}} {\begin{pmatrix} 1 &0 &0 &0 \\ -1 &0 &0 &0 \\ 0 &-1 &-1 &-1 \\ 1 &1 &1 &1 \end{pmatrix}} \\ = &\, {\begin{pmatrix} 1 &0 &0 &0 \\ -1 &0 &0 &0 \\ 0 &0 &0 &0 \\ 1 &0 &0 &0 \end{pmatrix}}, \end{aligned}} }$$

$${ {\begin{aligned} &\, (A - 1 I) ^3 \\ = &\, {\begin{pmatrix} 1 &0 &0 &0 \\ -1 &0 &0 &0 \\ 0 &-1 &-1 &-1 \\ 1 &1 &1 &1 \end{pmatrix}} {\begin{pmatrix} 1 &0 &0 &0 \\ -1 &0 &0 &0 \\ 0 &0 &0 &0 \\ 1 &0 &0 &0 \end{pmatrix}} \\ = &\, {\begin{pmatrix} 1 &0 &0 &0 \\ -1 &0 &0 &0 \\ 0 &0 &0 &0 \\ 1 &0 &0 &0 \end{pmatrix}}, \end{aligned}} }$$

$${ {\begin{aligned} &\, (A - 1 I) ^4 \\ = &\, {\begin{pmatrix} 1 &0 &0 &0 \\ -1 &0 &0 &0 \\ 0 &-1 &-1 &-1 \\ 1 &1 &1 &1 \end{pmatrix}} {\begin{pmatrix} 1 &0 &0 &0 \\ -1 &0 &0 &0 \\ 0 &0 &0 &0 \\ 1 &0 &0 &0 \end{pmatrix}} \\ = &\, {\begin{pmatrix} 1 &0 &0 &0 \\ -1 &0 &0 &0 \\ 0 &0 &0 &0 \\ 1 &0 &0 &0 \end{pmatrix}}, \end{aligned}} }$$

and so on.

Hence the stabilized subspaces

$${ {\begin{aligned} &\, \mathcal{K} := \text{ker}((A - 1I) ^2) = \text{ker}((A - 1I) ^3) = \ldots \\ &\, \mathcal{I} := \text{im}((A - 1I) ^2) = \text{im}((A - 1I) ^3) = \ldots \end{aligned}} }$$

Hence

$${ {\begin{aligned} &\, \mathcal{K} = \text{span}(e _2, e _3, e _4), \\ &\, \mathcal{I} = \text{span} \begin{pmatrix} 1 \\ -1 \\ 0 \\ 1 \end{pmatrix} . \end{aligned}} }$$

Note that by row reduction, the kernel

$${ \text{ker}(A - 1 I) = \text{span} \left\lbrace \begin{pmatrix} 0 \\ -1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ -1 \\ 0 \\ 1 \end{pmatrix} \right\rbrace . }$$

We can study the action of ${ (A - 1I) \vert _{\mathcal{I}} }$ on ${ \mathcal{I} . }$

Note that

$${ (A - 1I) \begin{pmatrix} 1 \\ -1 \\ 0 \\ 1 \end{pmatrix} = 1 \begin{pmatrix} 1 \\ -1 \\ 0 \\ 1 \end{pmatrix} . }$$

Hence the Jordan decomposition of ${ \mathcal{I} }$ with respect to ${ (A - 1I) \vert _{\mathcal{I}} }$ is

$${ \boxed{ \mathcal{I} = \text{span} \, \mathscr{B} _{A - 1I} \left( \begin{pmatrix} 1 \\ -1 \\ 0 \\ 1 \end{pmatrix} , 1 \right) }. }$$

We can study the action of ${ T := (A - 1I) \vert _{\mathcal{K}} }$ on ${ \mathcal{K}. }$

Note that

$${ {\begin{aligned} &\, \text{im}(T) \\ = &\, \text{im}((A - 1I) \vert _{\mathcal{K}}) \\ = &\, \left\lbrace {\begin{pmatrix} 1 &0 &0 &0 \\ -1 &0 &0 &0 \\ 0 &-1 &-1 &-1 \\ 1 &1 &1 &1 \end{pmatrix}} \begin{pmatrix} 0 \\ x _2 \\ x _3 \\ x _4 \end{pmatrix} : x _2, x _3, x _4 \in \mathbb{C} \right\rbrace \\ = &\, \left\lbrace \begin{pmatrix} 0 \\ 0 \\ - x _2 - x _3 - x _4 \\ x _2 + x _3 + x _4 \end{pmatrix} : x _2, x _3, x _4 \in \mathbb{C} \right\rbrace \\ = &\, \text{span} \begin{pmatrix} 0 \\ 0 \\ -1 \\ 1 \end{pmatrix} . \end{aligned}} }$$

We can study the action of ${ T \vert _{\text{im}(T)} }$ on ${ \text{im}(T) . }$

Note that

$${ (A - 1I) \begin{pmatrix} 0 \\ 0 \\ -1 \\ 1 \end{pmatrix} = 0 \begin{pmatrix} 0 \\ 0 \\ -1 \\ 1 \end{pmatrix} . }$$

Hence the Jordan decomposition of ${ \text{im}(T) }$ with respect to ${ T \vert _{\text{im}(T)} }$ is

$${ \text{im}(T) = \text{span} \, \mathscr{B} _{A - 1I} \left(\begin{pmatrix} 0 \\ 0 \\ -1 \\ 1 \end{pmatrix} , 0 \right) . }$$

We can pick a ${ v \in \mathcal{K} }$ such that ${ T(v) = \begin{pmatrix} 0 \\ 0 \\ -1 \\ 1 \end{pmatrix} . }$

For example, ${ v := e _2 }$ works.

Hence consider

$${ {\begin{aligned} &\, \mathscr{B} _{A - 1I} (v, 0) \\ = &\, \mathscr{B} _{A - 1I} (e _2, 0) \\ = &\, \left( e _2, \begin{pmatrix} 0 \\ 0 \\ -1 \\ 1 \end{pmatrix} \right) . \end{aligned}} }$$

We can pick a ${ z _1 \in \text{ker}(T) }$ such that ${ ( \mathscr{B} _{A - 1I} (v, 0); z _1) }$ is a basis of ${ \mathcal{K} . }$

For example, ${ z _1 := \begin{pmatrix} 0 \\ -1 \\ 1 \\ 0 \end{pmatrix} }$ works.

Hence the Jordan decomposition of ${ \mathcal{K} }$ with respect to ${ (A - 1I) \vert _{\mathcal{K}} }$ is

$${ \boxed{\mathcal{K} = \text{span} \, \mathscr{B} _{A - 1I} \left( e _2, 0 \right) \oplus \text{span} \, \mathscr{B} _{A - 1I} \left( \begin{pmatrix} 0 \\ -1 \\ 1 \\ 0 \end{pmatrix}, 0 \right)} . }$$

Hence the Jordan decomposition of ${ \mathbb{C} ^4 = \mathcal{K} \oplus \mathcal{I} }$ with respect to ${ (A - 1I) }$ is

$${ \boxed{\mathbb{C} ^4 = \text{span} \, \mathscr{B} _{A - 1I} \left( e _2, 0 \right) \oplus \text{span} \, \mathscr{B} _{A - 1I} \left( \begin{pmatrix} 0 \\ -1 \\ 1 \\ 0 \end{pmatrix}, 0 \right) \oplus \text{span} \, \mathscr{B} _{A - 1I} \left( \begin{pmatrix} 1 \\ -1 \\ 0 \\ 1 \end{pmatrix} , 1 \right) } . }$$

Verification:

Consider the basis

$${ {\begin{aligned} \mathscr{B} := &\, \left( e _2 , (A - 1I) e _2 ; \begin{pmatrix} 0 \\ -1 \\ 1 \\ 0 \end{pmatrix} ; \begin{pmatrix} 1 \\ -1 \\ 0 \\ 1 \end{pmatrix} \right) \\ = &\, \left( \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} , \begin{pmatrix} 0 \\ 0 \\ -1 \\ 1 \end{pmatrix} ; \begin{pmatrix} 0 \\ -1 \\ 1 \\ 0 \end{pmatrix} ; \begin{pmatrix} 1 \\ -1 \\ 0 \\ 1 \end{pmatrix} \right) . \end{aligned}} }$$

Note that

$${ (A - 1I) \mathscr{B} = \mathscr{B} \begin{pmatrix} 0 &0 &0 &0 \\ 1 &0 &0 &0 \\ 0 &0 &0 &0 \\ 0 &0 &0 &1 \end{pmatrix} , }$$

that is

$${ A \mathscr{B} = \mathscr{B} \begin{pmatrix} 1 &0 &0 &0 \\ 1 &1 &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 &2 \end{pmatrix} . }$$

Hence a Jordan normal form matrix similar to ${ A }$ is

$${ \boxed{A \sim \begin{pmatrix} 1 &0 &0 &0 \\ 1 &1 &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 &2 \end{pmatrix} } . }$$

Venkata Karthik Bandaru
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