I've been staring at a chapter in Bill Cooperstein's Advanced Linear ALgebra for some time now and one section is giving me trouble. It is about elementary divisors and invariant factors. My question is this:

Let $S$ be an operator on a finite dimensional real vector space and assume that $U=[S,\mathbf{u}_1] \oplus [S,\mathbf{u}_2] \oplus\cdots\oplus [S,\mathbf{u}_6]$ where $[S,\mathbf{u}_i]$ is the $S$-cyclic subspace generated by $\mathbf{u}_i$, where $\mu_{S,\mathbf{u}_i}(x)$ is the minimal polynomial of each $S$-cyclic subspace

and \begin{align*} \mu_{S,\mathbf{u}_1}(x) &= \mu_{S,\mathbf{u}_2}(x) = (x^2+1)^5,\\ \mu_{S,\mathbf{u}_3}(x) &= (x^2+1)^4,\\ \mu_{S,\mathbf{u}_4}(x) &= \mu_{S,\mathbf{u}_5}(x) = (x^2+1)^2,\\ \mu_{S,\mathbf{u}_6}(x) &= x^2+1. \end{align*} Set $U_i = \{\mathbf{u}\in U \mid (S^2+I_U)^i(\mathbf{u}) = 0\}$ for $i = 1,\ldots,6$.

Determine the dimension of each $U_i$.

We know each minimal polynomial is irreducible but they are not distinct from each other. What does this tell us about each $S$-cyclic subspace for vector $\mathbf{u}_i$?

  • 1
    $\begingroup$ I tried to edit but I really don't understand. Got the FAQ section for directions on how to use LaTeX in this site, otherwise this question may pass on without many people even reading it. $\endgroup$
    – DonAntonio
    Mar 29, 2013 at 13:30
  • $\begingroup$ Thanks for the help with using LaTeX. I actually edited it myself, but someone beat me to it.... $\endgroup$ Mar 29, 2013 at 13:51
  • $\begingroup$ I'm afraid your work isn't finished yet: what is $\,\mu\,$? For what I see it seems to be some polynomial (the characteristic one? But then of what?) Is $\,[S,u_i]:=Span\{u_i, Su_i,S^2u_i,...\}\,$ , or what? $\endgroup$
    – DonAntonio
    Mar 29, 2013 at 13:56
  • $\begingroup$ I guess, about $[S,u_i]$ you're right, but $\mu(S,u_i)$ is rather the minimal polynomial of $S|_{[S,u_i]}$. $\endgroup$
    – Berci
    Mar 29, 2013 at 14:12
  • $\begingroup$ @DonAntonio, [S,$u_i$] is an S-cyclic subspace of an arbitrary n-dimensional vector space, say U in this case. IN other words, I have this vector, $u_i$. The S-cyclic subspace is the subspace {f(S)($u_i$) | f(x) $\in$ F[x]}. It is usually denoted <S,$u_i$> but I am still working on my latex and am having trouble with some of the code. The minimal polynomials are the unique monic polynomials of smallest degree that annihilate the vector under transformation S. So using this information, I can I pull out the dimension of each $U_i$? $\endgroup$ Mar 30, 2013 at 4:28

1 Answer 1


Here is a summary of the answer to this perfectly standard exercise. Feel free to ask more questions.

Fundamental fact Let $C$ be a cyclic subspace, and let $P$ be the minimal polynomial of $S$ on $C$. Then the dimensionality of $C$ is exactly the degree of $P$.

Indeed, if we denote this degree by $d$ and if the cylic subspace $C$ is generated by a vector $u$, then $(u,Su,S^2u, \ldots ,S^{d-1}u)$ is a basis of $C$.

So if you denote by $P_i$ the minimal polynomial of $S$ on $[S,u_i]$, then the dimensionality of $[S,u_i]$ is the degree of $P_i$.

Next, for each $i$ we have $U_i=(U_i\cap[S,\mathbf{u}_1]) \oplus (U_i\cap[S,\mathbf{u}_2]) \oplus\cdots\oplus (U_i\cap[S,\mathbf{u}_6])$ and for each $j$ $(U_i\cap[S,\mathbf{u}_j])$ is again a cyclic subspace, so you can use the fundamental fact above again. Finally :

$$ \begin{array}{lccl} {\sf dim}(U_0) &=& 0+0+0+0+0+0 &=& 0 \\ {\sf dim}(U_1) &=& 2+2+2+2+2+2 &=& 12 \\ {\sf dim}(U_2) &=& 4+4+4+4+4+2 &=& 22 \\ {\sf dim}(U_3) &=& 6+6+6+4+4+2 &=& 28 \\ {\sf dim}(U_4) &=& 8+8+8+4+4+2 &=& 34 \\ {\sf dim}(U_5) &=& 10+10+8+4+4+2 &=& 38 \\ {\sf dim}(U_6) &=& 10+10+8+4+4+2 &=& 38 \\ \end{array} $$

UPDATE : Explanation on my second claim

Let $V_j=U_i \cap [S,u_j] $. Let us show that $U_i =\oplus_{j} V_j$. First, as $V_j \subseteq [S,u_j]$ we see that the sum $\sum_{j} V_j$ is direct, and as $V_j \subseteq U_i$ we see that the sum is contained in $U_i$.

Conversely, let $w\in U_i$. Because of $U=[S,\mathbf{u}_1] \oplus [S,\mathbf{u}_2] \oplus\cdots\oplus [S,\mathbf{u}_6]$, there are polynomials $Q_1,Q_2, \ldots ,Q_6$ such that

$$ w=\sum_{j=1}^6 w_j, \ \text{with} \ w_j=Q_j(S)u_j \in [S,u_j] $$

Then $$ O=(I+S^2)^i w= \sum_{j=1}^6 ((I+S^2)^i)w_j $$

Note that in the sum above, the summand number $j$ is in $[S,u_j]$. By the direct sum property, each of those summands must be zero. In other words, $w_j \in V_j$ as wished.

  • $\begingroup$ Everything up to your second to last step I had a grasp on but wasn't sure 100%. I saw the basis theorem of a minimal polynomial with degree d and was using during my first attempts. However, the second major claim of for each $i$, $U_i$ = ...., what drove you to that step? Why is each $U_i$ the direct sum of $U_i$ intersected with $[S,**u**_j]$ for $j=1..6$? That is what I am not understanding. By the say, that is the correct answer as the author has it, but change $dim(U_1) = 12$ not 6. $\endgroup$ Apr 1, 2013 at 12:17
  • $\begingroup$ @ChristopherErnst : please see my update above. $\endgroup$ Apr 1, 2013 at 13:05
  • $\begingroup$ So each $U_i$ annihilates a vector $u_j$. In the intersection of each $U_i$ and $[S,u_j]$, whichever polynomial is of least degree becomes the minimal for this new $V_i$ that you defined. Over all six cyclic subspaces then you simply sum the dimensions for each $i$ from 1 to 6. I think I see it. This has been by far the hardest section I've studied. $\endgroup$ Apr 1, 2013 at 13:18
  • $\begingroup$ is my prior statement correct? Do I have the correct interpretation? $\endgroup$ Apr 2, 2013 at 1:13
  • $\begingroup$ @ChristopherErnst To tell the truth, your prior statement looks rather nonsensical to me. Vectors are annihilated by linear maps, not subspaces. $\endgroup$ Apr 2, 2013 at 6:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .