6
$\begingroup$

I am trying to learn some linear algebra, and currently I am having a difficulty time grasping some of the concepts. I have this problem I found that I have no idea how to start.

Assume that $\bf A$ is an $n\times n$ complex matrix which has a cyclic vector. Prove that if $\bf B$ is an $n\times n$ complex matrix that commutes with $\bf A$, then ${\bf B}=p({\bf A})$ for some polynomial $p$.

All I know at this point is that ${\bf AB}={\bf BA}$.

$\endgroup$
4
  • 1
    $\begingroup$ Do you know what it means for a matrix to have a cyclic vector? $\endgroup$ – Gerry Myerson Aug 4 '12 at 0:02
  • $\begingroup$ @Gerry, I think you have hit it. See her other question. P.S. this is the first time I have heard of a cyclic vector. $\endgroup$ – Will Jagy Aug 4 '12 at 0:09
  • $\begingroup$ A cyclic vector of $A$ is an element $v \in C^{n}$ such that $A^{0}v,A{1}v,...,A^{n-1}v$ are linearly independent. @ Will...I went to your link, and I saw the Thm, but I do not understand how to show that (I) implies 2-4 statements. Yes, I am new to this, but I want to learn it. $\endgroup$ – Melky Aug 4 '12 at 0:44
  • $\begingroup$ @Beth, I give some links at the other answer, and mention at least one book. Any part of the equivalence is rather long for a website answer, but I give an easy example where an eigenvalue in two separate Jordan blocks makes a specific problem. $\endgroup$ – Will Jagy Aug 4 '12 at 1:18
10
$\begingroup$

Since $A^0v,A^1v,\dots,A^{n-1}v$ are linearly independent, they form a basis for ${\bf C}^n$. Thus, $$Bv=c_0A^0v+c_1A^1v+\cdots+c_{n-1}A^{n-1}v=p(A)v$$ where $$p(x)=c_0+c_1x+\cdots+c_{n-1}x^{n-1}$$ for some constants $c_0,c_1,\dots,c_{n-1}$. Since $B$ commutes with $A$, it commutes with all powers of $A$, so $$B(A^rv)=A^rBv=A^rp(A)v=p(A)(A^rv)$$ for $r=0,1,\dots,n-1$ (I've used $A^rp(A)=p(A)A^r$). But again the vectors $A^rv$ are a basis, so $B=p(A)$ ($Bx=Cx$ for all $x$ in a basis implies $Bx=Cx$ for all $x$ in the vector space, which implies $B=C$).

$\endgroup$
1
  • $\begingroup$ Nice solution... $\endgroup$ – copper.hat Aug 4 '12 at 3:43
1
$\begingroup$

see my answer at Given a matrix, is there always another matrix which commutes with it? and
cyclic vectors- cyclic subspaces

I will see what else there is here on cyclic vectors. Here is all that is needed. As Gerry points out, if you are a beginner at this it is hard to understand why you are asking about this topic, but what the hell: http://planetmath.org/?op=getobj&id=5690&from=objects

$\endgroup$

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.