3
$\begingroup$

I need to find a $n\times n$ real matrix whose minimal polynomial is $x^{n-1}$. I know this transformation with respect to the basis $\{v_1,\dots,v_n\}$ $$T(v_1)=v_2$$ $$T(v_2)=v_3$$ $$\dots\dots$$ $$T(v_{n-1})=v_n$$ $$T(v_n)=0$$ is nilpotent with characteristic polynomial $X^n$.

Could any one help to how to solve the above one?

$\endgroup$
3
  • $\begingroup$ Gentle tip for future writing: please keep abbreviations to a minimum. While they were not too hard to guess from context ("charpoly")), you should not be making your readers work that hard to understand eccentric abbreviations. $\endgroup$ – rschwieb May 22 '13 at 13:59
  • $\begingroup$ @N.S. I'm confused: the linked thread is marked as a duplicate of the present thread and now people are voting to close this one as a duplicate of its own duplicate? $\endgroup$ – Martin May 22 '13 at 16:18
  • $\begingroup$ @Martin Ups, my mistake, this is the problem with soo many similar questions. I remembered I answered the same question sometime, missed that it was closed later for duplication. $\endgroup$ – N. S. May 22 '13 at 16:22
7
$\begingroup$

For $n=1$ there is no solution (the only matrix with minimal polynomial $1$ is the $0\times 0$ matrix). Otherwise there is a solution, but not the $T$ in your question, which has $T^{n-1}\neq0$. Instead, you need the kernel of $T$ to be $2$-dimensional, which you can achieve for instance by changing the definition to have $T(v_1)=0$. The matrix of $T$ on the basis $v_1,\ldots,v_n$ then becomes $$ \begin{pmatrix} 0&0&0&\ldots&0&0\\0&0&0&\ldots&0&0\\ 0&1&0&\ldots&0&0\\0&0&1&\ldots&0&0\\\vdots&\vdots&\ddots&\ddots&0&0\\ 0&0&0&\ldots&1&0\\ \end{pmatrix}. $$ Another (equivalent) soultion is to define $T(v_{n-2})=0$ instead. In either case there is a vector $v_i$ with $T^{n-2}(v_i)\neq0$, but $T^{n-1}=0$.

To see that you must have $\dim(\ker T)=2$, consider the sequence $\dim(\ker T^i)$ for $i=0,1,\ldots,n-1$; it is weakly increasing, starts with $0$ and must end with numbers $d,n$ where $d<n$ (since $T^{n-2}\neq0$). Moreover, as a general fact, the "derived" sequence of its increments is weakly decreasing. It then follows that this derived sequence is $2,1,1,\ldots,1$ and the original sequence $0,2,3,4,\ldots,n-1,n$, in particular $\dim(\ker T^1)=2$.

The fact that the derived sequence is weakly decreasing follows because $T$ induces a map $\ker(T^{i+2})/\ker(T^{i+1})\to\ker(T^{i+1})/\ker(T^i)$ that can be checked to be always injective.

$\endgroup$
3
$\begingroup$

What about the following variant of the famous companion matrix ?:

$$\begin{pmatrix}0&1&0&0&\ldots&0\\0&0&1&0&\ldots&0\\...&...&...&...&...&...\\0&0&...&...&...&\;\;0\\0&0&0&0&\ldots&0\end{pmatrix}$$

which, of course, is almost the matrix of your linear operator: you just need to define $\,Tv_{n-1}=0\,$...

$\endgroup$
8
  • $\begingroup$ if this is the same as mine then its minpoly is $x^n$, isnt it? $\endgroup$ – miosaki Jan 29 '13 at 12:14
  • $\begingroup$ No, it is precisely $\,x^{n-1}\,$ ...its characteristic polynomial is $\,x^n\,$, though. $\endgroup$ – DonAntonio Jan 29 '13 at 12:19
  • $\begingroup$ okay so I was correct at the begining :-o? $\endgroup$ – miosaki Jan 29 '13 at 12:20
  • 1
    $\begingroup$ Hehe...well, yes you were, but I still am not sure whether you can actually prove it. Pay attention the fact that if in an upper triangular matrix we have that the main diagonal and $\,k\,$ subdiagonals over it are all zeros, then the resulting matrix is nilpotent of degree $\,n-k-1\,$ . If you can prove this result, or at least for the easier particular case with $\,k=0\,$ then you're done. $\endgroup$ – DonAntonio Jan 29 '13 at 12:24
  • 1
    $\begingroup$ This is not correct, this is actually (the transpose of) the companion matrix of the polynommial $X^n$, and therefore its minimal and characteristic are both $X^n$. Think of the case $n=2$, the matrix is not zero, but its square is. $\endgroup$ – Marc van Leeuwen Jan 29 '13 at 12:52
0
$\begingroup$

EDIT: following answer is invalid and is left here only to remind me to think before I answer.

Look up "companion matrix." Given any polynomial $p$, the companion matrix of $p$ has minimal polynomial $p$.

$\endgroup$
2
  • $\begingroup$ I think the exact companion matrix doesn't work here as in that case the matrix has both the same characteristic and minimal polynomial. In this case, the OP wants $\,x^{n-1}\,$ to be the minimal pol., but the companion matrix always gives a pol. of degree $\,n\,$ ... $\endgroup$ – DonAntonio Jan 29 '13 at 12:10
  • 1
    $\begingroup$ @Don, yes, I wasn't thinking. $\endgroup$ – Gerry Myerson Jan 29 '13 at 12:14
-2
$\begingroup$

Hint: If $A=\begin{bmatrix}{0}&{0}&{1}&0\\{0}&{0}&{0}&1\\{0}&{0}&{0}&0\\{0}&{0}&{0}&0\end{bmatrix}$ then, $\mu_A(x)=x^3$. Generalize.

$\endgroup$
1
  • 1
    $\begingroup$ Be careful: the minimal polynomial of this matrix is $\,x^2\,$ , not $\,x^3\,$ ... $\endgroup$ – DonAntonio Jan 29 '13 at 12:31

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.