# find a $n\times n$ real matrix whose minimal polynomial is $x^{n-1}$

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?

• 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. – rschwieb May 22 '13 at 13:59
• @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? – Martin May 22 '13 at 16:18
• @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. – N. S. May 22 '13 at 16:22

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.

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 $\,Tv_{n-1}=0\,$...

• if this is the same as mine then its minpoly is $x^n$, isnt it? – miosaki Jan 29 '13 at 12:14
• No, it is precisely $\,x^{n-1}\,$ ...its characteristic polynomial is $\,x^n\,$, though. – DonAntonio Jan 29 '13 at 12:19
• okay so I was correct at the begining :-o? – miosaki Jan 29 '13 at 12:20
• 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. – DonAntonio Jan 29 '13 at 12:24
• 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. – Marc van Leeuwen Jan 29 '13 at 12:52

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 $p$.

• 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 $\,n\,$ ... – DonAntonio Jan 29 '13 at 12:10
• @Don, yes, I wasn't thinking. – Gerry Myerson Jan 29 '13 at 12:14

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.

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