# Finding the minimal polynomial of an $n \times n$ matrix

How would one find the minimal polynomial of

$$A= \begin{pmatrix}0&-1&-2&\cdots&1-n\\1&0&-1&\cdots&2-n\\2&1&0&\cdots&3-n\\\vdots&\vdots&\vdots&\ddots&\vdots\\n-1&n-2&n-3&\cdots&0\end{pmatrix}$$

Where $$A$$ is an $$n \times n$$ matrix with $$n\ge 3$$?

• Not sure if it is helpful, but experimentally for first the few $n$ the characteristic polynomial is $(-1)^{n}x^{n-2}(a+x^2)$, where $a=n^2(n^2-1)/12$. The $a$ was found through OEIS. – Randy Savage Nov 14 '19 at 15:01
• And then the minimal polynomial is $x(a + x^2)$. – Robert Israel Nov 14 '19 at 15:44

Let $$U=\begin{pmatrix} 0 & -1 \\ 1 & -1 \\ 2 & -1 \\ \vdots & \vdots \\ n-1 & -1 \end{pmatrix} \;\;\;\text{and} \;\;\; V=\begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 0 & 1 & 2 & \cdots & n-1 \end{pmatrix}$$ Then $$A=UV.$$ Therefore, $$\mathrm{rank}(A)=2,$$ and $$A$$ has only $$2$$ non-zero eigenvalues. $$iA$$ is hermitian. This means $$A$$ is diagonalizable (in $$\mathbb{C}$$) and the minimal polynomial is square-free. As $$iA$$ has real eigenvalues ($$iA$$ is hermitian), the non-zero eigenvalues of $$A$$ are pure imaginary. Furthermore, the non-zero eigenvalues must appear in conjugate pairs, because $$A$$ is real.
If we put all this together, we get $$\mu_A(x) = x^3+ax$$ for a suitable $$a.$$
In order to find $$a$$, we take a look at $$\mu_A(A)$$ : $$\mu_A(A) = A^3+aA = UVUVUV+aUV = U\left((VU)^2+aI\right)V = 0$$ $$(VU)^2$$ can be computed using Faulhaber's formulas. We find: $$(VU)^2 = \begin{pmatrix} -\frac{n^2(n^2-1)}{12} & 0 \\ 0 & -\frac{n^2(n^2-1)}{12} \end{pmatrix}$$ Therefore, $$a=\frac{n^2(n^2-1)}{12}$$ and we have found our minimal polynomial.
• The first step is brilliant. Is this just an inspiration, or is there some general way to recognize that $A$ is of rank $2$? – saulspatz Nov 14 '19 at 16:17
• Wait, I see it now. Start from $a_{ij}=i-j.$ I noticed this, bu didn't think of factoring $A$. I still think you're brilliant, but I feel stupid. – saulspatz Nov 14 '19 at 16:23
• I tried a few other things (even thought about induction) before I noticed that each row can be obtained from the row above by adding $(1\,1\,\cdots\,1).$ Once you have this, the factorization is simple. – Reinhard Meier Nov 14 '19 at 16:27
• @David You know the characteristic polynomial, it is $x^{n-2}(x^2+a).$ So the eigenvalues are $0$ with algebraic and geometric multiplicity $n-2$ and $\sqrt{a}i$ and $-\sqrt{a}i.$ – Reinhard Meier Nov 14 '19 at 16:56