Given $\begin{pmatrix} 1 & 1 & \cdots & 1 \\ 2 & 2 & \cdots & 2 \\ \cdot & \cdot & \cdots & \cdot \\ n & n & \cdots & n \end{pmatrix}$, calculate the minimal polynomial.
I know that the characteristic polyomial is:
$$P_A(x) = x^{n-1} \cdot (x – \frac{n(n+1)}{2}) $$
But how do I exactly know the minimal polynomial? I am sensing that it is by using the Cayley-Hamilton theory!
Best Answer
If $$ A=\begin{pmatrix} 1 & 1 & ... & 1 \\ 2 & 2 & ... & 2 \\ . & . & ... & . \\ n & n & ...& n \end{pmatrix}, $$ then it is not hard to check that $$ A^2=\frac{n(n+1)}2\,A. $$ So $p(A)=0$, where $p(x)=x\left(x-\frac{n(n+1)}2\right)$. This polynomial is divided by the minimal polynomial $m_a(x)$. As the minimal polynomial has degree at least two (if $n\geq2$, because otherwise $A$ would be zero or a scalar multiple of the identity), we get $$ m_a(x)=x\left(x-\frac{n(n+1)}2\right). $$