Showing that the characteristic polynomial and the minimal polynomial of a certain matrix is the same. [duplicate]

Question: Consider some $n\times n$ matrix, $A$, with scalers $c_1,c_2,\dots,c_n$ along the bottom row, $1$'s directly above the diagonal, and $0$'s everywhere else including along the diagonal. We want to show that the characteristic and minimal polynomials are the same.

For the characteristic polynomial, I know I can find the value of $|\lambda I-A|$, where $\lambda$ is an eigenvalue of $A$, and $I$ is the identity matrix. Once I find the characteristic polynomial, I know that the minimal polynomial must divide it, so I can start taking the factors of the minimal polynomial and raising them to powers less than or equal to the degree of the the factors in the characteristic polynomial until I get $0$ when multiplying. I suppose I am just not getting the calculations to come out nicely, so I was wondering if this is the best way to do it, or if there is a better method. Thank you!

Solution 1:

Since you are already using the Cayley-Hamilton theorem (implying that the minimal polynomial divides the characteristic polynomial, it suffices to show that the degree of the minimal polynomial is that of the characteristic one, which is $n$. This amounts to showing that a nonzero polynomial of degree strictly less than$~n$ cannot annihilate your matrix$~A$. One can easily show by indication that $(1,0,0,\ldots,0)\cdot A^k$ with $k<n$ has as single nonzero entry a $1$ moved $k$ positions to the right of its initial position. Then one has by linearity $(1,0,0,\ldots,0)\cdot c_0I+c_1X+\cdots+c_{n-1}X^{n-1}=(c_0,c_1,\ldots,c_{n-1})$, which gives the result we need.