degree of $P(x)$ , the minimal polynomial of $A$

Question

Let $A$ be a $10 \times 10$ matrix defined by $A=(a_{ij})$ where $(a_{ij})=1-(-1)^{i-j}$. If $P(x)$ is the minimal polynomial of $A$ then

1) degree of $P(x)$?

2) coefficient of $x$ in $P(x)$?

My work: First of all I find the matrix it is of the form \begin{pmatrix} 0 & 2 & 0 & 2\:\cdots&0 &2 \\ 2 & 0 & 2 & 0\:\cdots&2 &0 \\ \vdots &\vdots &\vdots & &\vdots &\vdots\\ 0 & 2 & 0 & 2\:\cdots&0 &2\\ 2 & 0 & 2 & 0\:\cdots&2 &0\\ \end{pmatrix}

and I find $10$ is an eigen value with geometric multiplicity one and I also find solution of $Ax=0$ and see that geometric multiplicity is $8$ corresponding to the eigenvalue zero. From that I can say the matrix has zero as an eigenvalue at least $8$. But How do I find another eigenvalue or is there anything wrong? Please help. Thanks.

Answer 1

If your results are correct, always remind that if your matrix is $A$ $$ \text{Trace}\left(A\right)=\sum_{k=1}^{10}\lambda_k $$ where $\lambda_k$ is an eigenvalue of $A$ counted with multiplicity. Here you have $$ \text{Trace}\left(A\right)=10+\lambda=0 $$

Answer 2

Let me define the vectors $$ v_1=\pmatrix{1\\0\\1\\0\\\vdots}, \quad v_2=\pmatrix{0\\1\\0\\1\\\vdots}. $$ Then $$ Av_1 = 10 v_2, \ Av_2 = 10 v_1, $$ which implies $$ A^2v_1 = 100 v_1, \ A^2 v_2 = 100 v_2. $$ If $x$ is orthogonal to $v_1$ and $v_2$ then $Ax=0$. Since the columns of $A$ are multiples of $v_1,v_2$, we have $$ (A^2- 100 I)A = (A-10I)(A+10I)A=0. $$