Suppose A be a matrix with Diagonal entries are $\alpha+1$ then $\alpha $ is?

Question

Suppose $A=(a_{ij})$ be a $10\times 10$ order matrix such that $a_{ij}$=$1$ for $i\neq j$ and$ a_{ii}=\alpha+1$ for $\alpha\ge 0$.Let $ \lambda$ and $\mu $ be largest and smallest eigenvalues of $A$. If$\lambda+\mu=24$ then $\alpha=?$ here is my attempt -- according to the question diagonal elements are $\alpha+1$ and remaining entries are 1 then as sum of eigen value =sum of diagonal $10\alpha+10$=$\lambda+\mu+\sum_{i=2}^9\lambda_i $ suppose $\lambda_1=\mu$ and $\lambda_{10}=\lambda$ then $10\alpha$=$14+\sum_{i=2}^9\lambda_i $ as $\lambda+\mu=24$ but how can I find $\alpha$, is my process is correct? If not then what is the correct process?

Answer 1

Let $B$ be the $10\times10$ matrix such that all entries are equal to $1$. Then $B$ has $2$ eigenvalues: $10$ (multiplicity $1$) and $0$ (multiplicity $9$). Therefore, the eigenvalues of $A$ are $10+\alpha$ and $\alpha$, because if $P(\lambda)$ is characteristic polynomial of $B$, then the characteristic polynomial of $A$ is $P(\lambda-\alpha)$. So, the largest eigenvalue is $10+\alpha$ and the smallest one is $\alpha$. Since their sum is $10+2a=24$, $\alpha=7$.