Diagonalisability…without the characteristic polynomial

Question

Let us consider an $n\times n$ matrix $A$ defined as follows $$ A=\begin{pmatrix} 1+a&1&\cdots &1\\ 1&1+a&\ddots&\vdots\\ \vdots&\ddots&\ddots&1\\ 1&\cdots&1&1+a \end{pmatrix} $$ is diagonalizable without using the characteristic polynomial of $A$. My matrix contains $1+a$ ($a$ is a real number) along the diagonal and $1$ elsewhere. I just noticed that $a$ is eigenvalue but I don't know how to conclude with that.. Thanks.

Answer 1

Well, $A$ is real-symmetric and all real-symmetric matrices are diagonalizable.