Construct symmetric 3-by-3 matrix with 3 equal eigenvalues and nonzero offdiagonal components

Question

in order to test some function of mine I am looking for a symmetric 3-by-3 matrix with one single positive eigenvalue ($\neq 1$) with multiplicity 3. The problem is that at least 1 offdiagonal component of the matrix must be nonzero.

Does anyone know how to tackle this problem? Is it solvable at all?

For a starter I found

$$ \begin{pmatrix} 1 &a &0 \\ a &1 &0\\ 0&0&1-a \end{pmatrix} $$

as a matrix with just 2 same eigenvalues.

Thanks in advance

Answer 1

If $A\in\mathbb{R}^{3\times 3}$ is symmetric, there exists $P\in\mathbb{R}^{3\times 3}$ invertible such that $P^{-1}AP=D=\text{diag }(\lambda_1,\lambda_2,\lambda_3)$ ($\lambda_j$ eigenvalues of $A$). If $\lambda_j=\lambda$ for all $j$, necessarily: $$A=PDP^{-1}=P(\lambda I)P^{-1}=\lambda PP^{-1}=\lambda I.$$