Type of Matrix with only real eigenvalues

Question

Consider a parameter $0 \leq d \leq 1$ and a vector $\mathbf{q}$ of size $n$ made of real positive values. From this, construct the (real, non-negative) $n \times n $ matrix $\mathbf{M}$ such that $M_{i,j}=\frac{q_jd}{n-1}$ for $i \neq j$ and $M_{j,j}=(1-d)q_j$.
We can therefore also write $\mathbf{M}=\mathbf{D}.diag(\mathbf{q})$, with $D_{i,j}=\frac{d}{n-1}$ for $i \neq j$ and $D_{j,j}=(1-d)$

The sum of the elements of the jth column of $\mathbf{M}$ is therefore $\sum_i M_{i,j}=q_j$ and the sum of the elements of the jth column of $\mathbf{D}$ is therefore $\sum_i D_{i,j}=1$.

How can I show that all the eigenvalues of $\mathbf{M}$ are real (and positive) which seems to be the case, despite $\mathbf{M}$ not being symetrical ?
Is that a special case of matrices where I can get a formula for all the eigenvalues ?

Answer 1

$\mathbf{M}$ is similar to $\mathbf{A}=diag(\mathbf{q})^{1/2} \mathbf{M}diag(\mathbf{q})^{-1/2}=diag(\mathbf{q})^{1/2} \mathbf{D}diag(\mathbf{q})diag(\mathbf{q})^{-1/2}=diag(\mathbf{q})^{1/2}\mathbf{D}diag(\mathbf{q})^{1/2}$ which is symetrical therefore all eigenvalues of $\mathbf{M}$ are real.