Prove that if A is M-matrix then A is also a P-matrix $A \in \mathbb{R}^{n \times n}$ is a $P$-matrix if all its principal minors are positive.
Let $I$ be the identity matrix of rank $n$. $A \in \mathbb{R}^{n \times n}$ is a non-singular $M$-matrix if $A=I-B$ where $B \in \mathbb{R}^{n \times n}$ has only non negative entries and the largest eigenvalue of $B$ (or maximum of moduli) is strictly smaller than one.
Claim $1$: Nonsingular $M$-matrices are a subclass of $P$-matrices.
I am trying to prove claim $1$. If we call $\mathbb{M}$ the set of non-singular $M$-matrices and $\mathbb{P}$ the set of $P$-matrices, then proving the above claim comes down to taking one (any) element of the set $\mathbb{M}$ and showing that this element also belongs to $\mathbb{P}$.
It seems long but doable for $n=2$. Suppose $A=I-B \in \mathbb{M}$. Take a matrix $B$ with non-negative entries $b_{11}=a$, $b_{12}=b$,$b_{21}=c$, $b_{22}=d$. We compute the largest eigenvalue of $B$ (case where $\Delta$ is zero or strictly positive gives 6 different cases to consider). For example for the case $\Delta =0$ with $a=d$ and $b=c=0$ we get $$\lambda_1=\lambda_2 = \frac{a+d}{2}$$ Since $A$ is in $\mathbb{M}$ we have the condition: $$ a < 1$$
For A to also be in $\mathbb{P}$ the following must be satisfied:
$$ (1 - a)(1 -d) - bc \geq 0$$ $$\Leftrightarrow (1-a)^2 \geq 0$$ which is always true. Other cases get longer and messier. Then to suppose the hypothesis holds for n and show it for n+1 seems also long.
any suggestions of a straight forward way to prove the claim ?
Also for $n=2$ we can take $A \in \mathbb{P}$ and suppose that $A \not \in \mathbb{M}$. We get a contradiction directly for the case $a=d$ and $b=c=0$ (and with more effort we check it for the other cases)...
Thank you
 A: Your definition of an M-matrix is somewhat different from the usual one. Usually, $A$ is an M-matrix if $A=\sigma I-B$, where $B\geq 0$ (componentwise) and $\sigma>\rho(B)$. Anyway, you can of course assume that $\sigma=1$ without loss of generality. So let's assume it.
Since $\rho(B)<1$ and $B\geq 0$, we have
$$
A^{-1}=(I-B)^{-1}=\sum_{i=0}^\infty B^i\geq 0.
$$
Few observations are useful:


*

*$A$ has positive diagonal. This is because the dot product of the $i$th row of $A^{-1}$ (which is a nonzero nonnegative vector) with the $i$th column of $A$ (which has nonpositive off-diagonal entries) is one, so
$$
\underbrace{\color{blue}{a_{ii}^{(-1)}}}_{\color{blue}{\geq 0}}a_{ii}=1-\underbrace{\color{red}{\sum_{j\neq i}a^{(-1)}_{ij}a_{ij}}}_{\color{red}{\leq 0}}>0.
$$
Hence $a_{ii}^{(-1)}$ cannot be zero (and is positive) and $a_{ii}$ is positive as well.

*Let $x=A^{-1}e$, where $e:=[1,\ldots,1]^T$. Since $e>0$ and $A^{-1}\geq 0$, we have $x\geq 0$. If some component, say, $x_i=e_i^Tx$ ($e_i$ is the $i$th column of the identity matrix) would be zero, we would have $0=x_i=e_i^Tx=e_i^TA^{-1}e=(e_i^TA^{-1})e=0$, which would imply that the $i$th row of $A^{-1}$ is zero (contradiction). Hence, actually, $x>0$.

*Let $D:=\mathrm{diag}(x)$ (a positive definite diagonal matrix with $x$ on the diagonal). Then $ADe=Ax=e>0$, that is, the row sums of $AD$ are positive.
Since the row sums of $AD$ are positive and $AD$ has positive diagonal, by the Gershgorin theorem, the real parts of the eigenvalues of $AD$ are positive and hence $\det(AD)>0$. Since $D$ is positive definite, this gives $\det(A)=\det(AD)\det(D^{-1})>0$.
Now to translate this to the statement about all minors, note that $\det(A)>0$ follows from the strict diagonal dominance of $AD$. Hence take any principal submatrix of $AD$ and since this submatrix will have fewer negative off-diagonal entries, its row sums must be positive as well.
A: Possible answer i think using the following theorem:
Theorem $1$: Suppose $A$ is a square matrix and $\lambda$ is an eigenvalue of $A$. Let $q(x)$ be a polynomial in the variable $x$. Then $q(\lambda)$ is an eigenvalue of the matrix $q(A)$.
Proof: http://linear.ups.edu/html/section-PEE.html
claim: if $A=I-B$ is a non-singular $M$-$matrix$ then $A=I-B$ is a $P$-$matrix$.
proof:
The idea is to show that when $A$ is an $M$-$matrix$ then $A$ is positive (semi-) definite (same thing as having principal minors (strictly) positive).
suppose that $\lambda$ is an eigenvalue of $B$ and the largest eigenvalue of $B$ is strictly smaller than 1.
Using theorem $1$, $$q(B) = I - B$$ We have $q(\lambda) = 1 - \lambda $ an eigenvalue of $q(B)$. Since $q(B)=I - B =A$ it follows that $1 - \lambda $ is an eigenvalue of $A$. For $A$ to be positive definite we need $$ 1 - \lambda > 0 $$
This condition is satisfied since we supposed that the largest eigenvalue of $B$ is strictly smaller than 1. 
