Question: What is a good reference for a proof of Proposition 1 below? It is definitely a known result, appearing e.g. in Plemmons's M-matrix characterizations, but I have not managed to follow the chain of references to an actual proof of it (e.g., if it appears in Ostrowski's papers cited by Plemmons, then not in this exact form). Has anyone seen it appear in the literature explicitly and with a self-contained proof? I give such a proof further below in an answer to this very question, but I'd prefer to have a published reference I can cite as well.

Definition. Let $\ell$ be a nonnegative integer.

(a) In the following, $\mathbb{R}^\ell$ denotes the $\mathbb{R}$-vector space of column vectors of size $\ell$.

(b) If $w \in \mathbb{R}^\ell$ is a column vector, then the notation $w_i$ shall be used for the $i$-th entry of $w$ (for each $i \in \left\{1,2,\ldots,\ell\right\}$).

(c) For two column vectors $u \in \mathbb{R}^\ell$ and $v \in \mathbb{R}^\ell$, we write $u > v$ if and only if each $i \in \left\{1,2,\ldots,\ell\right\}$ satisfies $u_i > v_i$.

(d) If $Q \in \mathbb{R}^{n\times m}$ is any matrix, then $Q_{i,j}$ shall denote the $\left(i,j\right)$-th entry of $Q$ for each $i$ and $j$.

(e) A nonsingular $M$-matrix means a matrix $Q \in \mathbb{R}^{\ell \times \ell}$ satisfying the following two conditions:

  • The off-diagonal entries of $Q$ are nonpositive. In other words, $Q_{i,j} \leq 0$ for $i \neq j$.

  • There exists some $x \in \mathbb{R}^\ell$ such that both $x>0$ and $Qx > 0$.

Proposition 1. Let $Q \in \mathbb{R}^{\ell \times \ell}$ be a nonsingular $M$-matrix. Then, $Q$ is nonsingular (i.e., we have $\ker Q = 0$).

Proposition 1 is, of course, part of the reason for the suggestive name "nonsingular $M$-matrix". Nevertheless, it is not really obvious. I give a proof in an answer to this question, but my real question is: What is a good reference for Proposition 1 in published literature that doesn't send its reader on a wild goose chase? The notion of a nonsingular $M$-matrix is famous for its many equivalent definitions (R.J. Plemmons, M-matrix characterizations. I -- nonsingular M-matrices gives dozens of them), which is at the same time a blessing and a curse, the latter because it means that a result like Proposition 1 can be spread across several parts of the literature without ever being stated explicitly in one of them. My impression so far is that this is what has happened.


Let me give a self-contained proof of Proposition 1. The proof will rely on the following lemma:

Lemma 2. Let $\ell$ be a nonnegative integer. Let $x \in \mathbb{R}^\ell$ be a vector such that $x > 0$. Let $y \in \mathbb{R}^\ell$ be a nonzero vector. Then, there exists some $\mu \in \mathbb{R}$ such that the vector $x - \mu y$ is nonnegative but has at least one zero entry. (Here, a vector $z \in \mathbb{R}^\ell$ is said to be nonnegative if all its entries are nonnegative.)

Proof of Lemma 2. The vector $y$ is nonzero. Thus, we can WLOG assume that $y$ has at least one positive coordinate $y_i > 0$ (otherwise, replace $y$ by $-y$, and replace the resulting $\mu$ by $-\mu$). Hence, the set $\left\{ x_i / y_i \mid i \in \left\{1, 2, \ldots, \ell\right\} \mid y_i > 0 \right\}$ is nonempty. Moreover, all elements of this set are positive reals (because $x > 0$, so that $x_i > 0$ for all $i$). Now, let $\mu$ be the minimum of the set. Then, $\mu$ is a well-defined positive real (since it is the minimum of a nonempty finite set of positive reals). Moreover, $\mu$ is clearly an element of this set. In other words, there exists some $j \in \left\{1, 2, \ldots, \ell\right\}$ satisfying $y_j > 0$ and $\mu = x_j / y_j$. Consider this $j$.

The definition of $\mu$ shows that $\mu \leq x_i / y_i$ for all $i \in \left\{1, 2, \ldots, \ell\right\}$ satisfying $y_i > 0$. In other words, $x_i \geq \mu y_i$ for all $i \in \left\{1, 2, \ldots, \ell\right\}$ satisfying $y_i > 0$. But the same inequality $x_i \geq \mu y_i$ also holds for all $i \in \left\{1, 2, \ldots, \ell\right\}$ that do not satisfy $y_i > 0$ (because these $i$ satisfy $y_i \leq 0$ and thus $\underbrace{\mu}_{> 0} \underbrace{y_i}_{\leq 0} \leq 0$, and therefore from $x > 0$ we obtain $x_i > 0 \geq \mu y_i$). Therefore, $x_i \geq \mu y_i$ holds for all $i \in \left\{1, 2, \ldots, \ell\right\}$. In other words, the vector $x - \mu y$ is nonnegative. Furthermore, this vector has at least one zero entry (namely, its $j$-th entry is $x_j - \underbrace{\mu}_{= x_j / y_j} y_j = x_j - \left(x_j / y_j\right) y_j = 0$). Hence, we have found some $\mu \in \mathbb{R}$ such that the vector $x - \mu y$ is nonnegative but has at least one zero entry. This proves Lemma 2.

Proof of Proposition 1. Assume the contrary. Thus, there exists some nonzero vector $y \in \mathbb{R}^\ell$ such that $Qy = 0$. Consider this $y$.

Since $Q$ is a nonsingular $M$-matrix, we have

\begin{equation} Q_{i,j} \leq 0 \qquad \text{for any } \left(i,j\right) \in \left\{1,2,\ldots,\ell\right\}^2 \text{ satisfying } i \neq j. \tag{1} \label{pf.prop1.1} \end{equation}

Since $Q$ is a nonsingular $M$-matrix, there exists an $x \in \mathbb{R}^\ell$ such that both $x>0$ and $Qx > 0$. Consider this $x$.

There exists some $\mu \in \mathbb{R}$ such that the vector $x - \mu y$ is nonnegative but has at least one zero entry (by Lemma 2). Consider this $\mu$. Set $z = x - \mu y$. Thus, the vector $z$ is nonnegative but has at least one zero entry. Furthermore, $Q z = Q \left(x - \mu y\right) = Q x - \mu \underbrace{Q y}_{= 0} = Q x > 0$.

There exists an $i \in \left\{1,2,\ldots,\ell\right\}$ such that $z_i = 0$ (since the vector $z$ has at least one zero entry). Consider this $i$. Now, $$ \left(Q z\right)_i = \sum_{j=1}^\ell Q_{i,j} z_j = Q_{i,i} \underbrace{z_i}_{= 0} + \sum_{j \neq i} \underbrace{Q_{i,j}}_{\substack{\leq 0 \\ \text{(by \eqref{pf.prop1.1})}}} \underbrace{z_j}_{\substack{\geq 0 \\ \text{(since } z \text{ is nonnegative} \text{)}}} \leq 0 + \sum_{j \neq i} 0 = 0 . $$ This contradicts $Q z > 0$. This contradiction proves that our assumption was wrong. Hence, Proposition 1 is proven.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.