Skip to main content

[Community Question] Linear-algebra: non-negative matrix satisfying two conditions

One of our user asked:

A real matrix $B$ is called non-negative if every entry is non-negative. We will denote this by $B\ge 0$.

I want to find a non-negative matrix $B$ satisfying the following two conditions:

(1) $(I-B)^{-1}$ exists but not non-negative. Here $I$ is the identity matrix.

(2) There is a non-zero and non-negative vector $\vec{d}$ such that $(I-B)^{-1}\vec{d}\ge 0$.

I tried all the $2\times 2$ matrices, but it did not work. I conjecture that such a $B$ does not exist, but don't know how to prove it.


Labels:

Comments

Post a Comment

Popular posts from this blog

[Community Question] Calculus: Manifold with boundary - finding the boundary

One of our user asked: I have the manifold with boundary $M:= \lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1\geq 0, x_1^2+x_2^2+x_3^2=1\rbrace \cup\lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1= 0, x_1^2+x_2^2+x_3^2\leq1\rbrace$ and I need to find the boundary of this manifold. I think it is $\lbrace (x_1,x_2,x_3) \in \mathbb R^n : x_1= 0, x_2^2+x_3^2=1\rbrace$ , the other option is that the boundary is the empty set? I think the first is right? Am I wrong?
Post a Comment
Read more

[Community Question] Linear-algebra: Are linear transformations between infinite dimensional vector spaces always differentiable?

One of our user asked: In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).
Post a Comment
Read more