Skip to main content

[Community Question] Linear-algebra: Cardinality of infinite dimensional vector space of functions

One of our user asked:

Assume that F is an infinite field, k is an infinite cardinal, and V= F^k is a vector space. Can it be prove proved that |V|= dimV?
My thought was that we know that|V|= max {dimV,|F|}, so all I need is to prove is that if |V|=|F|, then also |F|=dimV. Now, from Konig theorem, if |V|=|F|, then k< cf(|F|), so I tried to prove that k< cf(|F|) is imposible, but i didn't know how to continue this line of thougt. Has sombodey know how to prove it?


Comments

Popular posts from this blog

[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.

[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).