MHB Prove A=B when A⊂span(B) and B⊂span(A)

toni07
Messages
24
Reaction score
0
Let A and B be subsets of a vector space V. Assume that A ⊂ span(B) and that B ⊂ span(A) Prove that A = B.
I don't know how to go about this question, any help would be appreciated.
 
Physics news on Phys.org
crypt50 said:
Let A and B be subsets of a vector space V. Assume that A ⊂ span(B) and that B ⊂ span(A) Prove that A = B.

That is not true. Choose for example $V=\mathbb{R}^2,$ $A=\{(1,0)\}$ and $B=\{(2,0)\}.$
 
crypt50 said:
Let A and B be subsets of a vector space V. Assume that A ⊂ span(B) and that B ⊂ span(A) Prove that A = B.
I don't know how to go about this question, any help would be appreciated.
Quite likely what you were actually after is the following:

If $A\subseteq\text{ span}(B)$ and $B\subseteq\text{ span}(A)$, then $\text{span}(A)=\text{span}(B)$.

Let $\text{span}(A)=U$ and $\text{span}(B)=W$.

Since $A\subseteq W$, we have $\text{span}(A)=U\subseteq W$. This is because $W$ is a subspace of $V$. Similarly $W\subseteq U$. We get $W=U$ and we are done.
 
The world of 2\times 2 complex matrices is very colorful. They form a Banach-algebra, they act on spinors, they contain the quaternions, SU(2), su(2), SL(2,\mathbb C), sl(2,\mathbb C). Furthermore, with the determinant as Euclidean or pseudo-Euclidean norm, isu(2) is a 3-dimensional Euclidean space, \mathbb RI\oplus isu(2) is a Minkowski space with signature (1,3), i\mathbb RI\oplus su(2) is a Minkowski space with signature (3,1), SU(2) is the double cover of SO(3), sl(2,\mathbb C) is the...
Back
Top