Prove 2 bases have the same cardinality

My attempt

Let V be a finite- dimensional vector space. Let B =$\displaystyle {{v_1},...{v_n}}$ and B'=$\displaystyle {{u_1,...,u_m}}$ be 2 bases for B. By the (one half of) exchange lemma, a spanning set has at least as many elements as a lin. independent set. Hence m is greater than or equal to n and n is greater than or equal to m. So m=n

When I set out to do this proof I thought I would have to use all the exchange lemma so I'm a bit unsure of my 'proof'. Thanks

Re: Prove 2 bases have the same cardinality

Quote:

Originally Posted by

**Duke** My attempt

Let V be a finite- dimensional vector space. Let B =$\displaystyle {{v_1},...{v_n}}$ and B'=$\displaystyle {{u_1,...,u_m}}$ be 2 bases for B. By the (one half of) exchange lemma, a spanning set has at least as many elements as a lin. independent set. Hence m is greater than or equal to n and n is greater than or equal to m. So m=n

When I set out to do this proof I thought I would have to use all the exchange lemma so I'm a bit unsure of my 'proof'. Thanks

Yes, that's a good proof.