# prove the following theorem Theorem: Assume $B=(v_1,…,v_k)$ is a maximal linearly independent subsequence of a vector space V then B is a basis.

Theorem: Assume $B=(v_1,\ldots,v_k)$ is a maximal linearly independent subsequence of a vector space $V$ then $B$ is a basis.

if $B$ is a basis, then we have to prove that $B$ is linearly independent and $B$ spans $V$

The theorem states that $B$ is linearly independent. If we add a $v_j \in V$ that is not in $B$ in $B$ then $B=(v_1,\ldots,v_k) \cup v_j$ becomes linearly dependent. This implies that $v_j \in \operatorname{span}(v_1,\ldots,v_k)$. So all $v_j \in V-B$ are in $\operatorname{span}(v_1,\ldots,v_k)$. Therefore, $V=\operatorname{span}(v_1,\ldots,v_k)$

My professor said that the exposition needs work. How can I make it perfect?

• What do you mean by "perfect"? – Luis Victoria May 22 '16 at 2:56
• like he doesn't put any comment on my work. – TheMathNoob May 22 '16 at 3:08
• There're a lot of theorems, properties and definitions that you can use to make your proof more "perfect" but this only depends on your professor. Maybe if you put more steps to show that "$\forall v_j \in V-B$, therefore, $V=\operatorname{span}(v_1,\ldots,v_k)$" makes him happy. – Luis Victoria May 22 '16 at 3:18

If your professor says "the exposition needs work", he is not criticising your mathematical arguments, he is criticising your writing. Here are some points which strike me. (I am assuming that what you have put in your question is exactly what you wrote for your professor - if not, then some of my points may be irrelevant.)

• Punctuation and grammar are important. In your statement of the theorem, the colon should be a full stop. (Or, leave it as a colon: then the following "a" should be lower case.) The sentence "if $B$ is a basis..." should begin with a capital letter and end with a full stop. The last sentence of your proof is missing a full stop. Note: a lot of people may tell you that these things are not mathematics and are not important. I do not agree. It's all about communicating clearly, and people will not understand what you are writing if your punctuation is all messed up. (I agree that there is little danger of misunderstanding in the present case. All the same, little improvements will always be helpful.)
• It is very helpful to write the word "Proof" so that it is clear to the reader where the statement stops and the reasoning begins.
• You begin by saying "if $B$ is a basis...". This sounds as if you are assuming that $B$ is a basis, which cannot be right as this is what you are to prove. I'm (reasonably) sure that's not what you meant, but the writing is misleading.
• Your next sentence is also a bit confusing. The theorem does not state that $B$ is independent, it states what happens if $B$ is independent (and maximal).
• Next sentence: "add" is ambiguous - do you mean append $v_j$ to the sequence, or do you mean something like $v_1+v_j$? Also, the notation $v_j$ is confusing - on the face of it, this should denote a vector which is already in $B$. It would be better to call it $v$ without a subscript: or perhaps even better, use a different letter entirely.

Here is what I would suggest. Note that the mathematical argument is pretty much the same as you have given and it is essentially only the writing that I have changed.

Theorem. Assume $B=(v_1,\ldots,v_k)$ is a maximal linearly independent subsequence of a vector space $V$. Then $B$ is a basis for $V$.

Proof. Let $B=(v_1,\ldots,v_k)$ be a maximal linearly independent subsequence of a vector space $V$. To show that $B$ is a basis for $V$, we have to prove that $B$ also spans $V$.

Suppose that $w$ is in $V$ and is not in ${\rm span}(B)$. Then $(v_1,\ldots,v_k,w)$ is linearly independent. But this contradicts the maximality of $B$. Therefore every element of $V$ is in ${\rm span}(v_1,\ldots,v_k)$. That is, $V={\rm span}(v_1,\ldots,v_k)$, and so $B$ is a basis for $V$.

Maybe, you should explicitly show that any vector can be written in terms of basis elements AND in a unique way. Maybe this is the part that is missing.