Determine if the set is a basis for the vector space There is a linearly independent set of vectors in the vector space $V$, given by  $\{v_1, v_2,...v_k\}$. We have to show that the set $\{v_1, v_2,...v_{k-1}\}$ cannot be a basis for V. 
It is clear that the statement is true since the initial set of vectors is a linearly independent it means that a basis of it has a dimension k--that may or may not be a basis for V. Using a set of vectors with k-1, is surely not a basis because it is less than k. 
My justification in more mathematical terms is below. I am wondering if it is the right way of justifying it:
$$\text{Basis}\{v_1, v_2,...v_k\} \subseteq \text{Basis}(V)$$
But, $$ \therefore \dim\{v_1, v_2,...v_k\}=k \implies \dim{V} \geq k$$
$$ \dim\{v_1, v_2,...v_{k-1}\}=k-1$$
Hence, it is impossible for the set, $\{v_1, v_2,...v_{k-1}\}$ , to be a basis for the vector space $V$ since $\dim(V)\geq k$
Is this a correct proof.
 A: I think it would be better to avoid the concept of dimension and work more directly with the definitions of the terms in the problem.
Show that the smaller set is not a basis by finding a vector that cannot be expressed as a linear combination of the $v_1, \dots, v_{k-1}$. 
A candidate for such a vector should not be hard to think of. 
Show that this vector cannot be written as a linear combination of the $v_1, \dots, v_{k-1}$ by using the definition of linear independence of the $v_1, \dots, v_k$ .
A: Yep, looks perfectly good to me.
Another way to prove the same would be to use the fact that any linear combination of the subset  $\{v_1, v_2,...v_{k-1}\}$ of $\{v_1, v_2,...v_k\}$ cannot give $\{v_k\}$ because if so, the given set would not be linearly independent, and hence the subset cannot be a basis.
A: Your ideas are correct but you need to communicate them clearly and precisely. For example, what do you mean when you write
$$\operatorname{Basis}\{v_1,\ldots,v_k\}\subseteq\operatorname{Basis}(V)?$$ Do you mean that there exists a basis for $V$ which contains $\{v_1,\ldots,v_k\}$? (This is true since $\{v_1,\ldots,v_k\}$ is linearly independent.) It's not clear what you mean by either of "$\operatorname{Basis}\{v_1,\ldots,v_k\}$" or "$\operatorname{Basis}(V)$".
Next, $$\operatorname{dim}\{v_1,\ldots,v_k\}=k$$ should say $$|\{v_1,\ldots,v_k\}|=k$$ or perhaps $$\dim(\operatorname{span}\{v_1,\ldots,v_k\})=k$$ as $\{v_1,\ldots,v_k\}$ is not itself a vector space, and so it doesn't have a "dimension." Likewise for the line after that.
