Is this a valid proof of the fact that a subspace of an $n$-dimensional vector space has a basis? Is the following proof of the statement &quot;Any vector subspace $M$ of a $n$-dimensional vector space $L$ has a finite basis&quot; valid ?
Proof: If $M$ consists of only zero vector then we are done because the empty set will be a basis. Now suppose there exists a nonzero vector $x_{1}\in M$. Now we do the following. Label set $A_{1}=\{x_{1}\}$
Step $1$: If  $\forall y\in M\setminus A_{1}: \{x_{1},y\} \text{is linearly dependent}$, then we conclude that $A_{1}=\{x_{1}\}$ forms a basis, because it is maximal linear independent set.
If it is not the case that $\forall y\in M\setminus A_{1}: \{x_{1},y\} \text{is linearly dependent}$, meaning $\exists y\in M\setminus A_{1}: \{x_{1},y\} \text{is linearly independent}$, then we label $x_{2}=y$ and form a set $A_{2}=A_{1}\cup\{x_{2}\}=\{x_{1},x_{2}\}$.
Step $2$: If  $\forall y\in M\setminus A_{2}: \{x_{1},x_{2},y\} \text{is linearly dependent}$, then we conclude that $A_{2}=\{x_{1},x_{2}\}$ forms a basis, because it is maximal linearly independent set.
If it is not the case that $\forall y\in M\setminus A_{2}: \{x_{1},x_{2},y\} \text{is linearly dependent}$, meaning $\exists y\in M\setminus A_{2}: \{x_{1},x_{2},y\} \text{is linearly independent}$, then we label $x_{3}=y$ and and form a set $A_{3}=A_{1}\cup\{x_{3}\}=\{x_{1},x_{2},x_{3}\}$.
Step $j$: If  $\forall y\in M\setminus A_{j}: \{x_{1},x_{2},..,x_{j},y\} \text{is linearly dependent}$, then we conclude that $A_{j}=\{x_{1},x_{2},..,x_{j}\}$ forms a basis, because it is maximal linearly independent set.
If it is not the case that $\forall y\in M\setminus A_{j}: \{x_{1},x_{2},..,x_{j},y\} \text{is linearly dependent}$, meaning $\exists y\in M\setminus A_{j}: \{x_{1},x_{2},y\} \text{is linearly independent}$, then we label $x_{j+1}=y$ and and form a set $A_{j+1}=A_{j}\cup\{x_{j+1}\}=\{x_{1},x_{2},..,x_{j},x_{j+1}\}$.
Because of the fact that $M$ is a subspace of a $n$-dimensional vector space $L$. If we will get to the step $n$, the set $A_{n}$ will neccesarily be maximal linearly independent set (and so will be a basis for $M$), because any set of $n+1$ vectors in $L$ is linearly dependent. Therefore the process will terminate and we will form a basis for $M$. Q.E.D.
My worries about this proof are that it does something like a proof by cases but the number of cases is not fixed and is dependent on $n$. So is this proof valid ?
Thanks in advance.
It looks okay but there's an easier way of doing this.
Let $V$ be a vector space such that $dim(V) = n$. Then, it has a basis $(x_1,\ldots,x_n)$. This basis generates $V$ and, therefore, it must generate $M$. So, $M$ is finitely generated. Then, there exists a finite list of vectors $(v_1,\ldots,v_m)$ that generate $M$. If this list is linearly independent, then we have a basis for $M$. Otherwise, there is a vector in the list that can be written as a linear combination of the other vectors. So, remove that vector and check if the resulting list is linearly independent.
If it is, then we're done. Otherwise, continue this process of removing vectors until a linearly independent list is found. Indeed, we are guaranteed to find a linearly independent list because if we reduce the list size to $1$, we will certainly have a linearly independent list. Moreover, this list is guaranteed to span $M$ at each stage of the process. Hence, $M$ has a basis. $\Box$
Actually, your argument seems to be a &quot;backwards&quot; version of the argument I've put forth above, where you've constructed a basis from ground up. What I've done is to take a list that generates $M$ (we are guaranteed that at least one such list exists) and then I've reduced it to a basis of $M$. You've also been rather formal and that's a good thing too.
It looks okay but there's an easier way of doing this.
Let $V$ be a vector space such that $dim(V) = n$. Then, it has a basis $(x_1,\ldots,x_n)$. This basis generates $V$ and, therefore, it must generate $M$. So, $M$ is finitely generated. Then, there exists a finite list of vectors $(v_1,\ldots,v_m)$ that generate $M$. If this list is linearly independent, then we have a basis for $M$. Otherwise, there is a vector in the list that can be written as a linear combination of the other vectors. So, remove that vector and check if the resulting list is linearly independent.
If it is, then we're done. Otherwise, continue this process of removing vectors until a linearly independent list is found. Indeed, we are guaranteed to find a linearly independent list because if we reduce the list size to $1$, we will certainly have a linearly independent list. Moreover, this list is guaranteed to span $M$ at each stage of the process. Hence, $M$ has a basis. $\Box$
Actually, your argument seems to be a &quot;backwards&quot; version of the argument I've put forth above, where you've constructed a basis from ground up. What I've done is to take a list that generates $M$ (we are guaranteed that at least one such list exists) and then I've reduced it to a basis of $M$. You've also been rather formal and that's a good thing too.