Question: Let $V$ be an $n$-dimensional vector space. Show that any set of $n$ vectors in $V$ spans $V$ if and only if it is linearly independent.
My attempt at the proof:
Claim #1: "If any set of $n$ vectors in $V$ spans $V$ then it is linearly independent."
Proof #1:
Suppose that the set $S = \{v_1,…,v_n\}$ spans $V$. Also, we know that $S \subset V$.
We further suppose that $c_1v_1+…+c_nv_n=0$ for some $c_1,…,c_n \in \mathbb{R}$. Note: Call this equation (1).
Want to show: $c_1=…=c_n=0$.
Since the vector space $V$ is $n$-dimensional, i.e. $dim(V) = n$ means that for some basis for $V$ contains $n$ elements by definition. Let such a basis be denoted by $W= \{w_1,…,w_n\}$.
This means that every vector in the set $S$ can be written as a linear combination of the vectors in $W$ and vice versa.
This means for some $ 1 \leq i \leq n,$ the vectors in $S$ can be represented as the following:
$v_i= a_{i1}w_1+…+a_{in}w_n$.
And we know that the vectors $v_i$'s are all unique since $W$ is a basis for $V$ and, every vector in $S$ is in $V$.
Then this means (1) can be re-written as the following:
$w_1(c_1a_{11}+…+c_na_{n1})+…+w_n(c_1a_{1n}+…+c_na_{nn})=0$ and since the set $W$ is linearly independent we know that the coefficients on the equation above is all $0$ where $c_i$ are the unknowns.
$c_1a_{11}+…+c_na_{n1}=0$
………..
$c_1a_{1n}+…+c_na_{nn}=0$
The above system of linear equations does not allow me to solve for $c_i$'s because I don't know anything about it…
I still haven't made use of the hypothesis that the set $S$ spans $V$ but I'm not sure where I can use it, i.e. $span(S)=V$. Which means every vector in $V$ can be written as a linear combination of the vectors in $S$. Also haven't used the fact that each $v_i$'s are uniquely expressed as a linear combination of the vectors in $w_i$'s.
Any suggestion would be appreciated! Thank you in advance.
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Edit #1: Thank you for Ron's suggestion to this question, and I have went somewhere but I'm not exactly sure if I proved it correctly, I'll show the proof below but please point out any small details that I might be missing. I would very much appreciate if anyone can point out small little details as I find them very important in proofs. Any explicit/implicit details that I might be missing.
Proof:
Suppose $S$ is a set that contains n vectors and spans $V$. Let this set be denoted as follows $S = \{v_1,…,v_n\}$.
We want to show that $S$ is linearly independent.
Further suppose that $c_1v_1+…+c_nv_n=0$. (1)
Also suppose $\exists c_i \neq 0$ such that equation (1) holds above.
This means (1) can be re-written as the following: $v_i= \frac{-1}{c_i}\sum_{j=1,j\neq i}^n c_jv_j$.
This means that we can take out the element $v_i$ in the set $S$ since $v_i$ can be written as a linear combination of the other vectors in the set $S$.
Call this new set without $v_i$ to be $S'= S \setminus \{v_i\}$.
We still know that $S'$ spans $V$ and now we know $S'$ is also linearly independent. This means $S'$ is a basis for $V$.
This means $dim(V) = n-1$ which contradicts one of our hypothesis that $V$ is a $n$-dimensional vector space. Thus, $S$ is linearly independent.
Question: One part I'm a little iffy on is where I stated "$S'$ is also linearly independent…" Is this really true? I only showed that one of the $v_i$ is linearly dependent. What if it turns out to be that more than just 1 vector in $S$ is linearly dependent.
Edit #2: Nevermind, I found the answer to the question right above I was being stupid. Still a proof verification would be appreciated thanks!
Best Answer
Since $V$ is $n$-dimensional , a set of $n$ vectors that spans $V$ and is linearly independent, is by definition a base of $V$, so the condition is sufficient. For the necessary part you could use the following Hint: Suppose $S$ is a set of $n$ vectors that spans $V$ but S is linearly dependent.