Does a set of basis vectors have to be linearly independent?
The definition for a set of vectors to be considered a basis for $R^n$ is that 1) this set spans $R^n$ - any vector in $R^n$ can be written as a combination of this set and 2) this set is linearly independent.
Extending this analogy to vector spaces, $V$, from the below article, it states "a set B of elements (vectors) in a vector space V is called a $\textbf{basis}$, if every element of V may be written in a unique way as a (finite) linear combination of elements of B."
Then, it goes on to say "B is a $\textbf{basis}$ if its elements are linearly independent and every element of V is a linear combination of elements of B. In more general terms, a basis is a linearly independent spanning set."
So, is a set $B$ considered a basis if everything in $V$ can be written with $B$ and $B$ must be linearly independent? Or, is $B$ a basis solely based on the fact that everything in $V$ can be written with $B$?
$\endgroup$ 55 Answers
$\begingroup$The section where it says "written in a unique way" is equivalent to linear independence. I think this is where some confusion may have arisen.
Your first concluding statement is correct. The second isn't. The basis must be linearly independent; either said explicitely or with the magic word "uniquely" added to your second statement to make it implicitely so.
$\endgroup$ $\begingroup$The key element in the definition is “in a unique way”.
A set of vectors by definition is linear dependent if there's a linear combination with non-zero coefficients of them that gives the zero vector. Since the linear combination with all coefficients equal zero also gives the zero vector, this quite directly says that the zero vector can be written in more than one way. And since you can add the zero vector to any other vector without changing it, the same is then true of any other vector as well.
$\endgroup$ 1 $\begingroup$To remove any confusion:
A Basis $B$ of a vector space $V$ is a subset $B \subset V$ such that
- $span(B) = V$ and
- $B$ is linearly independent
For $B$ to be a basis, linear independence of the elements of $B$ is required. That's the definition.
If you remove the independence condition, then it's no longer called a basis.
$\endgroup$ $\begingroup$Definition. A collection of vectors $\{v_1,\dotsc,v_d\}$ is a basis of a vector space $V$ if $\{v_1,\dotsc,v_d\}$ is linearly independent and $\operatorname{Span}\{v_1,\dotsc,v_d\}=V$.
Given this definition, there are a few things we can say.
Theorem 1. Suppose that $\{v_1,\dotsc,v_d\}$ and $\{w_1,\dotsc,w_{d^\prime}\}$ are two bases of a vector space $V$. Then $d=d^\prime$.
Theorem 1 allows us to define the dimension of $V$.
Definition. Suppose that $\{v_1,\dotsc,v_d\}$ is a basis of a vector space $V$. The dimension of $V$ is $\dim(V)=d$.
It turns out that we only need to check one of the two axioms defining a basis if we happen to know the dimension of our vector space.
Theorem 2. Suppose that $\dim(V)=d$ and consider a collection of vectors $\{v_1,\dotsc,v_d\}$ in $V$. Then $\{v_1,\dotsc,v_d\}$ is linearly independent if and only if $\operatorname{Span}\{v_1,\dotsc,v_d\}=V$.
For example, suppose that we know that $\dim(\Bbb R^3)=3$. Then to verify if the collection $$\{\left\langle2,\,-1,\,1\right\rangle, \left\langle0,\,2,\,1\right\rangle, \left\langle1,\,-2,\,0\right\rangle\}$$ is a basis of $\Bbb R^3$, we need only check that $$\operatorname{Span}\{\left\langle2,\,-1,\,1\right\rangle, \left\langle0,\,2,\,1\right\rangle, \left\langle1,\,-2,\,0\right\rangle\}=\Bbb R^3$$ and we don't have to check that the collection is linearly independent.
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