Vector space

A vector space is a set of objects that follows specific rules of addition and multiplication.

These objects are called vectors and the rules are:

1) Commutative and associative addition for all elements of the closed set.

$\boldsymbol{\mathit{v_{1}}}+\boldsymbol{\mathit{v_{2}}}=\boldsymbol{\mathit{v_{2}}}+\boldsymbol{\mathit{v_{1}}}$

$(\boldsymbol{\mathit{v_{1}}}+\boldsymbol{\mathit{v_{2}}})+\boldsymbol{\mathit{v_{3}}}=\boldsymbol{\mathit{v_{1}}}+(\boldsymbol{\mathit{v_{2}}}+\boldsymbol{\mathit{v_{3}}})$

2) Associativity and distributivity of scalar multiplication for all elements of the closed set

$c_1(c_2\boldsymbol{\mathit{v_{1}}})=(c_1c_2)\boldsymbol{\mathit{v_{1}}}$

$c_1(\boldsymbol{\mathit{v_{1}}}+\boldsymbol{\mathit{v_{2}}})=c_1\boldsymbol{\mathit{v_{1}}}+c_1\boldsymbol{\mathit{v_{2}}}$

$(c_1+c_2)\boldsymbol{\mathit{v_{1}}}=c_1\boldsymbol{\mathit{v_{1}}}+c_2\boldsymbol{\mathit{v_{1}}}$

where $c_1$ and $c_2$ are scalars.

3) Scalar multiplication identity.

$\boldsymbol{\mathit{1}}\boldsymbol{\mathit{v_{1}}}=\boldsymbol{\mathit{v_{1}}}$

4) Additive inverse.

$\boldsymbol{\mathit{v_{1}}}+(-\boldsymbol{\mathit{v_{1}}})=0$

5) Existence of null vector , such that

$\boldsymbol{\mathit{0}}+\boldsymbol{\mathit{v_{1}}}=\boldsymbol{\mathit{v_{1}}}$

In a vector space $V=\left \{\boldsymbol{\mathit{v_{1}}},\boldsymbol{\mathit{v_{2}}},\cdots ,\boldsymbol{\mathit{v_{k}}}\right \}$ , one vector can be expressed as a linear combination of other vectors in the set, e.g.:

$\boldsymbol{\mathit{z}}=c_1\boldsymbol{\mathit{v_{1}}}+c_2\boldsymbol{\mathit{v_{2}}}+\cdots+c_k\boldsymbol{\mathit{v_{k}}}$

The span of a set of vectors $V$ is the set of vectors that can be written as a linear combination of vectors in the set . For example, the span of the set of unit vectors $\boldsymbol{\mathit{\hat{i}}}$ and $\boldsymbol{\mathit{\hat{j}}}$ in the $\mathbb{R}^{2}$ space is the set of all vectors (including the null vector) in the $\mathbb{R}^{2}$ space. Alternative, we say that $\boldsymbol{\mathit{\hat{i}}}$ and $\boldsymbol{\mathit{\hat{j}}}$ span $\mathbb{R}^{2}$ .

If we vary $c_1,c_2,\cdots c_k$ (but not the trivial case where all scalars are zero) such that $\boldsymbol{\mathit{z}}$ is equal to $\boldsymbol{\mathit{0}}$ ,

$c_1\boldsymbol{\mathit{v_1}}+c_2\boldsymbol{\mathit{v_2}}+\cdots+c_k\boldsymbol{\mathit{v_k}}=\boldsymbol{\mathit{0}}\; \; \; \; \; \; \;\; 1$

the set of vectors $\boldsymbol{\mathit{v_1}},\boldsymbol{\mathit{v_2}},\cdots\boldsymbol{\mathit{v_k}}$ is said to be linearly dependent because any vector can be written as a linear combination of the others:

$\boldsymbol{\mathit{v_1}}=\frac{-c_2}{c_1}\boldsymbol{\mathit{v_2}}+\cdots+\frac{-c_k}{c_1}\boldsymbol{\mathit{v_k}}\; \; \; \; \; \; \; \; 2$

If the only way to satisfy eq1 is when $c_k=0$ for all $k$ , the set of vectors is said to be linearly independent. In this case, we can no longer express any vector as a linear combination of the other vectors (as $c_1=0$ , resulting in RHS of eq2 being undefined). An example of a set of linearly independent vectors is the set of unit vectors $\boldsymbol{\mathit{\hat{i}}}$ , $\boldsymbol{\mathit{\hat{j}}}$ in the $\mathbb{R}^{2}$ space.

Question

Can a set of linearly independent vectors include the zero vector?

Answer

No, because if $\boldsymbol{\mathit{v_{1}}}=0$ and $c_2,\cdots ,c_k=0$ in eq1, then $c_1$ can be any number. Since $c_1$ is not necessarily 0, it contradicts the definition of linear independence.

A set of $\textit{N}$ linearly independent vectors in an $\textit{N}$ -dimensional vector space $\textit{V}$ forms a set of basis vectors, $\boldsymbol{\mathit{e_1}},\boldsymbol{\mathit{e_2}},\cdots,\boldsymbol{\mathit{e_N}}$ . A complete basis set is formed by a set of basis vectors of $\textit{V}$ if any vector $\boldsymbol{\mathit{x}}$ in the span of $\textit{V}$ can be written as a linear combination of those basis vectors, i.e.