What is the difference between zero vector and null vector?

Question

What is the difference between zero vector and null vector?

I think it is not a duplicate.

Answer 1

Is my this argument correct?

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If all the components of $\vec{x}$ are zero, it is called the zero vector.

If the length of a vector $\vec{x}$ is zero then, it is called the null vector.

In $n$ dimensional Euclidean space $(E^{n})$, there is no distinction between zero vector and null vector. But it is not true in other than Euclidean space.

Explanation: In $n$ dimensional Euclidean space $(E^{n})$, for $\vec{x}=(x_1, x_2, ... , x_n) \in E^{n}$, $$ |\vec{x}| = \sqrt{{x_1}^2 + {x_2}^2 + . . . + {x_n}^2} . $$

If $\vec{x}$ is a zero vector then $x_1=x_2=. . . =x_n=0 $ and hence $|\vec{x}|=0$. Consequently the vector $\vec{x}$ is a null vector.

Again if $|\vec{x}|=0$, then we have $$\sqrt{x_1^2 + x_2^2 + . . . +x_n^2}=0$$ which yields $${x_1^2 + x_2^2 + . . . +x_n^2}=0$$ This implies that the equality holds only if $x_1=x_2=. . . =x_n=0$ and hence the vector is a zero vector. Thus in $E^{n}$ a null vector and a zero vector has no difference. That is, a vector in $(E^{n})$ is a zero vector if and only if it is a null vector.

But it is not true in other than Euclidean space.

For example we now consider the Riemannian space.

An $n$-dimensional space in which the distance between two neighbouring points $(x_1, x_2, ... , x_n)$ and $(x_1+dx_1, x_2+dx_2, ... , x_n+dx_n)$ is given by $$ds^2 = \sum_{i,j=1}^{n} {g_{ij} dx_i dx_j} $$ where $g_{ij}$ are positive definite bilinear arbitrary functions of $x_i$'s such that $|g_{ij}|\neq 0$ and $ds$ is assumed to be invariant is known as Riemannian space. $g_{ij}$'s are also known as the fundamental metric tensor.

In the case of Euclidean space $g_{ij}$ is defined by

$g_{ij} = 1 \quad \text{if} \ i=j$ and

$g_{ij} = 0 \quad \text{if} \ i \neq j $

The length of the vector $\vec{x}=(x_1, x_2, ... , x_n)$ for the Riemannian space is defined as follows: $$|\vec{x}|=\sqrt{\sum_{i,j=1}^{n} {g_{ij} x_i x_j}}$$

Now we consider a $4$-vector $\vec{x}=(x_1, x_2, x_3, x_4)=(1,0,0,1)$ with the fundamental metric $g_{11}=g_{22}=g_{33}=-1, g_{44}=1$ and $g_{ij}=0$ for $i \neq j$.

Thus the length of the considered vector is obtained as follows: $$|\vec{x}|=\sqrt{g_{11} x_1 x_1 + g_{22} x_2 x_2 + g_{33} x_3 x_3 + g_{44} x_4 x_4} =\sqrt{(-1).1.1 + (-1). 0. 0 +(-1). 0. 0 + 1 .1 .1 } =0$$

Hence $\vec{x}=(x_1, x_2, x_3, x_4)=(1,0,0,1)$ is a null vector, but is not a zero vector.

Hence the result holds.

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Ref.: "Vector Analysis with Applications" by A. A. Shaikh & S. K. Jana

Answer 2

Given a vector space V, the $\textbf{zero vector}$ is the vector $v\in V$ which is the identity for the sum operation, i.e., $v+u=u$ for any vector $u\in V$. Because a vector space is a group under the sum operation, this element is unique.

However, many times, you also have a way of measuring $\textit{magnitudes}$, usually a norm or a pseudo-norm, and in this case $\textbf{null vector}$ just means a vector of $0$ magnitude.

A norm requires that the only null vector is the zero vector, so they're still the same thing in this case.

However, a pseudo-norm relaxes this condition and only requires the zero vector to be a null vector, but not necessarily the only one.

A famous example comes from General Theory of Relativity, where pseudo-metrics are used and the null vectors are, not only the zero vector, but also light rays.