The space $\mathbb{R}^n$ has dimension $n$. Every basis for $\mathbb{R}^n$ consists of $n$ vectors. A plane through $0$ in $\mathbb{R}^3$ is two-dimensional, and a line through $0$ is one-dimensional.

Does somone mind explaining these concepts to me? I don't see why having a plane or line through 0 decreases the dimension of the space by 1. Also, why does every basis for $\mathbb{R}^n$ consists of $n$ vectors?

  • 4
    $\begingroup$ "decreases the dimension of the space"" of what space? $\Bbb R^3$ is three dimensional, regardless of whether or not you have planes or lines going through the origin. The plane itself is two-dimensional $\endgroup$ – JMoravitz Mar 29 '17 at 3:28

A line through $0$ can be thought of as all scalar multiples of a given (nonzero) direction vector, so that vector (or any nonzero multiple of it) constitutes a basis for the line. Given such a specific vector $v$, every other point $x$ on the line can be written uniquely in the form $x=\alpha v$ for some scalar $\alpha$.

Likewise, a plane through $0$ is determined by two nonparallel lines through $0$, so by two nonparallel vectors $u$ and $v$ through $0$. Every point $x$ in this plane can be written uniquely in the form $x=\alpha u+\beta v$ for some scalars $\alpha$ and $\beta$.

This is independent of the dimension of the surrounding space.

  1. The space $\mathbb{R}^n$ has $n$ dimensions

The canonical basis of $\mathbb{R}^n$ is $B = \{\vec{e}_1,\dots,\vec{e}_n\}$ where $\vec{e}_i := (0, \dots, \underbrace{1}_{i\text{-th element}}, \dots,0)$ and $\#B = n = $'number of elements in $B$' $= \dim(\mathbb{R}^n)$, for all $n \in \mathbb{N}$.

  1. Every base for $\mathbb{R}^n$ consists of $n$ vectors

The number of vectors of the basis $B$ above stated is $n$. Suppose that there is another base $C$ with one vector less. Then it is not a base because we can find a vector of $\mathbb{R}^n$ such that this vector is not a linear combination of vectors in $C$.

  1. A plane through $0$ in $\mathbb{R}^3$ is two-dimensional

A plane in $\mathbb{R}^3$ is a two-dimensional space. Every point in the plane can be described by a linear combination of two vectors that lie in the plane. This is not just for a plane that passes through the $(0,0,0)$ but it is correct for all planes in the space $\mathbb{R}^3$. This is the definition of a plane in $\mathbb{R}^3$

  1. A line through $0$ is one-dimensional

Every line is one-dimensional because you need just one vector and multiples of this vector in space to create all line. Same as the item 3.

The dimension of the space does not decreases if a plane pass through the zero, the plane has two-dimensions and the dimensions are related to a basis of the space. I suggest that you should learn about a basis of a vector space and this questions will be much more simplified. See those questions of math.SE: vector, basis, more vector

See also this other: Why study linear algebra, What is the difference between a point and a vector , Is linear algebra laying the foundation for something important? , Where to start learning Linear Algebra?.


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