# Unit vectors and unit basis vectors

I was reviewing basic linear algebra and saw these two statements about the relationship between unit vectors and unit basis vectors.

If $\chi$ is a unit vector, then $\chi$ is a unit basis vector.

If $\chi$ is a unit basis vector, then $\chi$ is a unit vector.

The first statement is false, and the second statement is true.

…Huh?

I will explain what they mean with simple examples in this post.

## Unit vector

For a vector to be a unit vector, the (Euclidean) length has to be 1.

It means this is a unit vector:

$\chi = \begin{bmatrix} 0\\ 1\\ \end{bmatrix} (\sqrt{\smash[b]{0^2 + 1^2}} = 1)$

This is also a unit vector:

$\chi = \begin{bmatrix} 0\\ 0\\ 1\\ \end{bmatrix} (\sqrt{\smash[b]{0^2 + 0^2 + 1^2}} = 1)$

## Unit basis vector

Unit basis vectors are also known as standard basis vectors. Each “direction” of the vector has a standard basis.

Let’s think about a simple vector space, a 2-dimensional vector:

$y = \begin{bmatrix} 2\\ 3\\ \end{bmatrix}$

There are 2 bases here: $e_x = \begin{bmatrix} 1\\ 0\\ \end{bmatrix} , e_y = \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}$

And $y$ can be expressed with these bases and scalars like this:

$\chi = 2 \cdot \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} + 3 \cdot \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}$

where $2$ and $3$ are scalars.

A special property of a standard basis is that only one of the entries is $1$ and all other entries are $0$. It means that there are 3 bases for a 3-dimensional vector.

$e_x = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} e_y = \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} e_z = \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix}$

## Unit vector and unit basis vector

Going back to the original statements:

If $\chi$ is a unit vector, then $\chi$ is a unit basis vector.

This is not always true. Suppose there is a unit vector:

$\chi = \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix}$

This is not a unit basis of, say, a 2-dimensional vector, because there’s no basis for the 3rd dimension. Think of this 2-dimensional vector:

$y = \begin{bmatrix} 2 \\ 3 \\ \end{bmatrix}$

$\chi$ is not a basis because the 2 bases are

$e_x = \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} , e_y = \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}$

If $\chi$ is a unit basis vector, then $\chi$ is a unit vector.

This is true. Think of a basis for a 3-dimensional vector:

$\chi = \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix}$

The length of $\chi$ is 1, and the length of any unit basis vector is 1. It means it is always a unit vector.

I hope these examples clarified a subtle confusion about unit vectors and unit basis vectors.

## Subscribe to my blog

* indicates required
Thank you for reading my blog post! If you liked it, please subscribe to receive new posts. I won't give your address to anyone else, won't send you any spam, and you can unsubscribe at any time.

Tags:

Updated: