# Unique volume element in a vector space

• yifli
In summary, the volume element is the unique f such that f(v_1,v_2,\cdots,v_n)=1 for any orthonormal basis v_1,v_2,\cdots,v_n given an orientation for V.
yifli
Given two orthonormal bases $v_1,v_2,\cdots,v_n$ and $u_1,u_2,\cdots,u_n$ for a vector space $V$, we know the following formula holds for an alternating tensor $f$:
$$f(u_1,u_2,\cdots,u_n)=\det(A)f(v_1,v_2,\cdots,v_n)$$
where A is the orthogonal matrix that changes one orthonormal basis to another orthonormal basis.

The volume element is the unique $f$ such that $f(v_1,v_2,\cdots,v_n)=1$ for any orthonormal basis $v_1,v_2,\cdots,v_n$ given an orientation for $V$.

Here is my question:
If $\det(A)=-1$, then $f(u_1,u_2,\cdots,u_n)=-1$, which makes $f$ not unique

Hi yifli!
I don't quite see why your reasoning implies that f is not unique. What's the other tensor that makes $f(v_1,...,v_n)=1$?

yifli said:
Here is my question:
If $\det(A)=-1$, then $f(u_1,u_2,\cdots,u_n)=-1$, which makes $f$ not unique

micromass said:
Hi yifli!
I don't quite see why your reasoning implies that f is not unique. What's the other tensor that makes $f(v_1,...,v_n)=1$?

Sorry for the confusion. Actually I mean $f(u_1,u_2,\cdots,u_n)$ is not necessarily equal to 1, it may be equal to -1 also, even though $f(v_1,v_2,\cdots,v_n)$ is made to be 1

Yes, but the point is that f of every basis with the same orientation will give you 1. If you take the opposite orientation, then you will get -1 of course.

micromass said:
Yes, but the point is that f of every basis with the same orientation will give you 1. If you take the opposite orientation, then you will get -1 of course.

So given a fixed orientation, it's impossible for $\det(A)$ to be -1?

What does it mean for $\det(A)$ to be -1?

yifli said:
So given a fixed orientation, it's impossible for $\det(A)$ to be -1?

What does it mean for $\det(A)$ to be -1?

The transition matrix between two orthonormal bases has det(A)=1 if and only if the bases have the same orientation and has det(A)=-1 if they have opposite orientation.

A orthonormal transformation matrix with det(A)=1 is a rotation.
A orthonormal transformation matrix with det(A)=-1 is a rotation combined with a reflection.

So the volume sign would be invariant.

## What is a unique volume element in a vector space?

A unique volume element in a vector space is a mathematical concept that represents the smallest unit of volume in a given vector space. It is also known as the determinant or Jacobian determinant.

## Why is a unique volume element important in vector spaces?

A unique volume element is important because it allows us to measure the volume of a given region in a vector space. It also helps us to understand the effects of linear transformations on the volume of a space.

## How is a unique volume element calculated?

A unique volume element is calculated by taking the determinant of the basis vectors of a given vector space. This can be done using various methods such as the cofactor expansion or the Gaussian elimination method.

## Can a unique volume element be negative?

Yes, a unique volume element can be negative. This happens when the basis vectors of a vector space are oriented in a way that results in a negative determinant. This does not affect the magnitude of the volume, but it indicates a change in orientation.

## How does a unique volume element relate to linear independence?

A unique volume element is closely related to linear independence. In fact, a set of vectors is linearly independent if and only if their determinant is non-zero, which means that they span a unique volume element in the vector space.

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