What is Volume?

Question

What i know is that "Volume" is a undefined notion in general, and the best approach to define volume is by Lebesgue measure.

I'm studying linear-algebra right now and the text says that "absolute value of determinant of n vectors is equal to n-dimensional volume formed by vectors". (Note that it is not defined to be, but it is equal to)

What is the definition of $n$-dimensional volume? and where can i see the proof that the Lebesgue measure of $n$-dimensional parallepiped is equal to the determinant of vectors forming the parallepiped?

It was not that easy to me to prove that the Lebesgue measure of $n$-dimensional rectagular parallelepiped is equal to the intuitive Volume i.e.$ \prod |I_k|$. So i guess the proof for equality of the Lebesgue measure and determinant would be really tedious and not that easy.. Where can i see this proof if there is, or what is the definition of volume in linear-algebra?

Answer 1

One way of thinking about this is the following. If you agree that the "volume" of the unit cube is equal to one (we measure the $n$-dimensional volume by counting how many unit cubes fit into the given space, right?), then the rest is more or less linear algebra.

Let $A$ be the matrix with the $n$ given vectors of $\mathbf{R}^n$ as rows. Recall that if the matrix $A$ is non-singular, then it is linked to $I_n$ by a finite sequence of elementary row operations. Let's study the effect of elementary row operation according to their type.

1. If we multiply a given row by a scalar, then the determinant is multiplied by that same scalar. The volume of $n$-dimensional parallelopiped is scaled accordingly (do take absolute values).
2. If we swap two rows, then the determinant is multiplied by $-1$. The parallelopiped is unchanged when we swap the edge vectors (its orientation is reversed, but we ignore that here).
3. Finally, if we add a multiple of one row to another, then the determinant does not change. For the parallelopiped this is essentially a 2-dimensional transformation taking place in the plane spanned by the two vectors involved. But geometrically this is a shear, i.e. a transformation that turns a rectangle to a parallelogram without changing either its base or its height. The area of the parallelogram is thus equal to that of the rectangle, and this is the $n$-dimensional analogue.

I leave the case of a singular $A$ as an exercise.

Answer 2

There is a proof in the book Lebesgue integration and measure by Alan Weir in section 6.3, called "The geometry of Lebesgue measure".

Answer 3

Suppose you have $n$ vectors $v_1,v_2,\dotsc,v_n$ forming some $n$-parallelotope $P \subset \mathbb R^n$ of non-zero volume. Then the matrix $A = v_1|v_2|\dotsb|v_n$ will be in $GL(\mathbb R^n)$ and $A([0,1]^n) = P$.

From the theory of translation invariant Radon measures on $\mathbb R^n$ it is known that any such measure $\mu$ must be equal to $cm_n$ for some $c \in (0,\infty)$. And $A^{-1}m_n$, the measure defined by $(A^{-1}m_n)(E) = m_n(A^{-1}(E))$, is indeed such a measure. Hence there is a constant $c_A$ such that $m_n(A(E)) = c_Am_n(E)$ for all Borel measurable $E \subseteq \mathbb R^n$. So what is this mysterious $c_A$? Why, it is exactly $m_n(A([0,1]^n))$, the volume of $P$.

But we can say more. The matrix $A$ above can be an arbitrary one in $GL(\mathbb R^n)$, so we can define a map $A \mapsto c_A$ and it turns out to be a homomorphism of groups. Moreover, it's easily seen for orthogonal matrices and for symmetric positive definite matrices that $c_A = |\det A|$. For a general matrix $A \in GL(\mathbb R^n)$ we can write $A=OS$ where $O$ is orthogonal and $S$ is symmetric positive definite, and we get that $c_A = |\det A|$ for all invertible $A$.