volume of a $n$-d parallelepiped with sides given by the row vectors of a matrix $A$ is the product of the singular values of this matrix $A$ Why the volume of a $n$-dimensional parallelepiped with sides given by the row vectors of a matrix $A$ can be seen as the product of the singular values of this matrix $A$?
I only know in 3-dimensinal vector space, the volume of a parallelepiped is equal to the determinant of the corresponding matrix.
 A: We should really start with the description for the volume of a parallelotope.
Definition: Let $\Pi$ be an $n$-dimensional parallelotope defined by edge vectors $\mathcal{B}=\left\{ \mathbf{v}_{1},\ \cdots,\ \mathbf{v}_{n}\right\}$. We define the volume of the parallelotope recursively as follows:


*

*For a 1-dimensional parallelotope, we have $\mathrm{vol}(\Pi)$.

*For an $n$-dimensional parallelotope with $n>1$, we define the base of the parllelotope to be the volume of the $(n-1)$-dimensional parallelotope defined by edges $\mathcal{B}\backslash\left\{\mathbf{v}_{n}\right\}$, and we define the height to be the length of the component of $\mathbf{v}_{n}$ orthogonal to every vector in the base. Then we define 
$$\mathrm{vol}\left(\Pi\right)=\mathrm{base}\times\mathrm{height}.$$
You should convince yourself that this is the just a simple definition for the intuitive notion of the volume for a paralleotope. Our main theorem relating the volume to the determinant follows.
Theorem: Let $\Pi$ be an $m$-dimensional parallelotope defined by edge vectors $\mathcal{B}=\left\{ \mathbf{v}_{1},\ \dots,\ \mathbf{v}_{m}\right\}$, where $\mathbf{v}_{i}\in\mathbb{R}^{n}$ for $n\ge m$. That is, we are looking at an $m$-dimensional parallelotope embedded inside $n$-dimensional space. Suppose that $A$ is the $m\times n$ matrix with row vectors $\mathcal{B}$, given by
$$A=\begin{pmatrix}\mathbf{v}_{1}^{\mathrm{T}}\\
\vdots\\
\mathbf{v}_{m}^{\mathrm{T}}\end{pmatrix}$$
Then the $m$-dimensional volume of the parallelotope is given by $$\left[\mathrm{vol}\left(\Pi\right)\right]^{2}=\det\left(AA^{\mathrm{T}}\right).$$
Proof: Note that $AA^{\mathrm{T}}$ is an $m\times m$ square matrix. Suppose that $m=1$. Then 
$$\det AA^{\mathrm{T}}=\det\left(\mathbf{v}_{1}^{\mathrm{T}}\mathbf{v}_{1}\right)=\mathbf{v}_{1}\cdot\mathbf{v}_{1}=\|\mathbf{v}_{1}\|^{2}=\left[\mathrm{vol}_{1}(\mathbf{v}_{1})\right]^{2},$$
so the proposition holds for $m=1$. Now we induct on $m$. Suppose that the proposition holds for $m\ge 1$ and consider $m+1$. Letting $A_{m}$ denote the matrix containing the rows $\mathbf{v}_{1}$ to $\mathbf{v}_{m}$, we can write $A=A_{m+1}$ as
$$A=\begin{pmatrix}A_{m}\\
\mathbf{v}_{m+1}^{\mathrm{T}}
\end{pmatrix}$$
We may decompose $\mathbf{v}_{m+1}$ orthogonally as 
$$\mathbf{v}_{m+1}=\mathbf{v}_{\perp}+\mathbf{v}_{\parallel},$$
where $\mathbf{v}_{\perp}$ lies in the orthogonal complement of the base (this is our height), i.e. $\mathbf{v}_{\perp}\cdot\mathbf{v}_{i}=0$ for $1\le i\le m$, and where $\mathbf{v}_{\parallel}\in\mathrm{span}\left\{ \mathbf{v}_{1},\ \cdots,\ \mathbf{v}_{m}\right\}$. Suppose that $$\mathbf{v}_{\parallel}=c_{1}\mathbf{v}_{1}+\cdots+c_{m}\mathbf{v}_{m}.$$ 
We apply a sequence of elementary row operations to $A$, adding a multiple $-c_{i}$ of row $i$ to row $m+1$ for each $1\le i\le m$. Write the resulting matrix as $B$, and we have
$$B=\begin{pmatrix}A_{m}\\
\mathbf{v}_{\perp}^{\mathrm{T}}
\end{pmatrix}=E_{m}\cdots E_{1}A,$$
where each $E_{i}$ is an elementary matrix adding a multiple of one row to another. Notice that the above operation corresponds to shearing the parallelotope so that the last edge is perpendicular to the base. We see that these operations do not change the determinant as
$$\det\left(BB^{\mathrm{T}}\right)=\det\left(E_{m}\cdots E_{1}\left(AA^{\mathrm{T}}\right)E_{1}^{\mathrm{T}}\cdots E_{m}^{\mathrm{T}}\right)=\det\left(AA^{\mathrm{T}}\right).$$
Through block-multiplication, we get $BB^{\mathrm{T}}$ as
$$BB^{\mathrm{T}}=\begin{pmatrix}A_{m}\\
\mathbf{v}_{\perp}^{\mathrm{T}}
\end{pmatrix}\begin{pmatrix}A_{m}^{\mathrm{T}} & \mathbf{v}_{\perp}\end{pmatrix}=\begin{pmatrix}A_{m}A_{m}^{\mathrm{T}} & A_{m}\mathbf{v}_{\perp}\\
\mathbf{v}_{\perp}^{\mathrm{T}}A_{m}^{\mathrm{T}} & \mathbf{v}_{\perp}^{\mathrm{T}}\mathbf{v}_{\perp}
\end{pmatrix}=\begin{pmatrix}A_{m}A_{m}^{\mathrm{T}} & A_{m}\mathbf{v}_{\perp}\\
\left(A_{m}\mathbf{v}_{\perp}\right)^{\mathrm{T}} & \|\mathbf{v}_{\perp}\|^{2}
\end{pmatrix}.$$
Now notice that 
$$A_{m}\mathbf{v}_{\perp}=\begin{pmatrix}\mathbf{v}_{1}^{\mathrm{T}}\\
\vdots\\
\mathbf{v}_{m}^{\mathrm{T}}
\end{pmatrix}\mathbf{v}_{\perp}=\begin{pmatrix}\mathbf{v}_{1}\cdot\mathbf{v}_{\perp}\\
\vdots\\
\mathbf{v}_{m}\cdot\mathbf{v}_{\perp}
\end{pmatrix}=\mathbf{0}.$$
Therefore we have
$$BB^{\mathrm{T}}=\begin{pmatrix}A_{m}A_{m}^{\mathrm{T}} & \mathbf{0}\\
\mathbf{0}^{\mathrm{T}} & \|\mathbf{v}_{\perp}\|^{2}
\end{pmatrix}$$
Taking the determinant, we therefore have
$$\det\left(BB^{\mathrm{T}}\right)=\|\mathbf{v}_{\perp}\|^{2}\det\left(A_{m}A_{m}^{\mathrm{T}}\right).$$
By definition, $\|\mathbf{v}_{\perp}\|$ is the height of the parallelotope, and by the induction hypothesis, $\det\left(A_{m}A_{m}^{\mathrm{T}}\right)$ is the square of the base. Therefore the result follows. $\square$
As a corollary, we can easily verify that the volume is independent of which $n-1$ vectors you choose to define your base, as it should be.
Corollary: The volume of a parallelotope is invariant of which base and height you choose.
Proof: Choosing a different base and height corresponds to permuting the rows of the matrix $A$. This does not change the magnitude of the resulting determinant. $\square$
Finally, note that the singular values are the absolute values of $A$ are precisely the square roots of the eigenvalues of $AA^\mathrm{T}$. Since the product of the eigenvalues is the determinant, it follows that the product of the singular values of $A$ gives the square root of $\det(AA^\mathrm{T})$, which is precisely the volume.
A: This is because the product of the singular values is the determinant. The determinant is the volume of the paralellipiped spanned by the rows of a square matrix (this is by itself an interesting proof which is probably easy to find). Now, we need to use the fact known as singular value decomposition, which says that a matrix can be decomposed into $M= U P V^*$, where $U$ and $V$ are unitary. (this means basically that they only rotate vectors, and don't change their length) The diagonal matrix's determinant is just the product of the elements of its diagonal, and these values are the singular values. Since the unitary matrices are basically just rotations, they have determinant 1 or negative 1, and since we are working with the volume of a paralellepiped positive or negative is irrelevant. So, we get $Det(M) = Det(U) Det(P) Det(V*) = Det(P) = \prod_{i=1}^n \sigma_i$.
A: I think the following is a nice argument establishing this. Assume that $A$ has at least as many columns (call this number $m$) as rows (of which we know there are $n$), for otherwise the n-volume inside $\mathbb{R}^m$ that we will be computing will automatically be zero. Now, writing SVD decomposition of $A$ we have $A=U\Sigma V^T$, and taking transpose $A^T=U^T \Sigma^T V$ is the SVD decomposition of $A^T$. We decompose $\Sigma^T$ as a composition of $(\Sigma^T)_n$ -- the diagonal $n$ by $n$ matrix which is composed of the first $n$ rows of $\Sigma^T$, with singular values of $A$ on the diagonal,-- and the $m$ by $n$ matrix $I_n^m$ with top $n$ rows forming the $n$ by $n$ identity matrix and the rest filled with zeros.
The matrix $A^T:\mathbb{R}^n \to \mathbb{R}^{m}$ takes standard basis vectors $e_1, \ldots, e_n$ to the row vectors of $A$, and so what we want to compute is the $n$-volume of the image of the standard $n$-dimensional cube with sides $e_1, \ldots, e_n$ under this map $A^T$. Now, $A^T$ is a composition of 4 maps: first, $V$ is a rotation of $R^n$, which does not change $n$-volume. Then, $(\Sigma^T)_n$ stretches each coordinate direction by corresponding singular value of $A$, thus making the $n$-volume of the image equal to (the absolute value of) the product of the singular values of $A$. Then the $m$ by $n$ matrix $I_n^m$ embeds $\mathbb{R}^n$ as the coordinate subspace in $\mathbb{R}^m$, without changing the $n$-dimensional volumes of any sets. And, finally, $U^T$ rotates this coordinate subspace to the image of $A^T$, also without changing $n$-dimensional volumes. Overall, the $n$-volume of the image of the standard cube changes (from the initial value of $1$) only once, and in the end is equal to (the absoulte value of) the product of singular values of $A$, as wanted.
