Why do we need two Unitary matrices instead of one in Singular Value Decomposition I am trying to understand Singular Value Decomposition from a intuitive point of view. As we know by applying SVD to any matrix, we get these three matrices as U, Sigma, V. And a matrix multiplication can be interpreted as a combination of rotation and scaling. Here U & V corresponds to the rotation performed by that matrix and Sigma the scaling factor.
My question is, why do we get or need two rotation matrices or unitary matrices (U & V), instead of one?
 A: You do not need two rotation matrices. The version with one rotation and "scaling" is called the polar decomposition, $A = QP$, where $Q$ is unitary and $P$ is positive semidefinite. It is related to the SVD as follows: 
$$A = U \Sigma V^H = \underbrace{U V^H}_{Q} \cdot \underbrace{V\Sigma V^H}_{P}.$$
Note, however, that the "scaling matrix", $P$, scales relative to a particular set of axes, namely, its eigenvectors $V$. The effect of multiplying a vector $x$ by the matrix $P$ can be interpreted as computing $x$'s projections on the axes $V$, $V^Hx$, scaling these projections, $\Sigma V^Hx$, and then building the vector again as the sum of the scaled projections, $V\Sigma V^Hx$. It is on this "scaled" vector that the rotation $Q$ is applied. Moreover, $V\Sigma V^Hx = \alpha x$, with $\alpha$ some scalar, iff $\Sigma = \alpha I$. Only in this case the effect of $P$ is to scale $x$. In general, $Px$ is a stretched version of $x$. 
So, you see, there are two important bases: $V$ is the "input" basis, and $U$ is the "output" basis, and the SVD interpretation of $A x$ is, you find the projections of $x$ on the axes of the "input" basis, multiply them by the singular values, and these projections are the coefficients of the output $Ax$ in the "output" basis $U$.
A: The Singular Value Decomposition theorem states that if  $V$ and $W$ are complex inner product spaces, and $L:V \to W$ is a linear map, then I can find an orthonormal basis $v_1,v_2,...,v_n$ for $V$ and an orthonormal basis $(w_1,w_2,...,w_m)$ of $W$ such that $L(v_i) = \sigma_i w_i$.
When you translate this into a statement about the matrix of the linear map, you need to apply a "change of basis" matrix to the domain and the codomain.  These are unitary since the basis is orthonormal.  This is where your two unitary bases come from.  The matrix for $L$ is diagonal WRT these bases, so that is where the diagonal matrix comes from.
A: Let $A = V\Sigma U^t$ over $\mathbb R$.
TL;DR: 1) Matrix $U$ defines the orthogonal directions along which the scaling will take place. This is one rotation whose angle is completely defined by the start direction of the scaling. 2) Matrix $V$ defines where these stretch directions end up afterwards.
$U$ rotates the input object (by an angle $\theta$ if $x\in\mathbb R^2$) such that the point $x$ to be stretched $\sigma_1$ will be along the $x_1$ direction (analogously for $x_i$).
$V$ rotates this (by an angle $\phi$ if V is $2\times 2$) to now position the final object arbitarily as a rotated version.
If you only had $U$, you would be able to stretch the object at arbitrary
directions with $U\Sigma U^t$ or to rotate the object to the stretch direction with $\Sigma U^t$ but not rotate it arbitrarily.
