# Understanding the decomposed matrices in Singular Value Decomposition [duplicate]

I am trying to get some intuition regarding the U and V matrices in SVD ($$M=UEV^T$$). I think these are orthonormal basis vectors, but I am struggling to get an intuition if they represent anything more than matrix transformations.

For example, in NMF, I understand that the decomposed matrices can be combined as a linear combination of basis vectors in one matrix with weights in the other matrix. In PCA, I understand that the eigenvectors have real significance in representing orthogonal axes of maximal variation, defined by the eigenvalues. But in SVD, I don't see any immediate connection.

Could someone enlighten me?

In a singular value decomposition, $$M$$ is not (necessarily) symmetric or even square; it's a transformation from one space ($$\mathbb{R}^M$$) to another ($$\mathbb{R}^n$$).
Let's supppose $$m>n$$. $$M$$ can be decomposed into a transformation into a convenient basis for $$\mathbb{R}^n$$ by $$U$$, then a projection and scaling by $$E$$ into a basis for $$\mathbb{R}^n$$ , then a rotation into the target basis in $$\mathbb{R}^n$$ by $$V^T$$.
(Alternatively, for the compact SVD, $$U$$ is $$m\times n$$ and includes the projection, and $$E$$ is just the scaling)
Also, just as in PCA, it's no loss of generality (and is standard) to organise $$E$$ from largest to smallest singular value, and $$U$$ and $$V$$ represent orthogonal axes of maximal variation in the two spaces. There are various efficient algorithms for giving you just the largest few singular values and corresponding vectors (either Lanczos-type or stochastic algorithms)