Is there some sort of relationship between SVD decomposition and diagonalizability? If $L$ is a linear operator acting on a hilbert space $H$ of dimension $n$ ( $L: H \to H$ ), then I know the following  


*

*If $L$ is normal operator then in some orthonormal basis $B$, matrix representation of $L$, $[L]_{B}$ will be a diagonal matrix.

*If sum of dimensions of eigen spaces of $L$ is equal to $n$ then in some non-orthonormal basis matrix representation of $L$ is diagonal.

*In general for any $L$, if its matrix representation in a basis $B$ is $[L]_B$ then $[L_B] = UDV$ where $U,V$ are unitary matrices and $D$ a diagonal matrix.


I can see that point $1$ and $2$ are the same thing ie if $L$ is normal operator. Is there any other relationship between SVD decomposition and diagonalizability ?  
 A: I think the best way to describe the relationship between SVD of a matrix (I'll just use $A$) and diagonalizability, and what makes it possible for every matrix to have a SVD, is because it is more closely related to the eigendecomposition of $AA^*$ and $A^*A$ (these matrices are positive semidefinite and is therefore always unitarily diagonalizable) than to $A$ itself. Notice that for $A=UDV$ we have \begin{equation} AA^*=UDVV^*D^*U^*=UD^2U^*\end{equation} and similarly we have $A^*A=V^*D^2V$. So $D$ is actually the square root of the eigenvalues of $AA^*$ and $A^*A$. Furthermore $U$ consists of eigenvectors for $AA^*$ and similarly $V$ consists of eigenvectors for $A^*A$.   
A: For any normal matrix, we can obtain an SVD from any spectral decomposition (i.e. orthonormal diagonalization).
Let $L$ be normal.  There exists a unitary matrix (orthonormal change of basis) $V$ such that $L = VDV^*$, where
$$
D = \pmatrix{\lambda_1 \\ & \ddots \\ & & \lambda_n}
$$
With $\lambda_k \in \Bbb C$.  We define the function
$$
\operatorname{sgn}(x) = 
\begin{cases}
\frac{x}{|x|} & x \neq 0\\
1 & x = 0
\end{cases}
$$
We then write $D = U'\Sigma$, where
$$
U' = 
\pmatrix{
\operatorname{sgn}(\lambda_1)\\
&\ddots\\
&& \operatorname{sgn}(\lambda_n)
}, \quad
\Sigma = 
\pmatrix{
|\lambda_1|\\
& \ddots \\
&&|\lambda_n|
}
$$
Thus, we have $L = VU'\Sigma V^*$.  Define $U = VU'$.  Note that $U$ is the product of unitary matrix, and is therefore itself unitary.  It follows that $L = U\Sigma V^*$ is a SVD of $L$.

If the matrix is not normal, then there is no such direct connection between diagonalization and SVD.  That is, only the things that we can do with orthonormal changes of basis allow us to deduce the SVD.
We do have some interesting general relationships between eigenvalues and singular values, though.  Suppose $s_1,\dots,s_n$ are the singular values in decreasing order and $\lambda_1,\dots,\lambda_n$ are the eigenvalues in order of decreasing magnitude.  Then Weyl's majorant theorem (Bhatia "Matrix Analysis", p. 42) states, among other things, that for any $p \geq 0$ and $k = 1,\dots,n$, we have 
$$
\prod_{j=1}^k |\lambda_j| \leq \prod_{j=1}^k s_j \\
\sum_{j=1}^k |\lambda_j|^p \leq \sum_{j=1}^k s_1^p
$$
