How Is Mixed Norm ($ {L}_{1, 2 }$) Better than $ {L}_{1} $ Norm for Sparse Representation?
The mixed norm allows you to impose some simple structure in the solution matrix. Using your example with $p=q=1$ then this means the solution could have arbitrary elements set to non-zero coefficients. This would not impose any structure in the solution. In this case the $L_{11} $ norm just sums the absolute value of all the matrix elements - similar to the Frobenius norm.

As an example, consider that $L_{11}$ norm of the following matrices are the same:

$\begin{bmatrix} 0&amp;2 \\ 0 &amp; 0\end{bmatrix}, \quad \begin{bmatrix} 0&amp;0 \\ 2 &amp; 0\end{bmatrix}, \quad \begin{bmatrix} 2&amp;0 \\ 0 &amp; 0\end{bmatrix}, \quad \begin{bmatrix} 0&amp;0 \\ 0 &amp; 2\end{bmatrix}$.

So there is no preference as to which element is set to non-zero 

Instead, if $p=2$ and $q=1$, then the objective function is minimizing the $l_2$ norm of the columns and then the $l_1$ norm over the that. This means you're looking for an $a$ matrix with a sparse number of column vectors and where each column vector has a small $l_2$ norm. Of course this assumes you are using the $l_1$ norm as a proxy for sparsity.

Consider the $L_{2,1}$ and $L_{1,2}$ norms of following matrices:

$\begin{bmatrix} 2&amp;0 \\ 2 &amp; 0\end{bmatrix}, \quad L_{2,1} = \sqrt{8} \approx 2.8, \quad L_{1,2} = 4
$

and

$\begin{bmatrix} 2 &amp; 0\\ 0 &amp; 2 \end{bmatrix}, \quad L_{2,1}= 4, \quad L_{1,2} = \sqrt{8}
$

So minimizing the $L_{2,1}$ norm tends to prefer the structure of the first matrix over the second one, while minimizing the $L_{1,2}$ norm tends to prefer the structure second matrix over the first.

Thus the mixed norm allows you impose a soft constraint on the structure of the sparse solution that you're looking for. It's relatively easy to see that be you could also set it up to look for a sparse set of row vectors, each with a small $l_2$ norm. Deciding which to use is really dependent on how you set up your equations.