I want to know how to compute the norm given by $\|\cdot\|_{L^{2}\left(\Omega\right)^{d\times d}}$ on the space $L^{2}\left(\Omega\right)^{d\times d}$ of $d\times d$ matrices whose entries are $L^{2}\left(\Omega\right)-$functions. Here $d\in\mathbb N_{0}$ and $\Omega\subseteq\mathbb R^{d}$ open. I have never understood how to compute any norm given on any space of matrices, and struggle to see how this particular norm should look like. What does $\|\cdot\|_{L^{2}\left(\Omega\right)^{d\times d}}$ look like?
I know that for the space $L^{2}\left(\Omega\right)^{d}$ the norm is just the sum of the $d-$copies of the standard $L^{2}-$norms on the respective entries of a given $f\in L^{2}\left(\Omega\right)^{d}$, and looks like $$\|f\|_{L^{2}\left(\Omega\right)^{d}}=\|f_{1}\|_{L^{2}\left(\Omega\right)^{}}+\dots+\|f_{d}\|_{L^{2}\left(\Omega\right)^{}}.$$
Is this norm in some way related to the norm $\|\cdot\|_{L^{2}\left(\Omega\right)^{d\times d}}$ which I want to compute? I have heard that the space $L^{2}\left(\Omega\right)^{d\times d}$ is unitarily equivalent to $L^{2}\left(\Omega\right)^{dd}$ (although I don't know how to prove this), which would perhaps provide such a relation.
The definition of $\|\cdot\|_{L^{2}\left(\Omega\right)^{d\times d}}$ is actually very similar to the one for $\|\cdot\|_{L^{2}\left(\Omega\right)^{d}}$ you gave.
Indeed, for $f \in L^{2}\left(\Omega\right)^{d\times d}$ as you defined it, $\|f\|_{L^{2}\left(\Omega\right)^{d\times d}}$ is given by $$\|f\|_{L^{2}\left(\Omega\right)^{d\times d}}^2 := \int_\Omega |f(x)|^2 d\mu (x) $$
Where $|f(x)|^2:=\sum_{1\le i,j\le d} f_{i,j}(x)^2 $ and $\mu$ is the Lebesgue measure. Also notice that, because all $f_{i,j}$ are square integrable and the sum is finite, we can write $$\begin{align*}\|f\|_{L^{2}\left(\Omega\right)^{d\times d}}^2 &=\int_\Omega |f(x)|^2 d\mu (x)\\ &= \int_\Omega\left( \sum_{1\le i,j\le d} f_{i,j}(x)^2\right) d\mu(x)\\ &= \sum_{1\le i,j\le d} \int_\Omega f_{i,j}(x)^2 d\mu(x)\\ &= \sum_{1\le i,j\le d} \|f_{i,j}\|_{L^2(\Omega)}^2 \end{align*} $$ Which is essentially the same expression as the one for vector-valued functions, and it can similarly be extended to higher orders too.