Let $\mu \in \mathbb R^d$ and $\sigma_1,\ldots,\sigma_d \in \mathbb R^k$. Let $W^{\mathbb P}$ be a $k$ dimensional brownian motion. We say that $S_t$ is a geometric brownian motion over $\mathbb R^d$ if the components $S_t^i$ of $S_t$ satisfy the SDE $$ \mathrm d S_t^i=S_t^i \mu_i \mathrm d t+S_t^i\sigma_i \cdot \mathrm dW^{\mathbb P} \qquad i=1,\ldots, d. \tag{$\dagger$} $$ I was trying to write such an expression in matricial form. If we indicate by $\Sigma \in \mathbb R^{d\times k}$ the matrix whose rows are $\sigma_1,\ldots \sigma_d$, this SDE can be written as $$ \mathrm dS_t=(\mu+\Sigma \mathrm dW^{\mathbb P})*S_t, $$ where $*$ denotes the Hadamard (or pointwise) product.
Question: Is there a simple way of writing $(\dagger)$ in matricial form without resorting to the Hadamard product? My guess is probably not. If the answer is negative, can we express it through tensors?