Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. Let $B= (B_t, t \ge 0)$ be a $d$-dimensional Brownian motion. Let $b:\mathbb R^d \to \mathbb R^d$ and $\sigma:\mathbb R^d \to \mathbb R^{d \times d}$ be Lipschitz. We consider the SDE $$ \mathrm d X_t = b(X_t) \, \mathrm d t + \sigma (X_t) \, \mathrm d B_t. $$
Let $X_t = (X_t^1, \ldots, X_t^d)$ and $B_t = (B_t^1, \ldots, B_t^d)$. Let $b = (b^1, \ldots, b^d)$ and $\sigma = (\sigma^{ij})_{1 \le i, j \le d}$. Then $$ \mathrm d X^i_t = b^i(X_t) \, \mathrm d t + \sum_{j=1}^d \sigma^{ij} (X_t) \, \mathrm d B^j_t. $$
Let $f:\mathbb R^d \to \mathbb R$ be twice continuously differentiable. By Itô's lemma, $$ \begin{align*} \mathrm d f(X_t) &= \sum_{i=1}^d \partial_i^1 f( X_t) \, \mathrm d X_t^i + \frac{1}{2} \sum_{i,j=1}^d \partial^2_{ij} f (X_t) \, \mathrm d \langle X^i, X^j \rangle_t \\ &= \sum_{i=1}^d \partial_i^1 f( X_t) b^i(X_t) \, \mathrm d t + \sum_{i,j=1}^d \partial_i^1 f( X_t) \sigma^{ij} (X_t) \, \mathrm d B^j_t \\ & \quad+ \frac{1}{2} \underbrace{\sum_{i,j=1}^d \partial^2_{ij} f (X_t) \sum_{k=1}^d \sigma^{ik} (X_t) \sum_{l=1}^d \sigma^{jl} (X_t) \, \mathrm d \langle B^k, B^l \rangle_t}_{A}. \end{align*} $$
I'm trying to obtain a more compact form of above equation as in these notes. Could you have a check on my below attempt?
As usual, a $d$-dimensional vector is considered as a $d \times 1$ matrix. Let $\langle \cdot, \cdot \rangle_F$ be the Frobenius inner product and $\cdot$ the matrix multiplication. Then $$ \begin{align*} \sum_{i=1}^d \partial_i^1 f( X_t) b^i(X_t) \, \mathrm d t &= \langle \nabla f( X_t), b (X_t) \rangle_F \, \mathrm d t, \\ \sum_{i,j=1}^d \partial_i^1 f( X_t) \sigma^{ij} (X_t) \, \mathrm d B^j_t&= (\nabla f( X_t))^\top \cdot \sigma(X_t) \, \mathrm d B_t. \end{align*} $$ and $$ \begin{align*} A &= \sum_{i,j=1}^d \partial^2_{ij} f (X_t) (\sigma \sigma^\top)^{ij}(X_t) \, \mathrm d t = \langle \operatorname{Hess} f (X_t), (\sigma \sigma^\top) (X_t) \rangle_F \, \mathrm d t. \end{align*} $$
It follows that $$ \mathrm d f(X_t) = \bigg [ \langle \nabla f( X_t), b (X_t) \rangle_F + \frac{1}{2} \langle \operatorname{Hess} f (X_t), (\sigma \sigma^\top) (X_t) \rangle_F \bigg ] \, \mathrm d t + (\nabla f( X_t))^\top \cdot \sigma(X_t) \, \mathrm d B_t. $$
In a similar manner, we can obtain a matrix form of Itô's lemma when the coefficients $b, \sigma$ are time-dependent and $f$ has a time variable.
Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. Let $B= (B_t, t \ge 0)$ be a $d$-dimensional Brownian motion. Let $b:\mathbb R_+ \times \mathbb R^d \to \mathbb R^d$ and $\sigma:\mathbb R_+ \times \mathbb R^d \to \mathbb R^{d \times d}$ be bounded Lipschitz. We consider the SDE $$ \mathrm d X_t = b(t, X_t) \, \mathrm d t + \sigma (t, X_t) \, \mathrm d B_t. $$
Let $X_t = (X_t^1, \ldots, X_t^d)$ and $B_t = (B_t^1, \ldots, B_t^d)$. Let $b = (b^1, \ldots, b^d)$ and $\sigma = (\sigma^{ij})_{1 \le i, j \le d}$. Then $$ \mathrm d X^i_t = b^i (t, X_t) \, \mathrm d t + \sum_{j=1}^d \sigma^{ij} (t, X_t) \, \mathrm d B^j_t \quad \forall i=1, \ldots,d. $$
Let $f: \mathbb R_+ \times \mathbb R^d \to \mathbb R, (t, x) \mapsto f(t, x)$ be of class $C^{1,2}$. By Itô's lemma [Theorem 3.6. at page 153 in Karatzas/Shreve's textboook Brownian Motion and Stochastic Calculus], $$ \begin{align*} \mathrm d f(t, X_t) &= \frac{\partial}{\partial t} f(t, X_t) \, \mathrm d t + \sum_{i=1}^d \frac{\partial}{\partial x_i} f(t, X_t) \, \mathrm d X_t^i + \frac{1}{2} \sum_{i,j=1}^d \frac{\partial^2}{\partial x_i \partial x_j} f (t,X_t) \, \mathrm d \langle X^i, X^j \rangle_t \\ &= \frac{\partial}{\partial t} f(t, X_t) \, \mathrm d t + \sum_{i=1}^d \frac{\partial}{\partial x_i} f(t, X_t) b^i (t, X_t) \, \mathrm d t + \sum_{i,j=1}^d \frac{\partial}{\partial x_i} f(t, X_t) \sigma^{ij} (t, X_t) \, \mathrm d B^j_t \\ & \quad+ \frac{1}{2} \underbrace{\sum_{i,j=1}^d \frac{\partial^2}{\partial x_i \partial x_j} f (t, X_t) \sum_{k=1}^d \sigma^{ik} (t, X_t) \sum_{l=1}^d \sigma^{jl} (t, X_t) \, \mathrm d \langle B^k, B^l \rangle_t}_{A}. \end{align*} $$
Then $$ \begin{align*} \sum_{i=1}^d \frac{\partial}{\partial x_i} f(t, X_t) b^i (t, X_t) \, \mathrm d t &= \langle \nabla_x f(t, X_t), b (t, X_t) \rangle_F \, \mathrm d t, \\ \sum_{i,j=1}^d \frac{\partial}{\partial x_i} f(t, X_t) \sigma^{ij} (t,X_t) \, \mathrm d B^j_t&= (\nabla_x f(t, X_t))^\top \cdot \sigma(t,X_t) \, \mathrm d B_t. \end{align*} $$ and $$ \begin{align*} A &= \sum_{i,j=1}^d \frac{\partial^2}{\partial x_i \partial x_j} f (t, X_t) (\sigma \sigma^\top)^{ij} (t,X_t) \, \mathrm d t = \langle \operatorname{Hess}_x f (t, X_t), (\sigma \sigma^\top) (t, X_t) \rangle_F \, \mathrm d t. \end{align*} $$
It follows that $$ \begin{align*} \mathrm d f(t,X_t) &= \bigg [ \frac{\partial}{\partial t} f(t, X_t)+ \langle \nabla_x f(t, X_t), b (t,X_t) \rangle_F + \frac{1}{2} \langle \operatorname{Hess}_x f (t, X_t), (\sigma \sigma^\top) (t,X_t) \rangle_F \bigg ] \, \mathrm d t \\ &\qquad + (\nabla_x f(t, X_t))^\top \cdot \sigma(t, X_t) \, \mathrm d B_t. \end{align*} $$