Gram–Schmidt process: matrix form

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Let $$ \operatorname{proj}_{\mathbf{u}}(\mathbf{v})=\frac{\langle\mathbf{u}, \mathbf{v}\rangle}{\langle\mathbf{u}, \mathbf{u}\rangle} \mathbf{u} $$ be the projection operator, where $\langle\mathbf{u}, \mathbf{v}\rangle$ denotes the inner product of the vectors $\mathbf{u}$ and $\mathbf{v}$. This operator projects the vector $\mathbf{v}$ orthogonally onto the line spanned by vector $\mathbf{u}$. The Gram-Schmidt process works as follows: $$ \begin{aligned} \mathbf{u}_1 &=\mathbf{v}_1, & \mathbf{e}_1 &=\frac{\mathbf{u}_1}{\left\|\mathbf{u}_1\right\|} \\ \mathbf{u}_2 &=\mathbf{v}_2-\operatorname{proj}_{\mathbf{u}_1}\left(\mathbf{v}_2\right), & \mathbf{e}_2 &=\frac{\mathbf{u}_2}{\left\|\mathbf{u}_2\right\|} \\ \mathbf{u}_3 &=\mathbf{v}_3-\operatorname{proj}_{\mathbf{u}_1}\left(\mathbf{v}_3\right)-\operatorname{proj}_{\mathbf{u}_2}\left(\mathbf{v}_3\right), & \mathbf{e}_3 &=\frac{\mathbf{u}_3}{\left\|\mathbf{u}_3\right\|} \\ \mathbf{u}_4 &=\mathbf{v}_4-\operatorname{proj}_{\mathbf{u}_1}\left(\mathbf{v}_4\right)-\operatorname{proj}_{\mathbf{u}_2}\left(\mathbf{v}_4\right)-\operatorname{proj}_{\mathbf{u}_3}\left(\mathbf{v}_4\right), & \mathbf{e}_4 &=\frac{\mathbf{u}_4}{\left\|\mathbf{u}_4\right\|} \\ & \vdots & & \vdots \\ \mathbf{u}_k &=\mathbf{v}_k-\sum_{j=1}^{k-1} \operatorname{proj}_{\mathbf{u}_j}\left(\mathbf{v}_k\right), & \mathbf{e}_k &=\frac{\mathbf{u}_k}{\left\|\mathbf{u}_k\right\|} \end{aligned} $$ The sequence $\mathbf{u}_1, \ldots, \mathbf{u}_k$ is a system of orthogonal vectors, and the normalized vectors $\mathbf{e}_1, \ldots, \mathbf{e}_k$ form an orthonormal set.

My question is: is there a matrix form of the GS process? In other words, I'm looking for a matrix $A$ such that $e_j=A v_j$ for all j. In this case $A$ is the matrix that orthogonalizes the set $\{v_j\}$.