I wanted to get some clarification on a linear algebra operation I've not seen before.
In demonstrating how subgradients are found, a lecuturer did the following:
$$\begin{align} 0 &\geq \langle g, y-x\rangle + (-1)(v - f(x))\\ &= \big\langle (g,-1) , (y,v)-(x,f(x))\big\rangle \end{align} $$
Looking at this geometrically, it makes sense. I just want to know what law/axiom this is employing to convert the first inequality to the second. I'm only really familiar with Cauchy-Schwarz, and basic stuff like $\langle a,b\rangle + \langle a,c \rangle = \langle a, b+c \rangle$.
Appreciate the help!
For any $v, w \in \mathbb{R}^n$ and $x, y \in \mathbb{R}^m$, it holds that $\langle v, w \rangle + \langle x, y \rangle = \langle \begin{bmatrix}v\\x\end{bmatrix},\begin{bmatrix}w\\y\end{bmatrix}\rangle$ by the definition of dot product (here $\begin{bmatrix}v\\x\end{bmatrix}$ is the vector the first $n$ components of which are the components of $v$ and the other $m$ components are the components of $x$). In your case, the lecturer used $(v, x)$ notation instead of $\begin{bmatrix}v\\x\end{bmatrix}$. This, together with linearity of dot product in each argument and with the definition of dot product, trivially gives us the desired equality (with $m=1$):
$$\langle \begin{bmatrix}g\\-1\end{bmatrix}, \begin{bmatrix}y\\v\end{bmatrix}-\begin{bmatrix}x\\f(x)\end{bmatrix}\rangle = \langle g, y-x \rangle + \langle \begin{bmatrix} -1 \end{bmatrix}, \begin{bmatrix} v \end{bmatrix} - \begin{bmatrix} f(x) \end{bmatrix} \rangle = \langle g, y-x \rangle + (-1)(v-f(x)).$$