How to eliminate vector from equation This is probably a super simple question.
Let's say I have an equation
$$
\mathbf{v}^T(\mathbf{a} + \mathbf{b}) = \mathbf{v}^T(\mathbf{c}).
$$
Does this imply that
$$
(\mathbf{a} + \mathbf{b}) = \mathbf{c}
$$?
Intuitively, I would say that it does. On the other hand, there is no left hand side operation I can come up with that eliminates $\mathbf{v}$. I tried with:
$$
\mathbf{v} = \begin{pmatrix}v_1 \\ v_2\end{pmatrix}
$$
$$
\mathbf{x} = \begin{pmatrix}1/v_1\\1/v_2\end{pmatrix}
$$
And then 
$$\mathbf{x}\mathbf{v}^T(\mathbf{a} + \mathbf{b}) = \mathbf{x}\mathbf{v}^T(\mathbf{c})
$$ 
But this doesn't result in the identity matrix on the left side as I had hoped:
$$
\begin{pmatrix}1/v1\\1/v2\end{pmatrix}\begin{pmatrix}v1\ v2\end{pmatrix} = \begin{pmatrix}1 \ \ v_2/v_1 \\ v_1/v_2 \ 1 \end{pmatrix}
$$
 A: Generally speaking: no!
Imagine $\mathbf{v}$ being the zero vector. Or $\mathbf{v}$ being a vector both orthogonal to $\mathbf{c}$ and $\mathbf{a}+\mathbf{b}$.
A whole different story is the following:
$\mathbf{v}\cdot\mathbf{c}=\mathbf{v}\cdot(\mathbf{b}+\mathbf{a})$ for all $\mathbf{v}$ $\Rightarrow$ $\mathbf{c}=\mathbf{b}+\mathbf{a}$
A: Consider $\mathbf{v} = \begin{bmatrix}1\\0\\0\end{bmatrix}$, $\mathbf{x} = \begin{bmatrix}0\\1\\0\end{bmatrix}$, $\mathbf{y} = \begin{bmatrix}0\\0\\1\end{bmatrix}$.
Then $\mathbf{v}^T\mathbf{x} = \mathbf{v}^T\mathbf{y} = 0$ but $\mathbf{x} \neq \mathbf{y}$.
In general, $\mathbf{u}^T\mathbf{v}$ is the dot product of $\mathbf{u}$ and $\mathbf{v}$, which measures the angle between them:
$$\mathbf{u}\cdot\mathbf{v} = \lVert \mathbf{u} \rVert \lVert \mathbf{v} \rVert \cos \theta$$
so if $\mathbf{u}\cdot\mathbf{v} = \mathbf{u}\cdot\mathbf{w}$, then all you can conclude is that $\mathbf{v}$ and $\mathbf{w}$ are the same angle from $\mathbf{u}$, provided they have the same length.
A: No. In fact, $v^T(a+b) = v^Tc$ precisely when $v$ is orthogonal to $a + b - c$ ($v^T(a+b-c)=0$, so $v^T(a+b) = v^Tc$). In all other cases, $v^T(a + b - c)$ is nonzero, and $v^T(a+b) = v^Tc + v^T(a+b-c) \neq v^Tc$. The key observation here is that multiplication by a transpose vector is the same thing as a linear map to the base field, and some such linear maps have non-trivial kernels. 
