# 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:

$$\begin{pmatrix}1/v1\\1/v2\end{pmatrix}\begin{pmatrix}v1\ v2\end{pmatrix} = \begin{pmatrix}1 \ \ v_2/v_1 \\ v_1/v_2 \ 1 \end{pmatrix}$$

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 $$\mathbf{v}$$ $$\Rightarrow$$ $$\mathbf{c}=\mathbf{b}+\mathbf{a}$$

• Thanks Max. I don't understand the last part of your answer though? Are you saying c = b + a if v is nonzero? Nov 15, 2018 at 11:55
• Max is saying that if the equation is true for all vectors $v$, then indeed $c=a+b$. This is a consequence of Gram Schmidt Nov 15, 2018 at 11:58
• If you know that the equation holds true for all vectors $\mathbf{v}$ of your vector space the implication is true. But it is not enough that you have it for one such vector $\mathbf{v}$. Nov 15, 2018 at 11:58
• Okay, in my application, the $\mathbf{v}$ vector is the random variables of a multivariate normal distribution, and the equation is derived from multiplying a prior Gaussian distribution with a Gaussian likelihood and then comparing the quadratic terms of the exponents with the posterior distribution. Since $\mathbf{v}$ is any collection of possible values of the random variables, can I conclude that my equation holds for all $\mathbf{v}$? Nov 15, 2018 at 12:06

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.