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:

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


2 Answers 2


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

  • $\begingroup$ Thanks Max. I don't understand the last part of your answer though? Are you saying c = b + a if v is nonzero? $\endgroup$
    – Sandi
    Nov 15, 2018 at 11:55
  • $\begingroup$ 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 $\endgroup$
    – b00n heT
    Nov 15, 2018 at 11:58
  • 1
    $\begingroup$ 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 $\mathbf{v}$. $\endgroup$
    – maxmilgram
    Nov 15, 2018 at 11:58
  • $\begingroup$ 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 $\mathbf{v}$? $\endgroup$
    – Sandi
    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.


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