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If $\mathbf{v}\cdot\mathbf{A}=\mathbf{v}\cdot\mathbf{B}$ for all $\mathbf{v}$, does it imply $\mathbf{A}=\mathbf{B}$? Can there be a simple proof for this?

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    $\begingroup$ $\mathbf v\cdot(\mathbf A-\mathbf B)=0$ $\endgroup$
    – J. W. Tanner
    Jul 27, 2020 at 16:11
  • $\begingroup$ But can $(A-B)$ not always be perpendicular to $v$? $\endgroup$
    – Indeterminate
    Jul 27, 2020 at 16:12

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$\mathbf{v}\cdot\mathbf{A}=\mathbf{v}\cdot\mathbf{B}$ implies $\mathbf v\cdot(\mathbf A-\mathbf B)=0$. Then take $\mathbf v = (\mathbf A-\mathbf B)$ to obtain $\Vert \mathbf A-\mathbf B \Vert^2=0$ and hence $\mathbf A=\mathbf B$.

Provided of course that by $ \cdot $, you mean an inner product!

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