I have some questions about the uniqueness of matrices when post- and pre-multiplied with vectors (inner product).

Say we have two vectors $\vec{a}$ and $\vec{b}$, whose inner product is a scalar, known to satisfy the following equation involving matrix $\left[C\right]$:

$$ \vec{a} \cdot \vec{b} = \vec{a}^T \vec{b} = \vec{a}^T \left[C\right] \vec{b} $$

In this case, is $\left[C\right]$ guaranteed to be the identity matrix? Can it be anything else? Why?

Along the same lines, is it ever possible to "eliminate" vectors from an equation? For example, if we also have a matrix $\left[D\right]$ that satisfies the equation:

$$ \left[C\right] \vec{b} = \left[D\right] \vec{b} $$

Could we just post-multiply each side by $\vec{b}^{-1}$ to obtain $\left[C\right]$ = $\left[D\right]$? Is this valid under any set of conditions?



2 Answers 2


For any two specific vectors, no $C$ does NOT have to be the identity matrix. For example if one of the vectors is $0$ then $C$ can be anything.

The only way to talk about $b^{-1}$ is if $b$ is one dimensional and not $0$.

  • $\begingroup$ Thanks. Would this still be the case if neither of the vectors is $\vec{0}$? $\endgroup$ Commented Jul 27, 2016 at 15:39
  • 4
    $\begingroup$ @StuartBarth Yes. Consider $a=(1,1)^T$, $b=(1,0)^T$ and $C=\pmatrix{\frac 12 & 0 \\ \frac 12 & 0}$. $\endgroup$
    – user137731
    Commented Jul 27, 2016 at 15:54

First question, let $\{v_i\}$ be a basis, then $\vec{a}=\sum_i a_i v_i$ and $\vec{b}=\sum_i b_i v_i$.

Let $C_{ij}=v_i\cdot v_j$, this is for example $\delta_{ij}$ if the basis is orthonormal, but we're not assuming that.

Then by bilinearity of the dot product:

$\vec{a}\cdot \vec{b}=\sum_i\sum_j a_ib_j v_i\cdot v_j=\sum_i\sum_j a_ib_j C_{ij}$.

The $C_{ij}$ is just a collection of numbers. $C_{ij}$ has some restrictions, as all entries can't be zero, $C_{ii}>0$ if the dot product is an inner product, and so on.

For the second question consider $\vec{b}=(1,0)^T$, $D$ the zero matrix and $C=\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}$,


$$C\vec{b}=\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}\begin{pmatrix}1\\ 0\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}=\begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}\begin{pmatrix}1\\ 0\end{pmatrix}=D\vec{b}$$


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