Information loss when taking the dot product of vector equations

[Van Ladmon]

Homework Statement

Is the following conclusion correct? Assume there's an equation with vectors on both sides. Taking the dot product of this equation with vectors on both sides loses information, but information will not lose when taking dot products with higher rank tensors on both sides. What's the reason behind this?

Homework Equations

$$\mathbf{a}=\mathbf{b}\Rightarrow\mathbf{a}\cdot\mathbf{x}=\mathbf{b}\cdot\mathbf{x}$$
$$\mathbf{a}=\mathbf{b}\Longleftrightarrow\mathbf{a}\cdot\mathbf{T}=\mathbf{b}\cdot\mathbf{T}$$

The Attempt at a Solution

In the second situation, it is always feasible to take the dot product of the inverse tensor on both sides and the equation with be restored.

 

[fresh_42]

Homework Statement

Is the following conclusion correct? Assume there's an equation with vectors on both sides. Taking the dot product of this equation with vectors on both sides loses information, but information will not lose when taking dot products with higher rank tensors on both sides. What's the reason behind this?

Homework Equations

$$\mathbf{a}=\mathbf{b}\Rightarrow\mathbf{a}\cdot\mathbf{x}=\mathbf{b}\cdot\mathbf{x}$$
$$\mathbf{a}=\mathbf{b}\Longleftrightarrow\mathbf{a}\cdot\mathbf{T}=\mathbf{b}\cdot\mathbf{T}$$

The Attempt at a Solution

In the second situation, it is always feasible to take the dot product of the inverse tensor on both sides and the equation with be restored.

In the first case, we have $(a\cdot x - b\cdot x)=(a-b)\cdot x =0$ which means, all $x$ perpendicular to $a-b$ fulfill this equation. This means, that there is no unique $x$, which thus cannot be canceled to get back $a-b=0$. Exception: dimension one.

In the second case, $T$ represents (implicitly assumed by you) a regular transformation. Regularity means, we can invert it. Or with the argument above: $(a-b)\cdot T = 0$ and injectivity of $T$ allows us to conclude $a-b=0$.
More elementary said, the second equation are actually at least as many equations as the dimension of $a,b$ is, and thus we can uniquely solve a system of linear equations. In the previous case, we had dimension many unknowns and only one equation, which yields to an underdetermined system.