I did an Algebra on a vector equation and got what seems to be a contradiction.

Question

This was an edit on a previos question that I don't think anyone saw. In the previous question, I did algebra to an equation with vectors $\vec u, \vec v \in \mathbb R^2$ where I got a function equal to $\frac{\vec u}{\vec v}$ and was trying to figure out if this was defined or sensible in anyway.

Tinkering around with with the problem, I tried out this simplified problem (I dropped the units for readability, so assume velocities are all meters per second, distances are all meters, and time is all seconds):

Object $q$ at time $t = 0 $ is located at $x(0) = \begin{bmatrix} 0 \\0 \end{bmatrix}$ and traveling at a velocity $\dot x = \begin{bmatrix} 10 \\12 \end{bmatrix}$. At what time does $x(t) = \begin{bmatrix} 5 \\6 \end{bmatrix}$?

Our function is:

$$ x(t) = x(0) + \dot x(t) $$

Plugging in the knowns:

$$\begin{bmatrix} 0 \\0 \end{bmatrix} + \begin{bmatrix} 10 \\12 \end{bmatrix}t = \begin{bmatrix} 5 \\6 \end{bmatrix}$$

We can get rid of the zero vector:

$$\begin{bmatrix} 10 \\12 \end{bmatrix}t = \begin{bmatrix} 5 \\6 \end{bmatrix}$$

Now to isolate the $t$, we do the following operation:

$$\begin{bmatrix} \frac{1}{10} && \frac{1}{12} \end{bmatrix}\begin{bmatrix} 10 \\12 \end{bmatrix}t = \begin{bmatrix} \frac{1}{10} && \frac{1}{12} \end{bmatrix}\begin{bmatrix} 5 \\6 \end{bmatrix} $$

$$2t = 1$$

$$t = \frac{1}{2}$$

All well and good and what we expect. It takes half a second with that velocity vector to reach the final position.

The trouble comes if we make $\dot x = \begin{bmatrix} 10 \\ 13 \end{bmatrix}$. If that is our velocity vector starting from the origin, we miss $x(t) = \begin{bmatrix} 5 \\ 6 \end{bmatrix}$.

Nonetheless, I wanted to see what would happen if I blindly applied the above algebra to this equation that should have no solutions.

$$\begin{bmatrix} 10 \\13 \end{bmatrix}t = \begin{bmatrix} 5 \\6 \end{bmatrix}$$ $$\begin{bmatrix} \frac{1}{10} && \frac{1}{13} \end{bmatrix}\begin{bmatrix} 10 \\13 \end{bmatrix}t = \begin{bmatrix} \frac{1}{10} && \frac{1}{13} \end{bmatrix}\begin{bmatrix} 5 \\6 \end{bmatrix} $$

$$2t = \frac{1}{2} + \frac{6}{13} = \frac{25}{26}$$

$$t = \frac{25}{52}$$

But:

$$\frac{25}{52} \begin{bmatrix} 10 \\13 \end{bmatrix} \approx \begin{bmatrix} 4.81 \\6.25 \end{bmatrix} \ne \begin{bmatrix} 5 \\6 \end{bmatrix}$$

What just happened? My understanding from a bit of the math I've learned in quantum mechanics and doing vector operations in Hilbert spaces was that:

$$ | \psi \rangle = | \phi \rangle \implies \langle \xi | \psi \rangle = \langle \xi | \phi \rangle $$

Given that these are real vectors and I don't need to complex conjugate them to get their inner products, it seems like this should hold. And it seems like I got (maybe) the point in time that is closest to the target position vector, but it still nonetheless seems like this should be a contradiction. What happing?

Answer 1

The equation $\begin{bmatrix}10\\12\end{bmatrix}t = \begin{bmatrix}5\\6\end{bmatrix}$ is a system of 2 linear equations in $t$, namely $10t = 5$ and $12t = 6$. Multiplying on the left by $\begin{bmatrix}1/10 & 1/12\end{bmatrix}$ transforms this system of 2 linear equations into a single linear equation by adding $1/10$ times the first to $1/12$ times the second, yielding the equation $$ \frac{1}{10}(10t) + \frac{1}{12}(12t) = \frac{1}{10}(5)+\frac{1}{12}(6). $$ Any solution to the original system will be a solution to this new equation. But the converse need not hold in general. That is, there is no reason to expect a solution to the new equation to be a solution to the original system of equations. That said, the converse does happen to hold here because the two equations in the original system are just scalar multiples of one another, so any solution to one is a solution to the other.

But in your modified scenario, with $13$ in place of $12$, this is no longer the case: the system has no solutions, even though the equation obtained by adding together the multiples of the two equations does have a solution.

In terms of inner products, it is true that for any vectors $a,b,c$, we have $a = b$ implies $\langle a,c\rangle=\langle b, c\rangle$. But what you’re trying to use here is the converse, which is false: $\langle a,c\rangle=\langle b, c\rangle$ for some given $c$ does not mean that $a=b$. Intuitively, taking the inner product with $c$ collapses a multi-dimensional vector $a$ down into a one-dimensional scalar $\langle a,c\rangle$, and thus loses some information. Put more formally, the map $a\mapsto \langle a,c\rangle$ is not injective.

EDIT

As for your follow-up questions. Geometrically, what you’ve done is found the vector parallel to $\begin{bmatrix}10\\13\end{bmatrix}$ whose projection onto (the line spanned by) $\begin{bmatrix}1/10\\1/13\end{bmatrix}$ is the same as that of $\begin{bmatrix}5\\6\end{bmatrix}$:

This isn’t quite the closest vector to $\begin{bmatrix}5\\6\end{bmatrix}$ that lies parallel to $\begin{bmatrix}10\\13\end{bmatrix}$, although it happens to be close. But if your lines were steeper (i.e., the blue line were steeper, which would then make the orange line much less steep), then by looking at the picture above you might be able to see that the resulting “approximation” (in green) would end up being much further from the actual closest point on the (blue) line (because the black line would itself be much further from being perpendicular to the blue line).

If you had multiplied by $\begin{bmatrix}10&13\end{bmatrix}$ instead of $\begin{bmatrix}1/10&1/13\end{bmatrix}$, you would have gotten the actual closest point on the line, since then you would have computed the vector $v$ parallel to $\begin{bmatrix}10\\13\end{bmatrix}$ whose projection onto $\begin{bmatrix}10\\13\end{bmatrix}$ (which is just $v$, since it’s already parallel) is the same as that of $\begin{bmatrix}5\\6\end{bmatrix}$.

As for the question about not finding out algebraically that there is really no solution: it is just much easier to be a solution to the single equation obtained by summing the original equations than it is to be a solution to both of the original equations. You do indeed find out that you don’t have a solution when you test it out by plugging in to the original equations. Also note that if you had multiplied by $\begin{bmatrix}-1/10&1/13\end{bmatrix}$, you indeed would have found a contradiction, since the left-hand side of the resulting equation would be $0$ whereas the RHS would be non-zero.

I think the moral of the story is that the choice of $\begin{bmatrix}1/10&1/13\end{bmatrix}$, while it makes the algebra nice, is not geometrically well-motivated, so it’s not reasonable to expect nice things to happen in general when you pick the vector to multiply by just by inverting the elements of your original vector.

Answer 2

You're taking the (weighted) projection of the $x(t)$ vector on $\left(\begin{matrix} \frac{1}{10} \\ \frac{1}{13} \end{matrix}\right)$ for two non-parallel vectors. Lets say you have two arbitrary non-parallel vectors. Their (unweighted) projection onto the x-axis is going to simply be their x coordinate respectively. Seeing that the vector components generally scale at different rates (x-rate vs. y-rate) as the vector magnitude increases (this is akin to $\sin$ = opposite/hypotenuse, $\cos$ = adjacent/hypotenuse), taking the proportion of two vectors in one direction will result differently if we took the proportion in another direction.

Lets say we have vectors $a=\left(\begin{matrix} 1 \\ 2 \end{matrix}\right)$ and $ b=\left(\begin{matrix} 2 \\ 1 \end{matrix}\right)$. Projecting them onto the standard basis vectors we have $a_1=\frac{1}{2}b_1$ and $a_2 = 2b_2$. Multiplying $b$ by the first scaling gives $\hat a=\left(\begin{matrix} 1 \\ \frac{1}{2} \end{matrix}\right)$, multiplying $b$ by the second scaling gives $\hat a=\left(\begin{matrix} 4 \\ 2\end{matrix}\right)$.

From this I think you'll be able to see the cause.