by Sother
Last Updated July 09, 2017 00:20 AM

I am watching a lecture from my professor at MIT, and he says that we can solve this system of linear equations using a projection. I’m not able to see how it done.

We are solving for $dx$ and $ds$ which are vectors:

$\mu dx + ds = 1 \mu$ where $\mu$ is a scalar, real number.

$ A dx = 0$

$(dy) A + ds = 0$

He says that since we know:

$dx \in A$

$ds \in A^\perp$

…then we can use a projection to solve for $dx$ and $ds$. (we will get $dy$ too I think, but we don’t care about its value). He say:

“Find a vector, that’s the all ones vector, and break it up into a part that is inside of A and a part that is perpendicular to A. This is a projection”.

I have not had any luck going about solving this. But I have a few confusions about this.

1) The equation $ A dx = 0$ does not mean that $dx$ is inside of A. It means it is in the nullspace of A, so I think that the professor is wrong here in saying that.

2) The professor is also wrong in saying that it is the “all ones” vector we are projecting. It is not the “all ones” vector that we are projecting, it is a scaled version of the all ones vector, which admittedly doesn’t change the direction, but still changes the solution to the linear system.

3) No matter what I use for the matrix A, when I solve this using a projection, I always get the $ds$ vector as the zero vector, and the $dx$ vector as having all non-zero values. This causes most of the equations to be violated. E.g., I get values for $dx$ s.t. $A dx \ne 0$

How do I solve this using a projection of the all ones vector?

*P.S.* I am watching https://www.youtube.com/watch?v=78sNnf3pOYs at approximately minute 53.

Starting at 53:26, the speaker has swapped the symbols $A$ and $A^\perp$. If $A \cdot \overline{dx} = \overline{0}$, (where I use overlines to indicate vectors because the MathJax "$\vec{dx}$" is awful) then $dx$ is perpendicular to each row vector in $A$, so $dx \in A^\perp$.

Note that $\mu \, \overline{dx} + \overline{ds} = \overline{1} \, \mu$ can be rescaled to $\overline{dx} + \frac{1}{\mu} \overline{ds} = \overline{1}$, so we can just as easily think of this as projecting $\overline{1}$ into a "perpendicular to the $Ax = b$ constraint ($\overline{dx}$) component" and a "scaled copy of the parallel component". Some prefer the equation with only one mention of $\mu$ and some prefer not having the division. Both represent equivalent computations.

When I try the easiest possible version of solving this system, with $A = 1$ (the $1 \times 1$ matrix containing $1$), I get $\overline{dx} = 0$, $\overline{dy} = -\mu$, and $\overline{ds} = \mu$. Are you sure you're solving the system correctly?

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