Understanding rectangular matrices

[Urmi Roy]

So if I have a system of equations:
$$x_1+x_2+x_3=1$$
and $$x_4+x_5+x_6=1$$

Then they can be put into a matrix representation
\begin{equation*}
\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 & 1\\
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5 \\
x_6 \\
\end{bmatrix}
=
\begin{bmatrix}
1 \\
1 \\
\end{bmatrix}
\end{equation*}

So I know that columns in a matrix represent vectors. Is it true that in this matrix we therefore have 6 2D vectors?

Also it looks like there are only 2 vectors, and 3 each of them.

It's just surprising to me that there are 6 variables and only 2D vectors.

If I imagine them to be 2D vectors in x-y plane, then they are also mutually perpendicular. So eventually, the equations I mentioned above, although they look like planes, are they just really the x and y axes?

I guess I'm confused in going from the vector interpretation to the equation interpretation of the matrix.

I would appreciate help in understanding how to interpret the matrix form

 

[Erland]

Does this originate in some textbook problem? Could you then please state it from the book...

It is not clear to me which 2D-vectors you are talking about. Also, your two equations are independent in the sense that they can be solved separately, and I see no reason to put them together in a system of equations.

 

[Urmi Roy]

It's from a linear programming question in standard form. It's a part of a worksheet we got, but I didn't want help with the whole question (so I didn't put it in the homework help forum), just understanding the meaning of the rectangular matrix. I've attached the question here for context.

By 2D vectors I mean
\begin{pmatrix}
1\\
0
\end{pmatrix}
and
\begin{pmatrix}
0\\
1
\end{pmatrix}

There are 3 each of them in the matrix.

 

[BvU]

You are mapping 6 variables (so a 6 dimensional space) onto 2 numbers (so a 2 dimensional space).
The 6 2D vectors you mention are the images of the 6 unit vectors in the 6D domain space.
So $(1,0,0,0,0,0,0)\rightarrow(1,0)$ etc.
These images are clearly not perpendicular: $(0,1,0,0,0,0,0)\rightarrow(1,0)$ also, etc.

With 6 DImensions and 2 constraints (your 2 equations) you have four degrees of freedom.
6D vectors that satisfy your equations look like e.g. $(x-1, x_2, 1-x_1-x_2,x_4,x_5,1-x_4-x_5)$ so they are definitely not the axes.

 

[Urmi Roy]

You are mapping 6 variables (so a 6 dimensional space) onto 2 numbers (so a 2 dimensional space).
The 6 2D vectors you mention are the images of the 6 unit vectors in the 6D domain space.
So $(1,0,0,0,0,0,0)\rightarrow(1,0)$ etc.
These images are clearly not perpendicular: $(0,1,0,0,0,0,0)\rightarrow(1,0)$ also, etc.

With 6 DImensions and 2 constraints (your 2 equations) you have four degrees of freedom.
6D vectors that satisfy your equations look like e.g. $(x-1, x_2, 1-x_1-x_2,x4,x_5,1-x_4-x_5)$ so they are definitely not the axes.

Can you elaborate a bit more? In a regular Ax=b formulation, where A is, say, a 3x3 matrix, the columns are vectors in 3D space. the elements of x are 'weighting factors' which multiply the columns (vectors) and the result b is the vector sum.

So we have

\begin{equation*}
\begin{pmatrix}
1 & 2 & 3\\
5 & 7 & 8 \\
3 & 1 & 12\\
\end{pmatrix}
\begin{pmatrix}
x\\
y \\
z\\
\end{pmatrix}
=
\begin{pmatrix}
1x+2y+3z\\
5x+7y+8z \\
3x+1y+12z\\
\end{pmatrix}
=
x\begin{pmatrix}
1\\
5 \\
3\\
\end{pmatrix}
+
y\begin{pmatrix}
2\\
7 \\
1\\
\end{pmatrix}
+
z\begin{pmatrix}
3\\
8 \\
12\\
\end{pmatrix}
\end{equation*}

Why is this not the same for my example? Why are the columns in my rectangular matrix not 2D vectors?

 

[BvU]

Why are the columns in my rectangular matrix not 2D vectors?

Sounds weird when I just wrote

The 6 2D vectors you mention are the images of the 6 unit vectors in the 6D domain space.
So $\ (1,0,0,0,0,0,0)\rightarrow(1,0)\$etc.

thereby calling them vectors, isn't it ?

Exactly analogous to your 3 x 3 example, where $\begin{pmatrix} 1\\ 5 \\ 3\\ \end{pmatrix}$ is the image of unit vector $\begin{pmatrix} 1\\ 0 \\ 0\\ \end{pmatrix}$

 

[Urmi Roy]

You are mapping 6 variables (so a 6 dimensional space) onto 2 numbers (so a 2 dimensional space).
The 6 2D vectors you mention are the images of the 6 unit vectors in the 6D domain space.
So $(1,0,0,0,0,0,0)\rightarrow(1,0)$ etc.
These images are clearly not perpendicular: $(0,1,0,0,0,0,0)\rightarrow(1,0)$ also, etc.

With 6 DImensions and 2 constraints (your 2 equations) you have four degrees of freedom.
6D vectors that satisfy your equations look like e.g. $(x-1, x_2, 1-x_1-x_2,x4,x_5,1-x_4-x_5)$ so they are definitely not the axes.

Ok so thinking about this a bit more, I'm getting most of it. You say that since $(1,0,0,0,0,0,0)\rightarrow(1,0)$
and also that $(0,1,0,0,0,0,0)\rightarrow(1,0)$,
The 2D vectors are not perpendicular? I'm just struggling with the idea of why the 2D vectors in their 2D system aren't perpendicular. As I said, it's just easiest for me to imagine them as the x and y axes. Thanks for your help!

 

[BvU]

Yes, $(1,0)$ and $(1,0)$ are clearly one and the same and therefore not perpendicular. You can't find more than two mutually perpendicular vectors in 2D. As it happens the images of $(1,0,0,0,0,0)$ and $(0,0,0,1,0,0)$ are perpendicular, but you can't say: "the 6 unit vectors in the 6D domain of this matrix have images in the 2D range space that are perpendicular"