# Understanding rectangular matrices

• I
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.

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.

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BvU
Homework Helper
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.

Last edited:
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
Homework Helper
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}##

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