Exploring Canonical Form of Linear Programming

[evinda]

Hello! (Wave)

A linear programming problem is in canonical form if it's of the following form:

$$\pm \max (c_1 x_1+ \dots + c_n x_n) , c_1, \dots, c_n \in \mathbb{R} \\ Ax=b, A \in F^{m \times n}, x=\begin{bmatrix}
x_1\\
\dots\\
\dots \\
x_n
\end{bmatrix}, b=\begin{bmatrix}
b_1\\
\dots\\
\dots \\
b_m
\end{bmatrix}$$

$(A=[a_{ij}], a_{ij} \in \mathbb{R})$

$b_j \geq 0, j=1, \dots, m \\ x_i \geq 0, i=1, \dots, n$

$r(A)=m<n$

where $r(A)$ is the rank of the matrix $A$.

$r(A):=$ maximum number of linearly independent rows of the matrix $A$.First of all, could you explain me why it has to hold that $A \in F^{m \times n}$ where $F$ is the set of feasible solutions?
Doesn't $F$ contain only the $x$ s for which it holds that $Ax=b$ ? Or am I wrong?

Also, does the condition $r(A)=m<n$ imply that all the $m$ equations that we get from $Ax=b$ are not multiple the one of the other? Or have I understood it wrong? (Thinking)

 

[I like Serena]

First of all, could you explain me why it has to hold that $A \in F^{m \times n}$ where $F$ is the set of feasible solutions?
Doesn't $F$ contain only the $x$ s for which it holds that $Ax=b$ ? Or am I wrong?

Hey evinda! (Smile)

It seems to me that the definitions are a bit mixed up.
I think $F$ is just a field, typically $\mathbb R$.
The set of feasible solutions is the set of all $x$ that satisfy $Ax=b, x_i \geq 0$, which is a subset of $F^n$.
And the matrix $A$ is an $m\times n$ matrix with elements of $F$, so $A \in F^{m\times n}$. (Mmm)

Also, does the condition $r(A)=m<n$ imply that all the $m$ equations that we get from $Ax=b$ are not multiple the one of the other? Or have I understood it wrong? (Thinking)

Even stronger, anyone of the equations cannot be a linear combinations of any of the other equations.
The equations have to be linearly independent.
And there have to be infinitely many solutions $x$ for $Ax=b$ (Nerd)

 

[evinda]

Hey evinda! (Smile)

It seems to me that the definitions are a bit mixed up.
I think $F$ is just a field, typically $\mathbb R$.
The set of feasible solutions is the set of all $x$ that satisfy $Ax=b, x_i \geq 0$, which is a subset of $F^n$.
And the matrix $A$ is an $m\times n$ matrix with elements of $F$, so $A \in F^{m\times n}$. (Mmm)

Ah, I see... (Nod)

Even stronger, anyone of the equations cannot be a linear combinations of any of the other equations.

Any of the $m$ equations, right? (Thinking)

And there have to be infinitely many solutions $x$ for $Ax=b$ (Nerd)

Why does this hold?

 

[I like Serena]

Any of the $m$ equations, right? (Thinking)

Exactly. (Nod)

Why does this hold?

Because $\operatorname{rank} A = m < n$.
That means more columns than rows, while the rows are independent.
In turn that means more variables than equations, which guarantees infinitely many solutions.

As an example, let's pick $F=\mathbb R, m=1, n=2$ and $2x+3y = 6$.
How many solutions does it have? (Wondering)

 

[evinda]

As an example, let's pick $F=\mathbb R, m=1, n=2$ and $2x+3y = 6$.
How many solutions does it have? (Wondering)

Infinitely many... We can write the above equivalently as follows, right?

$\begin{bmatrix}
2 & 3
\end{bmatrix} \begin{bmatrix}
x\\
y
\end{bmatrix}=6$

How can we justify then that there are infinitely many solutions? (Thinking)

 

[I like Serena]

Infinitely many... We can write the above equivalently as follows, right?

$\begin{bmatrix}
2 & 3
\end{bmatrix} \begin{bmatrix}
x\\
y
\end{bmatrix}=6$

How can we justify then that there are infinitely many solutions? (Thinking)

Right. (Nod)

If we have $m$ independent equations with $m$ variables, there is a unique solution.
We can tell because if $A$ is square, the fact that the rows are independent means that it's invertible.
So $Ax=b \Rightarrow x = A^{-1}b$.

Now that we have more than $m$ variables, we can for starters set them to zero, giving us a first solution.
And since we have a linear dependency in the columns, we can shift them around, giving us infinitely many solutions.
This corresponds to moving along a boundary line of the feasible region. (Thinking)