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While looking at the derivation of Linear optimization, I see that it starts by considering standard form; $$min(c^T x)$$ $$Ax = b ; \qquad x \ge 0 \qquad b \ge 0$$ Then, it proceeds saying $Without\ Loss\ of\ generality\ rank(A) = m < n $, where, $A$ can be represented as $$ [ B\ N ]\ [x_b\ x_n]^T = b $$ $B = m\ x\ m$ and $ N = m\ x\ (n-m)$. Then we consider $B^{-1}$ and so on to proceed towards Simplex Algorithm.

I wanted to know how do we exclude the fact that $A$ can have $m >n $. Or is it possible to convert it back to standard form?

Also, why is it necessary that $A$ has rank $m$?

EDIT: $m$ is number of rows and $n$ is number of columns.

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    What do you mean by "thick" ? – 2017-02-02
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    that number of rows are less than number of columns. – 2017-02-02
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    Not usual, nor satisfying, because the transpose of a thick matrix would be a thin matrix. Usually, when a person is thick, he is always thick either on his legs or lying on his bed (in which case, he could be sick as well). I have sometimes seen "portrait matrix" or "landscape matrix" that convey some poetry in an area which is somewhat deprived of humanity. – 2017-02-02
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    Thanks for pointing it out. I would avoid using the term now onwards. – 2017-02-02

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I assume $x \in \mathbb{R}^n$ and $b \in \mathbb{R}^k$ (and therefore $A \in \mathbb{R}^{k \times n}$).

If $A$ has rank $n$ (and therefore $k\geq n$), there is only one point $x$, that satisfies the linear constraints. It comes down to solving a linear system of equation, which is hardly worthy to be called optimization.

It is impossible to have $rank(A) >n$, since $A$ has a maximum of $n$ collumns.

The only option, that leaves us with an optimization procedure is $rank(A)

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    Oh, yes you are right. I was thinking something else. It seems elementary now. Thanks. – 2017-02-02