Let us write x = (x$_0$, x$_s$)$^T$, where x$_0$ contains the original variables and x$_s$ contains the $m$ slack variables. Then it is obvious that by setting x$_0$ = 0, we have Ix$_s$ = b.
The slack variables x$_s$ = b and they form an initial basic feasible solution.
I am self-teaching myself the simplex method and I don't understand this statement. Would like an explanation.
Also, what notation is this? Is it vector-matrix notation?
(x$_0$, x$_s$)$^T$ $\implies$ [x$_0$, x$_s$]$^T$
I suppose, the original problem is of the form: maximize $x \cdot c$ subject to $A x \leq b$, where $b$ is a vector with $b \geq 0$. If not, you'd have to specify. The slack variable is defined as $x_s := b - A x$; it is nonnegative for any feasible $x$ and vice versa. Rename $x$ to $x_0$. The problem can then be converted into: maximize $(x_0, x_s) \cdot (c, 0)$ subject to $A x_0 + I x_s = b$ and $x_s \geq 0$, where $I$ is the identity matrix of appropriate size, and $0$ in $(c, 0)$ is a zero vector of the same size as $x_s$. Is there at least one feasible (not necessarily optimal, just to start the method) solution? Yes: just take $x_0 = 0$ and $x_s = b$. The point of this transformation is that the inequality, which is the difficult part, has an easy form in this formulation.