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# 数学代写|优化理论代写Optimization Theory代考|CSC4512 Preliminaries

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## 数学代写|优化理论代写Optimization Theory代考|Preliminaries

Matrices of full rank
Recall that if $A$ is an $m \times n$ matrix with $m \leq n$, and $\operatorname{rank}(A)=m$, then $A$ is said to be of full rank. (Reduction of $A$ to row-echelon form will reveal whether or not $A$ has full rank.) In discussing LP algorithms, it is very common to assume that the problem is in standard form and that the coefficient matrix $A$ has full rank. In that case, a basis in $A$ will consist of $m$ linearly independent columns of $A$ which together form an $m \times m$ matrix $B$.
A common notational convention
If $m \leq n$ and $A$ is an $m \times n$ matrix of full rank, it possesses at least one $m \times m$ submatrix $B=A \cdot \beta$ which is a basis. The columns of $B$ are linearly independent columns of $A$. Let $N=A_{\bullet} \nu$ denote the matrix formed by the rest of the columns of $A$. Then (by permuting the columns of $A$ if necessary), we may assume
$$A=[B N] .$$
The vector $x$ has an analogous decomposition and is commonly written as $\left(x_B, x_N\right)$, instead of $\left(x_\beta, x_\nu\right)$. Accordingly $x_B \in R^m$ and $x_N \in R^{n-m}$ are subvectors of $x$ corresponding to the submatrices $B$ and $N$ of $A$. The equation $A x=b$ is then expressed as
$$A x=[B N]\left[\begin{array}{l} x_B \ x_N \end{array}\right]=B x_B+N x_N=b .$$
This style of representation has the advantage of being easy to read. Its disadvantage is that the use of $B$ and $N$ as subscripts – when in fact they are matrices can lead to confusion.

## 数学代写|优化理论代写Optimization Theory代考|Basic solutions again

If $A=[B N]$ and $B$ is a basis in $A$, we get the corresponding basic solution of $A x=b$ as follows: we set $x_N=0$ and find the unique solution of the equation
$$B x_B=b,$$

to obtain
$$\left[\begin{array}{l} x_B \ x_N \end{array}\right]=\left[\begin{array}{c} B^{-1} b \ 0 \end{array}\right]$$
This can also be written as $\left(x_B, x_N\right)=\left(B^{-1} b, 0\right)$. It should be noted that computationally it can be a poor idea to find $x_B$ using $B^{-1}$; however, we will continue to write $x_B=B^{-1} b$ for notational convenience.
Basic feasible solutions
In the LP
$$\begin{array}{ll} \operatorname{minimize} & c^{\mathrm{T}} x \ \text { subject to } & A x=b \ & x \geq 0, \end{array}$$
where $A$ has full rank, the basic solution, $\left(x_B, x_N\right)=\left(B^{-1} b, 0\right)$ corresponding to $B$, is feasible if and only if $x_B=B^{-1} b \geq 0$. When this is the case, we say that $B$ is a feasible basis and the corresponding solution is called a basic feasible solution, or BFS. Note that saying that “B is a basis in $A$ ” is independent of the vector $b$, whereas saying ” $B$ is a feasible basis for the LP (3.1)” does depend on $b$.

## 数学代写|优化理论代写Optimization Theory代考|Preliminaries

$A$ 有满秩。）在讨论 LP 算法时，通常假设问题是标准形式并且系数矩阵 $A$ 有满级。在这种情况下，根据 $A$ 将包括 $m$ 的线性独立列
$A$ 它们一起形成了一个 $m \times m$ 矩阵 $B$.

If $m \leq n$ 和 $A$ 是一个 $m \times n$ 满秩矩阵，它至少有一个 $m \times m$ 子矩阵 $B=A \cdot \beta$ 这是一个基础。的列 $B$ 是线性独立的列 $A$. 让
$N=A \bullet \nu$ 表示由 的其余列形成的矩阵 $A$. 然后（通过排列 $A$ 如有必要），我们可以假设
$$A=[B N] .$$

$$A x=[B N]\left[x_B x_N\right]=B x_B+N x_N=b .$$

## 数学代写|优化理论代写Optimization Theory代考|Basic solutions again

$$B x_B=b,$$

minimize $\quad c^{\mathrm{T}} x$ subject to $A x=b \quad x \geq 0$

## MATLAB代写

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