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# 数学代写|交换代数代写Commutative Algebra代考|Generalized Cramer Formula

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## 数学代写|交换代数代写Commutative Algebra代考|Generalized Cramer Formula

We study in this subsection some generalizations of the usual Cramer formulas. We will exploit these in the following paragraphs.

For a matrix $A \in \mathbf{A}^{m \times n}$ we denote by $A_{\alpha, \beta}$ the matrix extracted on the rows $\alpha=\left{\alpha_1, \ldots, \alpha_r\right} \subseteq \llbracket 1 . . m \rrbracket$ and the columns $\beta=\left{\beta_1, \ldots, \beta_s\right} \subseteq \llbracket 1 . . n \rrbracket$.

Suppose that the matrix $A$ is of rank $\leqslant k$. Let $V \in \mathbf{A}^{m \times 1}$ be a column vector such that the bordered matrix $[A \mid V]$ is also of rank $\leqslant k$. Let us call $A_j$ the $j$-th column of $A$. Let $\mu_{\alpha, \beta}=\operatorname{det}\left(A_{\alpha, \beta}\right)$ be the minor of order $k$ of the matrix $A$ extracted on the rows $\alpha=\left{\alpha_1, \ldots, \alpha_k\right}$ and the columns $\beta=\left{\beta_1, \ldots, \beta_k\right}$. For $j \in \llbracket 1 . . k \rrbracket$ let $\nu_{\alpha, \beta, j}$ be the determinant of the same extracted matrix, except that the column $j$ has been replaced with the extracted column of $V$ on the rows $\alpha$. Then, we obtain for each pair $(\alpha, \beta)$ of multi-indices a Cramer identity:
$$\mu_{\alpha, \beta} V=\sum_{j=1}^k \nu_{\alpha, \beta, j} A_{\beta_j}$$
due to the fact that the rank of the bordered matrix $\left[A_{1 . . m, \beta} \mid V\right]$ is $\leqslant k$. This can be read as follows:
\begin{aligned} \mu_{\alpha, \beta} V & =\left[A_{\beta_1} \ldots A_{\beta_k}\right] \cdot\left[\begin{array}{c} \nu_{\alpha, \beta, 1} \ \vdots \ \nu_{\alpha, \beta, k} \end{array}\right] \ & =\left[A_{\beta_1} \ldots A_{\beta_k}\right] \cdot \operatorname{Adj}\left(A_{\alpha, \beta}\right) \cdot\left[\begin{array}{c} v_{\alpha_1} \ \vdots \ v_{\alpha_k} \end{array}\right] \ & =A \cdot\left(\mathrm{I}n\right){1 . . n, \beta} \cdot \operatorname{Adj}\left(A_{\alpha, \beta}\right) \cdot\left(\mathrm{I}m\right){\alpha, 1 . . m} \cdot V \end{aligned}
This leads us to introduce the following notation.

## 数学代写|交换代数代写Commutative Algebra代考|A Magic Formula

An immediate consequence of the Cramer’s identity (12) is the less usual identity (17) given in the following theorem. Similarly the equalities (18) and (19) easily result from (15) and (16).
5.14 Theorem Let $A \in \mathbf{A}^{m \times n}$ be a matrix of rank $k$. We thus have an equality $\sum_{\alpha \in \mathcal{P}{k, m}, \beta \in \mathcal{P}{k, n}} c_{\alpha, \beta} \mu_{\alpha, \beta}=1$. Let
$$B=\sum_{\alpha \in \mathcal{P}{k, m}, \beta \in \mathcal{P}{k, n}} c_{\alpha, \beta} \operatorname{Adj}_{\alpha, \beta}(A)$$

1. We have
$$A \cdot B \cdot A=A$$
Consequently $A B$ is a projection matrix of rank $k$ and the submodule $\operatorname{Im} A=$ $\operatorname{Im} A B$ is a direct summand in $\mathbf{A}^m$.
2. If $k=m$, then
$$A \cdot B=\mathrm{I}_m$$
3. If $k=n$, then
$$B \cdot A=\mathrm{I}n$$ The following identity, which we will not use in this work, is even more miraculous. 5.15 Proposition (Prasad and Robinson) With the assumptions and the notations of Theorem 5.14 , if we have $$\forall \alpha, \alpha^{\prime} \in \mathcal{P}{k, m}, \forall \beta, \beta^{\prime} \in \mathcal{P}{k, n} \quad c{\alpha, \beta} c_{\alpha^{\prime}, \beta^{\prime}}=c_{\alpha, \beta^{\prime}} c_{\alpha^{\prime}, \beta}$$
then
$$B \cdot A \cdot B=B$$

## 数学代写|交换代数代写Commutative Algebra代考|Generalized Cramer Formula

$$\mu_{\alpha, \beta} V=\sum_{j=1}^k \nu_{\alpha, \beta, j} A_{\beta_j}$$

\begin{aligned} \mu_{\alpha, \beta} V & =\left[A_{\beta_1} \ldots A_{\beta_k}\right] \cdot\left[\begin{array}{c} \nu_{\alpha, \beta, 1} \ \vdots \ \nu_{\alpha, \beta, k} \end{array}\right] \ & =\left[A_{\beta_1} \ldots A_{\beta_k}\right] \cdot \operatorname{Adj}\left(A_{\alpha, \beta}\right) \cdot\left[\begin{array}{c} v_{\alpha_1} \ \vdots \ v_{\alpha_k} \end{array}\right] \ & =A \cdot\left(\mathrm{I}n\right){1 . . n, \beta} \cdot \operatorname{Adj}\left(A_{\alpha, \beta}\right) \cdot\left(\mathrm{I}m\right){\alpha, 1 . . m} \cdot V \end{aligned}

## 数学代写|交换代数代写Commutative Algebra代考|A Magic Formula

5.14定理设$A \in \mathbf{A}^{m \times n}$为秩为$k$的矩阵。因此我们有一个等式$\sum_{\alpha \in \mathcal{P}{k, m}, \beta \in \mathcal{P}{k, n}} c_{\alpha, \beta} \mu_{\alpha, \beta}=1$。让
$$B=\sum_{\alpha \in \mathcal{P}{k, m}, \beta \in \mathcal{P}{k, n}} c_{\alpha, \beta} \operatorname{Adj}_{\alpha, \beta}(A)$$

$$A \cdot B \cdot A=A$$

## MATLAB代写

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