📂行列代数

基本行列

定理: 可逆行列の同値条件

$A$ を大きさが $n\times n$ の正方行列とする。すると、以下の命題はすべて同値である。

(a) $A$ は可逆行列である。

(b) 同次線形システム $A\mathbf{x}=\mathbf{0}$ には自明解のみが存在する。

(c) $A$ の簡約行階段形 は $I_{n}$ である。

(d) $A$ は基本行列の積として表現できる。

Proof

(a) $\implies$ (b)

Assume that $A$ is invertible. And let $\mathbf{x}_{0}$ be any solution of $A\mathbf{x} = \mathbf{0}$. Then the following equation holds.

$$A \mathbf{x}_{0} = \mathbf{0}$$

Multiplying both sides of the above equation by $A^{-1}$ results in:

$$\begin{align*} && \left( A^{-1} A \right)\mathbf{x}_{0} &= A^{-1} \mathbf{0} \\ \implies && \mathbf{x}_{0} &= \mathbf{0} \end{align*}$$

Therefore, $A\mathbf{x} = \mathbf{0}$ has only the trivial solution.

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(b) $\implies$ (c)

Assume that the homogeneous system $A\mathbf{x} = \mathbf{0}$ has only the trivial solution. Then solving this homogeneous system yields:

$$\begin{alignat*}{} x_{1} & & & &=0 \\ & x_{2} & & &=0 \\ & & \ddots & &=0 \\ & & & x_{n} &=0 \end{alignat*}$$

Representing this with the augmented matrix results in:

$$\begin{bmatrix} 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \end{bmatrix}$$

Therefore, the reduced row echelon form of $A$ is $I_{n}$.

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(c) $\implies$ (d)

Assume that the reduced row echelon form of $A$ is $I_{n}$. Then, due to the properties of elementary matrices, there exists an elementary matrix $E_{1}, E_{2}, \dots, E_{k}$ such that the following equation holds:

$$E_{k} \dots E_{2} E_{1} A = I_{n}$$

Then, by the properties of the inverse matrix, the following holds:

$$A = E_{k}^{-1} \dots E_{2}^{-1} E_{1}^{-1} I_{n} = E_{k}^{-1} \dots E_{2}^{-1} E_{1}^{-1}$$

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(d) $\implies$ (a)

Assume that $A$ can be represented as a product of elementary matrices. Since an elementary matrix is an invertible matrix, and the product of invertible matrices is also an invertible matrix, $A$ is an invertible matrix.

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