Row Operation has Inverse

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Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.

Let $\mathbf A \in \map \MM {m, n}$ be a matrix.

Let $\Gamma$ be a row operation which transforms $\mathbf A$ to a new matrix $\mathbf B \in \map \MM {m, n}$.

Then there exists another row operation $\Gamma'$ which transforms $\mathbf B$ back to $\mathbf A$.


Let $\sequence {e_i}_{1 \mathop \le i \mathop \le k}$ be the finite sequence of elementary row operations that compose $\Gamma$.

Let $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$ be the corresponding finite sequence of the elementary row matrices.

From Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices, we have:

$\mathbf R \mathbf A = \mathbf B$

where $\mathbf R$ is the product of $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$:

$\mathbf R = \mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1$

By Elementary Row Matrix is Invertible, each of $\mathbf E_i$ is invertible.

By Product of Matrices is Invertible iff Matrices are Invertible, it follows that $\mathbf R$ is likewise invertible.

Thus $\mathbf R$ has an inverse $\mathbf R^{-1}$.


\(\ds \mathbf R \mathbf A\) \(=\) \(\ds \mathbf B\)
\(\ds \leadsto \ \ \) \(\ds \mathbf R^{-1} \mathbf R \mathbf A\) \(=\) \(\ds \mathbf R^{-1} \mathbf B\)
\(\ds \leadsto \ \ \) \(\ds \mathbf A\) \(=\) \(\ds \mathbf R^{-1} \mathbf B\)

We have:

\(\ds \mathbf R^{-1}\) \(=\) \(\ds \paren {\mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1}^{-1}\)
\(\ds \) \(=\) \(\ds {\mathbf E_1}^{-1} {\mathbf E_2}^{-1} \dotsb {\mathbf E_{k - 1} }^{-1} {\mathbf E_k}^{-1}\) Inverse of Matrix Product

From Elementary Row Matrix for Inverse of Elementary Row Operation is Inverse, each of ${\mathbf E_i}^{-1}$ is the elementary row matrix corresponding to the inverse $e'_i$ of the corresponding elementary row operation $e_i$.

Let $\Gamma'$ be the row operation composed of the finite sequence of elementary row operations $\tuple {e'_k, e'_{k - 1}, \ldots, e'_2, e'_1}$.

Thus $\Gamma'$ is a row operation which transforms $\mathbf B$ into $\mathbf A$.

Hence the result.


Also see