Column Operation has Inverse

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Theorem

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


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


Proof

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

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

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

$\mathbf A \mathbf K = \mathbf B$

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

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

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

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

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


Hence:

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


We have:

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


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

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


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


Hence the result.

$\blacksquare$


Also see