Elementary row operation is matrix pre-multiplication
Dependencies:
- Field
- Every elementary row operation has a unique inverse
- Identity matrix
- Comparing coefficients of a polynomial with disjoint variables
- Inverse of a matrix
Let $A$ be a $m$ by $n$ matrix over a field. Let $f$ be an elementary row operation. Then there exists a unique $m$ by $m$ matrix $R$ such that $f(A) = RA$. $R$ is also invertible.
By plugging in $A = I_m$ (identity matrix), we get $R = f(I_m)$. Therefore, the maxtrix associated with a row operation is the one obtained by applying that operation on the identity matrix.
Proof
We will show that there is a unique matrix $R$ which satisfies $f(A) = RA$ for all $A$.
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$\langle p \rangle \leftarrow c\langle p \rangle$: \[ (RA)[i, j] = \sum_k R[i, k] A[k, j] = \begin{Bmatrix} c & i = p \\ 1 & i \neq p \end{Bmatrix}A[i, j] \] On comparing coefficients of $A[*, j]$, we get \[ R[i, j] = \begin{cases} c & i = j = p \\ 1 & j = i \neq p \\ 0 & i \neq j \end{cases} \]
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$\langle p \rangle \leftarrow \langle p \rangle + c\langle q \rangle$: \[ (RA)[i, j] = \sum_k R[i, k] A[k, j] = \begin{cases} A[i, j] & i \neq p \\ A[p, j] + cA[q, j] & i = p \end{cases} \] On comparing coefficients of $A[*, j]$, we get \[ R[i, j] = \begin{cases} 0 & i \neq p \wedge i \neq j \\ 1 & i \neq p \wedge i = j \\ 0 & i = p \wedge j \neq p \wedge j \neq q \\ 1 & i = p \wedge j = p \\ c & i = p \wedge j = q \end{cases} \]
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$\langle p \rangle \leftrightarrow \langle q \rangle$: \[ (RA)[i, j] = \sum_k R[i, k] A[k, j] = \begin{cases} A[i, j] & i \neq p \wedge i \neq q \\ A[q, j] & i = p \\ A[p, j] & i = q \end{cases} \] On comparing coefficients of $A[*, j]$, we get \[ R[i, j] = \begin{cases} 0 & i \neq p \wedge i \neq q \wedge i \neq j \\ 1 & i \neq p \wedge i \neq q \wedge i = j \\ 0 & i = p \wedge j \neq q \\ 1 & i = p \wedge j = q \\ 0 & i = q \wedge j \neq p \\ 1 & i = q \wedge j = p \end{cases} \]
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Invertibility:
We know that every elementary row operation has a unique inverse. Let $R$ be the matrix corresponding to an elementary row operation and let $S$ be the matrix of its inverse operation.
- Since $S$ represents the inverse operation of $R$, $S(RA) = A$ for all $A$. For $A = I_m$, we get $SR = I_m$.
- Since $R$ represents the inverse operation of $S$, $R(SA) = A$ for all $A$. For $A = I_m$, we get $RS = I_m$.
Therefore, $R$ is invertible and its inverse is $S$.
Dependency for:
Info:
- Depth: 5
- Number of transitive dependencies: 13