$\det(A) = \det(A^T)$ for elementary matrix.

Question

We proofed in class that for any matrix $\det(A) = \det(A^T)$.
I was asked to prove the same, only for elementary matrices. Though repeating the proof for any matrix would do the work, it's like using heavy machinery and I guess it's not the intention of this exercise.

There are three row operations.

For the operation $R_i \rightarrow \alpha\cdot R_i$. We know that $I=I^T$. And so, $E = E^T$. Hence, $\det E = \det E^T$.

For the opertaion, $R_i \leftrightarrow R_j$. I've noticed a symmetry which sustained after transposing. I guess that's the key for this operation and for the third one.

I'm having trouble with formulating it into a rigorous proof.

Answer 1

Do you know this definition? ($ϵ(σ):=$ the sign of the permutation)

$$ \det A = \sum_\sigma {\epsilon(\sigma)} \prod_i A_{i,\sigma (i)} $$ If you do: $$ \det A^T = \sum_\sigma {\epsilon(\sigma)} \prod_i A_{\sigma (i),i} \\\forall\sigma\ \ \prod_i A_{\sigma(i),i} = \prod_j A_{j,\sigma^{-1}(j)} $$ As $\sigma \sigma^{-1} = {\rm id}, 1 = \epsilon( {\rm id})= \epsilon(\sigma \sigma^{-1} ) =\epsilon(\sigma)\epsilon(\sigma^{-1}) $ so $$\epsilon(\sigma)=\epsilon(\sigma^{-1})$$

Now as $\sigma \to \sigma^{-1}$ is a bijection: $$ \det A^T = \sum_\sigma {\epsilon(\sigma^{-1})}\prod_j A_{j,\sigma^{-1}(j)} = \sum_\tau {\epsilon(\tau)}\prod_j A_{j,\tau(j)} = \det A $$

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If you want a simplier proof for elementary matrices:

• you already proved it for $L_i \to aL_i$.
• for the operation $L_i \to L_i + bL_j$, it really depends on you definition of the determinant. But every definition should get to the conclusion that the $\det$ of such a matrix is $1$. And the operation associated with the transposition of such a matrix is of the same kind (probably $L_j \to L_j + bL_i$, but I didn't check), hence both determinants are $1$.
• the operation $L_i \leftrightarrow L_j$ is just $$ L_i \to L_i - L_j \\ L_j \to L_i + L_j \\ L_i \to L_i - L_j \\ L_i \to -Li $$ and just use the result for the two first kinds of elementary matrices (and $(AB)^T = B^TA^T$) to conclude.