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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.

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    $\begingroup$ Regarding the first paragraph, one of the proofs of $\det(A)=\det\left(A^T\right)$ relies on knowing how to find determinants of elementary matrices. $\endgroup$
    – Git Gud
    Commented Apr 13, 2014 at 17:03

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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: $$ \det A^T = \sum_\sigma {\epsilon(\sigma^{-1})}\prod_j A_{j,\sigma^{-1}(j)} = \sum_\tau {\epsilon(\tau)}\prod_j A_{j,\tau(j)} = \det A $$

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.
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  • $\begingroup$ It may be helpful to remark that $\;\epsilon(\sigma):=$ the sign of the permutation $\;\sigma\;$, and if the OP already knows this then the equality follows at once from the first two lines, since $\;\forall\,\sigma\in S_n\;,\;\;\epsilon(\sigma)=\epsilon(\sigma^{-1})\;$ ... and, of course, because being $\;S_n\;$ a group, it is the same to run over elements of it or over the inverses of elements of it. +1 $\endgroup$
    – DonAntonio
    Commented Apr 13, 2014 at 17:13
  • $\begingroup$ I am familiar with this proof but I was asked to prove it only for elementary matrices in order to infer the claim is right for every matrix. $\endgroup$
    – AnnieOK
    Commented Apr 13, 2014 at 17:22
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    $\begingroup$ I edited with what you want. $\endgroup$
    – mookid
    Commented Apr 13, 2014 at 17:32
  • $\begingroup$ Thanks a lot, but how did you come to the conclusion that for the second operation $\det(A)=1$? $\endgroup$
    – AnnieOK
    Commented Apr 13, 2014 at 17:38
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    $\begingroup$ It really depends on the definition of the determinant. You can define it using decomposition into a product of elementary matrices and saying that the determinant of the second operation is 1m, and in this case this is part of the definition of the determinant. Otherwise, the matrix has a shape of a triangle with ones on the diagonal, for example. $\endgroup$
    – mookid
    Commented Apr 13, 2014 at 17:40

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