# Thread: Prove that detA = detA^T by showing that detE=detE^T

1. ## Prove that detA = detA^T by showing that detE=detE^T

Greetings! I'm trying to prove that detA = detA^T by showing that detE = detE^T. I'm not sure how to prove that detE = detE^T, where E is an elementary matrix and that E^T is the transpose of E. I plan to approach this by taking the determinant of a generic elementary matrix, for each type of elementary matrix (row swap, scale, and combining rows). Is there a faster/better way to do this? Thanks!

2. If you do the cofactor expansion of A along the first row and the cofactor expansion of A^T along the first column, you obtain the same determinant.

If you need to know how to do the cofactor expansion, refer to this thread: http://www.mathhelpforum.com/math-he...tml#post518256

3. Originally Posted by dwsmith
If you do the cofactor expansion of A along the first row and the cofactor expansion of A^T along the first column, you obtain the same determinant.

If you need to know how to do the cofactor expansion, refer to this thread: http://www.mathhelpforum.com/math-he...tml#post518256
This works perfectly fine, and is probably the most straightforward way to prove $\mathrm{det}(A)=\mathrm{det}(A^T)$.

But the way in the original post works fine as well. Taking the determinant of an elementary matrix should be easy to do, strictly from the definition of determinant. The definition I mean is:

If $A=(a_{ij})\in M_n(R)$ (whatever field/ring $R$ the entries come from), then

$\mathrm{det}(A)=\sum_{\sigma \in S_n}\mathrm{sgn}(\sigma)a_{1,\sigma(1)}\cdots a_{n,\sigma(n)}$

For each elementary matrix, this summation vastly simplifies; there should only be one nonzero term for each.

For example, say we have the elementary matrix $E_{12}=\left(\begin{array}{ccc}0&1&0\\1&0&0\\0&0&I _{n-2}\end{array}\right)$ corresponding to switching the first and second rows.

The only POSSIBLE way of picking one entry out of each row and column, without picking a 0, is to take entries (2,1), (1,2), and (k,k), for $k\geq 3$. In the formula for determinant, this corresponds to the term whos permutation is $(1\,2)\in S_n$, switching 1 and 2, and fixing everything else. (Do you understand the definition/formula for determinant?)

Since $(1\,2)$ is an odd permutation, $\mathrm{sgn}((1\,2))=-1$. We get $\mathrm{det}(E_{12})=(-1)1\cdot 1\ldots \cdot 1=-1$.

After computing what the determinants of arbitrary elementary matrices are, use the fact that $\mathrm{det}(M)\mathrm{det}(N)=\mathrm{det}(MN)$.

Though I'm drawing a complete blank on how to compute the transpose using elementary matrices LOL.

4. Originally Posted by topspin1617

Though I'm drawing a complete blank on how to compute the transpose using elementary matrices LOL.
Why does it matter? Once you've proven that $\det\left(E\right)=\det\left(E^{\top}\right)$ for all elementary matrices then we just see that

\begin{aligned}\det\left(A^{\top}\right)&=\det\lef t(\left(E_1\cdots E_n\right)^{\top}\right)\\ &=\det\left(E_n^{\top}\cdots E_1^{\top}\right)\\ &=\det\left(E_n^{\top}\right)\cdots\det\left(E_1^{ \top}\right)\\ &=\det\left(E_n\right)\cdots\det\left(E_1\right )\\ &=\det\left(E_1\right)\cdots\det\left(E_n\right )\\ &=\det\left(E_1\cdots E_n\right)\\ &=\det\left(A\right)\end{aligned}

5. Originally Posted by Drexel28
Why does it matter? Once you've proven that $\det\left(E\right)=\det\left(E^{\top}\right)$ for all elementary matrices then we just see that

\begin{aligned}\det\left(A^{\top}\right)&=\det\lef t(\left(E_1\cdots E_n\right)^{\top}\right)\\ &=\det\left(E_n^{\top}\cdots E_1^{\top}\right)\\ &=\det\left(E_n^{\top}\right)\cdots\det\left(E_1^{ \top}\right)\\ &=\det\left(E_n\right)\cdots\det\left(E_1\right )\\ &=\det\left(E_1\right)\cdots\det\left(E_n\right )\\ &=\det\left(E_1\cdots E_n\right)\\ &=\det\left(A\right)\end{aligned}
Well that's what I meant. How to use the elementary matrices to write the transpose.

Was thinking too much about the transpose in particular; wasn't thinking that you could simply apply the elementary row operations on the identity matrix to get anything you want haha.