Why the determinant of a matrix is equal to its transpose

[Jennifer1990]

Homework Statement

I don't understand why the determinant of a matrix is equal to its transpose...how is this possible?

Homework Equations

The Attempt at a Solution

 

[Dick]

If you mean the determinant of a matrix is equal to the determinant of the transpose, why do you think that's so impossible? Try and prove it.

 

[Jennifer1990]

ohhh it works o.o
can i see a rigorous proof of this?

 

[Mark44]

Yes, if you have functioning eyes and someone shows you such a proof.

 

[Jennifer1990]

lol that's funny. What i meant is May you show me such a proof, please?

 

[Mark44]

I might, if you prove it to yourself with a 2 x 2 matrix first.
A = [a b; c d] (this is in row-major order).

 

[Jennifer1990]

A = [a b; c d]
det (A) = ad-bc

A transpose = [a c; b d]
det A transpose = ad - bc = det A

 

[aostraff]

In this case, a visual interpretation and the fact that you can take determinants along and row/column is usually a fair explanation.

For a possible proof, you can prove the determinant inductively for matrices of size n. Then you take the idea of permutations to swap rows/columns to show that it does not change the determinant. It should be straight forward from then on.

 

[Count Iblis]

The proof is trivial:

If A is an n by n matrix, then:

$$\det(A) =\sum_{\pi}\operatorname{sign}(\pi)\prod_{i=1}^{n}A_{i,\pi(i)}$$ (1)

The determinant of the transpose can thus be written as:

$$\det(A^{T}) =\sum_{\pi}\operatorname{sign}(\pi)\prod_{i=1}^{n}A_{\pi(i),i}$$

So, to prove that the determinant of the transpose is the same, we have move the permutation from the second index to the first in (1). We can do this as follows.

In the product, it doesn't matter in which order the matrix elements are multiplied:

$$\prod_{i=1}^{n}A_{i,\pi(i)} = \prod_{i=1}^{n}A_{\pi^{-1}(i),i}$$


Using that the sign of a permutation is the same as the sign of its inverse, gives:


$$\det(A) =\sum_{\pi}\operatorname{sign}(\pi^{-1})\prod_{i=1}^{n}A_{\pi^{-1}(i),i}$$


The set of all inverse permutations is the same as the set of all permuations, so we can write this as


$$\det(A) =\sum_{\pi}\operatorname{sign}(\pi)\prod_{i=1}^{n}A_{\pi(i),i}$$