# Prove that $\det(A)=\det(A^T)$ algebraically

If we use row operations to turn matrix $A$ into an upper triangular matrix then the $\det(A)$ is equal to the product of the entries on its main diagonal. So if we transpose $A$, then those row operations can be made column operations and we would have the same upper triangular matrix where $\det(A^T)$ is equal to the product of the entries on its main diagonal. So $\det(A)=\det(A^T)$.

Is there an algebraic way to show this.

For an $n\times n$ matrix $A=[A(i,j)]$ we define Determinant as$$\det A = \sum _{\sigma\in S_n} A(1,\sigma(1))\cdot A(2,\sigma(2))\cdots A(n,\sigma(n))$$

Above formula for $A$ would imply that $$\det A^T = \sum _{\sigma\in S_n} A^T(1,\sigma(1))\cdot A^T(2,\sigma(2))\cdots A^T(n,\sigma(n))$$ $$=\sum _{\sigma\in S_n} A(\sigma(1),1)\cdot A(\sigma(2),2)\cdots A(\sigma(n),n)$$

I guess you should do the remaining now which should not take much time to realize..

Caution : I am not saying that $$A(\sigma(1),1)\cdot A(\sigma(2),2)\cdots A(\sigma(n),n)=A(1,\sigma(1))\cdot A(2,\sigma(2))\cdots A(n,\sigma(n))$$ but their combined sum run over all permutations is same.

i.e., May be for some $\sigma_1, \sigma_ \in S_n$ you have $$A(\eta(1),1)\cdot A(\eta(2),2)\cdots A(\eta(n),n)=A(1,\sigma(1))\cdot A(2,\sigma(2))\cdots A(n,\sigma(n))$$

If you use the definition of determinant where you take the sum of all signed products of n elements where each element is chosen from a distinct row and distinct column, then you get that $\det(A) = \det(A^T)$ for free, and I guess you could call it an algebraic proof because you get the same algebraic formula for both $\det(A)$ and $\det(A^T)$.

There are at least three ways to show this, all very good exercise: (i) SVD, which takes one line (ii) show it for diagonalizable matrices and use density of diagonalizable matrices (iii) show this for invertible and singular matrices where the former is by decomposition to elementary matrices and latter is just 0.