demonstration of the transposed determinant equality

Question

I wonder how can you prove that Det (A) = Det (A^t) ?

i search for a demonstration without induction :) this one is obvious. I really want to find something alternative,

thank you !

Answer 1

Using the definition of determinants as:

$$\det(A)=\sum\limits_{\sigma\in S_n}sgn(\sigma)\prod\limits_{i=1}^na_{i,\sigma(i)}$$

we first recognize what this notation means in the first place.

It means that for every "pattern" in the matrix which uses $n$ entries with exactly one entry in each row and exactly one entry in each column, we will either add or subtract the product of the entries in the pattern depending on the "sign" of the pattern.

As a smaller example:

$$\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}$$ has the following six patterns:

$$\begin{bmatrix}1&&\\&5&\\&&9\end{bmatrix}~\begin{bmatrix}1&&\\&&6\\&8\end{bmatrix}~\begin{bmatrix}&2&\\4&&\\&&9\end{bmatrix}~\begin{bmatrix}&2&\\&&6\\7\end{bmatrix}~\begin{bmatrix}&&3\\4\\&8\end{bmatrix}~\begin{bmatrix}&&3\\&5\\7\end{bmatrix}$$

These have the signs $1,-1,-1,1,1,-1$ respectively. (The sign has to do with the parity of the corresponding permutation. Even permutations give sign of $+1$ while odd permutations give sign of $-1$. You can figure out which it is by how many "swaps" are necessary to get the entries into a topleft-to-bottomright diagonal formation, for example the second could be made diagonal by switching the second and third row)

The determinant is then $1\cdot5\cdot9-1\cdot6\cdot8-2\cdot4\cdot9+2\cdot6\cdot7+3\cdot4\cdot8-3\cdot5\cdot7=0$, the same result as what you should have gotten via other methods.

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As for what this has to do with the question of whether or not $\det(A)=\det(A^T)$, notice that any product that occurs in the summation of the calculation of $\det(A)$ as a result of a pattern $\sigma$ will also occur during the calculation of $\det(A^T)$ except instead of corresponding to the pattern $\sigma$ it corresponds to the pattern $\sigma^{-1}$.

For example, during the calculation of the determinant of $\begin{bmatrix}1&2&3&4\\5&6&7&8\\9&10&11&12\\13&14&15&16\end{bmatrix}$ there will be a term with the pattern

$$\begin{bmatrix}1&&&\\&&&8\\&10\\&&15&\end{bmatrix}$$

Similarly, during the calculation of the determinant of the transpose $\begin{bmatrix}1&5&9&13\\2&6&10&14\\3&7&11&15\\4&8&12&16\end{bmatrix}$ there will be a term with the pattern

$$\begin{bmatrix}1\\&&10\\&&&15\\&8\end{bmatrix}$$

Further, these both have the same sign since $sgn(\sigma)=sgn(\sigma^{-1})$

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This is admittedly a bit hand-wavy, but can be formalized

$$\det(A)=\sum\limits_{\sigma\in S_n}sgn(\sigma)\prod\limits_{i=1}^na_{i,\sigma(i)}=\sum\limits_{\sigma\in S_n}sgn(\sigma^{-1})\prod\limits_{i=1}^na_{\sigma^{-1}(i),i}=\sum\limits_{\sigma^{-1}\in S_n}sgn(\sigma^{-1})\prod\limits_{i=1}^n(A^T)_{i,\sigma^{-1}(i)}=\det(A^T)$$

The idea being, rather than running through the product row by row, we run through the product instead column by column

Answer 2

Over an algebraically closed field, $A$ can be triangularised: there is an invertible $M$ with $MAM^{-1}=T$ upper triangular. Then $\det A=\det T$ is the product of the diagonal entries of $T$. Also then ${M^t}^{-1}A^t M^t=T^t$ which is lower triangular but with the same diagonal entries as $T$....

Answer 3

If $\det A\neq 0$, we can write $A$ as the product of elementary matrices: $A=E_1\ldots E_n\,.$

For any elementary matrix $E$, you can show that $\det E = \det E^T$.

Therefore, $$\det A^T=\det (E_1\ldots E_n)^T=\det (E_n^T\ldots E_1^T)=\det(E_n^T)\ldots\det(E_1^T)=\det(E_n)\ldots\det(E_1)=\det(E_1)\ldots\det(E_n)=\det(E_1\ldots E_n)=\det A$$