Determinant of a transposed matrix

Therefore, the determinant of the transpose is equal to the determinant of the original matrix. In summary, the determinant of a matrix A is equal to the determinant of its transpose, as shown by the fact that each summand in the determinant of A is also present in the determinant of A^T. This is due to the fact that each row ordering in the determinant of A can be mapped to a row ordering in the determinant of A^T, resulting in the same term being present in both determinants.
  • #1
Eclair_de_XII
1,083
91
TL;DR Summary
Let ##A## be a matrix of size ##(n,n)## where ##n\in\mathbb{N}##. Then the determinant of ##A## is equal to the determinant of ##A## transposed, to be denoted ##A^T##.
By definition, ##\det A=\sum_{p_j\in P}\textrm{sgn}(p_j)\cdot a_{1j_1}\cdot\ldots\cdot a_{nj_n}##, where ##P## denotes the set of all permutations of the ordered sequence ##(1,\ldots,n)##. Denote the number of permutations needed to map the natural ordering to ##p_j## as ##N_j##.

Now consider ##\det A^T## which is equal to:
\begin{align}\sum_{p_i\in P}\textrm{sgn}(p_i)\cdot a_{i_11}\cdot\ldots\cdot a_{i_nn}\end{align}

Note: ##i_k## denotes the element of ##p_i## at the k-th index.

To show equality, we must show that each summand in ##\det A## is also in ##\det A^T##. In other words, we must show that there is a permutation ##p_l## s.t.:
\begin{align}
\textrm{sgn}(p_l)\cdot a_{l_11}\cdot\ldots\cdot a_{l_nn}=\textrm{sgn}(p_j)\cdot a_{j_11}\cdot\ldots\cdot a_{j_nn}
\end{align}

Consider the ordered list:
\begin{align}(i_1,1),\ldots,(i_n,n)\end{align}

For each element in the list, there is an integer ##m## s.t. ##j_m=k##. It will take ##N_j## permutations in order to map this ordering to an ordering of the form:
\begin{align}(i_1',j_1),\ldots,(i_n',j_n)\end{align}

where ##(i_1',\ldots,i_n')## is the ordering obtained from permutating ##p_i## wrt the ordering ##p_j##. Bearing in mind that ##\prod_{k=1}^n a_{i_kk}\equiv \prod_{k=1}^n a_{i_k'j_k}##, we have:
\begin{align}
\textrm{sgn}(p_i)\cdot\prod_{k=1}^n a_{i_kk}=\textrm{sgn}(p_j)\cdot\textrm{sgn}(p_{i'})\cdot\prod_{k=1}^n a_{i_k'j_k}
\end{align}

It will take ##N_{i'}## permutations to map ##p_{i'}## to the natural ordering. This corresponds to ##N_{i'}## sign changes:
\begin{align}
\textrm{sgn}(p_j)\cdot\prod_{k=1}^n a_{kj_k}=\textrm{sgn}(p_{i'})\cdot\left[\textrm{sgn}(p_j)\cdot\textrm{sgn}(p_{i'})\cdot\prod_{k=1}^n a_{i_k'j_k}\right]
\end{align}

% I am asking for critique on this proof. Is it accurate? Is it understandable? Is there any unnecessary notation I used?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Generally one uses [itex]S_N[/itex] for the group of permutations of [itex]\{1, \dots, N\}[/itex] and [itex]\sigma[/itex] and [itex]\rho[/itex] for arbitrary permutations. You can simplify your notation and argument significantly by using the fact that [itex]S_N[/itex] is a group acting on [itex]\{1, \dots, N\}[/itex].

By definition [tex]
\begin{split}
\det A &= \sum_{\sigma \in S_N} \operatorname{sgn}(\sigma) a_{1\sigma(1)} \cdots a_{N\sigma(N)}, \\
\det A^T &= \sum_{\rho \in S_N} \operatorname{sgn}(\rho) a_{\rho(1)1} \cdots a_{\rho(N)N}.
\end{split}[/tex] Your claim is then that there exists a bijection from [itex]S_N[/itex] to itself such that if [itex]\sigma \mapsto \rho_\sigma[/itex] then for each [itex]\sigma \in S_N[/itex] we have [tex]
\operatorname{sgn}(\sigma) a_{1\sigma(1)} \cdots a_{N\sigma(N)} = \operatorname{sgn}(\rho_\sigma) a_{\rho_\sigma(1)1} \cdots a_{\rho_\sigma(N)N}.[/tex] (We need a bijection because we want each summand of [itex]\det A[/itex] to appear exactly once in the sum for [itex]\det A^T[/itex] and vice versa). Ordering the factors on the left hand side by the second index rather than the first is the way to proceed, but because [itex]\sigma[/itex] is a permutation this can be done in one step: The factor with second index [itex]i[/itex] has first index [itex]\sigma^{-1}(i)[/itex]. Two basic facts about permutation groups complete the proof.
 
  • #3
Unfortunately, I've not taken abstract algebra during my undergraduate career, so I am unfamiliar with groups and would be uncomfortable attempting to implement them in my explanation/proof.

I meant to say in the last two lines that given any row-ordering ##p_i## in a summand of the determinant of the transpose, there is a row-ordering ##p_{i'}## that can be applied to ##p_i##. The row-ordering ##p_{i'}## is how I described it earlier. This resulting ordering corresponds to an arbitrary term in the determinant of the original matrix.
 
Last edited:
  • #4
What I would write is: For every permutation ##\sigma## in the set of permutations (denoted ##S_N##) there exists a unique inverse permutation ##\sigma^{-1}## in the set of permutations and where no two distinct permutations have the same inverse.

Say ##\sigma (i) = j## then

$$
a_{i \sigma(i)} = a_{ij} = a_{\sigma^{-1} (j) j}
$$

so that

$$
\prod_{i=1}^N a_{i \sigma (i)} = \prod_{i=1}^N a_{\sigma^{-1} (i) i} .
$$
If ##\sigma## is an even permutation then obviously ##\sigma^{-1}## is also an even permutation. If ##\sigma## is an odd permutation then obviously ##\sigma^{-1}## is also an odd permutation. Therefore,

$$
\text{sgn} (\sigma) = \text{sgn} (\sigma^{-1})
$$

and

$$
\text{sgn} (\sigma ) \prod_{i=1}^N a_{i \sigma (i)} = \text{sgn} ( \sigma^{-1} ) \prod_{i=1}^N a_{\sigma^{-1} (i) i} .
$$

Then

\begin{align*}
\det (A) & = \sum_{\sigma \in S_N} \text{sgn} (\sigma ) \prod_{i=1}^N a_{i \sigma (i)}
\nonumber \\
& = \sum_{\sigma \in S_N} \text{sgn} ( \sigma^{-1} ) \prod_{i=1}^N a_{\sigma^{-1} (i) i}
\nonumber \\
& = \sum_{\sigma \in S_N} \text{sgn} ( \sigma ) \prod_{i=1}^N a_{\sigma (i) i} = \det (A^T)
\end{align*}

where we have arrived at the last line by noting that for every permutation ##\sigma## in ##S_N## there exists a unique inverse permutation ##\sigma^{-1}## in ##S_N## and where no two distinct permutations have the same inverse, and so we are in effect summing over all permutations in ##S_N## in the second line.
 
Last edited:

Related to Determinant of a transposed matrix

1. What is the determinant of a transposed matrix?

The determinant of a transposed matrix is equal to the determinant of the original matrix. This means that transposing a matrix does not change its determinant value.

2. How do you calculate the determinant of a transposed matrix?

To calculate the determinant of a transposed matrix, you can use the same methods as you would for a regular matrix. This includes using cofactor expansion or the row reduction method.

3. Can the determinant of a transposed matrix be negative?

Yes, the determinant of a transposed matrix can be negative. The determinant value is determined by the sign of the product of the eigenvalues of the matrix, which can be positive or negative.

4. How does the determinant of a transposed matrix affect the matrix's invertibility?

The determinant of a transposed matrix does not affect its invertibility. A matrix is invertible if and only if its determinant is non-zero, and transposing a matrix does not change its determinant value.

5. What is the significance of the determinant of a transposed matrix?

The determinant of a transposed matrix is an important mathematical concept that has various applications in fields such as linear algebra, physics, and engineering. It can be used to determine the volume of a parallelepiped, solve systems of linear equations, and find the inverse of a matrix.

Similar threads

  • Linear and Abstract Algebra
Replies
15
Views
4K
  • Linear and Abstract Algebra
Replies
4
Views
1K
  • Linear and Abstract Algebra
Replies
15
Views
1K
Replies
27
Views
1K
  • Linear and Abstract Algebra
Replies
8
Views
912
  • Linear and Abstract Algebra
Replies
10
Views
2K
  • Linear and Abstract Algebra
Replies
10
Views
2K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Math Proof Training and Practice
Replies
8
Views
1K
Back
Top