Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given $N$ $n \times n$ matrices $\mathsf{A}^{1}, \dots, \mathsf{A}^{N}$, \begin{align} \det \left( \sum_{i = 1}^{N} \mathsf{A}^{i} \right) = \sum_{\sigma \in S} \det \mathsf{A}^{\sigma}, \end{align} where $S = \{ \sigma \colon \{ 1, \dots, n \} \mapsto \{ 1, \dots, N \} \}$ and $(\mathsf{A}^{\sigma})_{ij} = \mathsf{A}_{ij}^{\sigma(i)}$.

Where is this beautiful result published?

share|improve this question
does this apply to complex matrices also? –  Benjamin Sep 10 '14 at 17:44
also, is $S$ the symmetric group? –  Benjamin Sep 10 '14 at 17:58

1 Answer 1

up vote 11 down vote accepted

This is a straight-forward application of the linearity of the determinant on the rows of the matrix. Call the sum of the matrices $\Sigma_A$. Notice that each row of $\Sigma_A$ is a sum of the corresponding rows of each $A^i$. If we split the determinant along the first row, then we have $$\det\left(\Sigma_A\right) = \sum_{i=1}^N\det\left(\Sigma_A^{(i)}\right)$$ where each $\Sigma_A^{(i)}$ is the matrix obtained from $\Sigma_A$ by replacing the first row with the first row of $A^i$. Notice now that the second row (and on wards) of each $\Sigma_A^{(i)}$ remains a sum of rows. Splitting each $\Sigma_A^{(i)}$ along the second row now gives $$\det\left(\Sigma_A^{(i)}\right) = \sum_{j=1}^N\det\left(\Sigma_A^{(i,\,j)}\right)$$ where $\Sigma_A^{(i,\,j)}$ has first row of $A^i$ and second row of $A^j$. Substituting this new expression into our original summation, we have that $$\det\left(\Sigma_A\right) = \sum_{i=1}^N\sum_{j=1}^N\det\left(\Sigma_A^{(i,\,j)}\right)$$ or more compactly $$\det\left(\Sigma_A\right) = \sum_{(k_1,\ k_2)}\det\left(\Sigma_A^{(k_1,\,k_2)}\right)$$ Where the sum ranges over all $2$-tuples with $1\le k_1,\ k_2 \le N$. Continuing in the obvious fashion, we eventually reach $$\det\left(\Sigma_A\right) = \sum_{(k_1,\ \cdots, k_n)}\det\left(\Sigma_A^{(k_1,\,\cdots,k_n)}\right)$$ where the sum ranges over all $n$-tuples. Of course $\Sigma_A^{(k_1,\,\cdots,k_n)}$ is the matrix with row $i$ from $A^{k_i}$. This is exactly your desired sum.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.