I wish to explicitly compute the determinant of $$ K = \begin{pmatrix} I & A \\ B & 0 \end{pmatrix} $$ where $A,B$ are $n\times n$ matrices and $I$ is the $n\times n$ identity, using the permutation definition. If I write $$ \det K = \sum_{\sigma\in S_{2n}} \text{sgn}(\sigma) \prod_{i=1}^{2n} k_{i,\sigma(i)}, $$ then you can check that if $\sigma(i) > n$ for some $i>n$, then the product is zero, so we only care about permutations $\sigma$ that factorise as $\sigma=\sigma_1\sigma_2$, where $\sigma_1$ maps $\{1,2,\dots, n\}$ to $\{n+1 , n+2, \dots, 2n\}$ and $\sigma_2$ maps $\{n+1, n+2, \dots 2n\}$ to $\{1,2,\dots,n\}$.

For each $\sigma_1$, I can construct a permutation $f(\sigma_1)$ of $\{1,2,\dots, n\}$ by $f(\sigma_1)(t) = \sigma_1(t) - n$, and $f$ is a bijection from all such $\sigma_1$ to $S_n$. Similarly, I can a bijection $g$ mapping possible $\sigma_2$ to the permutations of ${n+1,\dots, 2n}$, and $$\text{sgn}(\sigma_1\sigma_2) = \text{sgn}(g(\sigma_1))\text{sgn}(f(\sigma_2)).$$ So

$$ \det K = \sum_{\sigma_1} \sum_{\sigma_2} \text{sgn}(f\sigma_1) \text{sgn}(g\sigma_2) \prod_{i=1}^{n} k_{i,\sigma_1(i)} \prod_{i=n+1}^{2n} k_{i,\sigma_2(i)} = \left( \sum_{\sigma_1} \text{sgn}(f\sigma_1) \prod_{i=1}^{n} k_{i,f\sigma_1(i)} \right) \left( \sum_{\sigma_2} \text{sgn}(g\sigma_2) \prod_{i=n+1}^{2n} k_{i,g\sigma_2(i)} \right). $$

But this is just $\det A\cdot \det B$, and I know that the answer is $\det K = -\det A \cdot \det B$. Where have I gone wrong?

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    $\begingroup$ Are you sure that the answer you "know" is correct? Shouldn't the sign be $(-1)^n$? Also, I think your final product formula has errors in it. You should end up with $\text{sgn}\,f$ and $\text{sgn}\,g$ pulling out of the correct formula, no? $\endgroup$ Commented Nov 11, 2023 at 18:05
  • $\begingroup$ @TedShifrin I was told that the answer was correct since one can transform $$ \begin{pmatrix} I & A \\ B & 0 \end{pmatrix} $$ to $$ \begin{pmatrix} I & A \\ 0 & -AB \end{pmatrix} $$ by subtracting the relevant multiple of each of the top $n$ rows from each of the bottom $n$ to 'clear out' the entries of $B$. I'm not sure what you mean by 'pulling out of' in your final comment. $\endgroup$
    – RDL
    Commented Nov 11, 2023 at 18:34
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    $\begingroup$ In general, $\det(-A) = (-1)^n \det(A) \color{red}{\neq} -\det(A)$. $\endgroup$ Commented Nov 11, 2023 at 18:36
  • $\begingroup$ I say you should have $\text{sgn}\,f \sum \text{sgn}\,\sigma_1\prod k_{i\sigma_1(i)}$, for example. You surely do not want the $f\sigma_1$ messing up the entries of the matrix. (Here you think of $f$ and $g$ as elements of $S_{2n}$.) $\endgroup$ Commented Nov 11, 2023 at 18:39
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    $\begingroup$ $$ \begin{pmatrix} I & A \\ B & 0 \end{pmatrix} \begin{pmatrix} I & -A \\ 0 & I \end{pmatrix} = \begin{pmatrix} I & 0 \\ B & -AB \end{pmatrix} $$ and the determinant of the middle matrix, which is genuinely upper triangular (not just as blocks), is one. $\endgroup$
    – Will Jagy
    Commented Nov 11, 2023 at 18:50

1 Answer 1


Accounting for misbracketing and the fact that $\det(-A) \not\equiv \det A$, the steps above lead to the right solution. Thanks to those who commented.


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