The Question:

Let $M$ be an $n \times n$ matrix and suppose that $M =\begin{bmatrix}I&B\\0&A\end{bmatrix} ,$ where $A$ is a $j \times j$ matrix with $j < n$; where $I$ is the $(n − j) \times (n − j)$ identity matrix; and where $0$ a matrix of zeros. Prove that $\det(M) = \det(A)$.

I would like to prove this using Leibniz's formula of the determinant: For an $n \times n$ matrix $A$,

$$\det(A)=\sum_{\sigma\in S_n} \operatorname{sgn}(\sigma)\cdot a_{1,\sigma(1)}\cdot...\cdot a_{n,\sigma(n)} = \sum_{\sigma\in S_n} \operatorname{sgn}(\sigma)\cdot \prod_{i=1}^na_{i,\sigma(i)}$$

Any help would be appreciated, thank you.

  • $\begingroup$ There is something like if the upper or lower triangle of a matrix consist of only zeros, then the product of the values on the main diagonal is the determinant of the matrix. What values do you have on the main diagonal? $\endgroup$
    – imranfat
    Commented Nov 22, 2016 at 18:01
  • $\begingroup$ @imranfat I am not assuming we know what the values are. what you are saying is true, in fact we know this to be true whenever we have an upper triangular matrix. $\endgroup$ Commented Nov 22, 2016 at 18:04

2 Answers 2


Hint: Consider the partition $S_n =A\bigcup A^C$ of the set $S_n$, where $$A=\{\sigma \in S_n: \sigma ([n-j])\cap [n-j]\neq \emptyset\},$$ What can you conclude from $\prod_{i=1}^na_{i,\sigma(i)}$ if $\sigma \in A$?
Then you will have to play with identity matrix. So you will have to do a projection from $S_n$ to $S_j.$
Hope it helps.


If you go that way, you have to prove that any permutation $\sigma$ whose associated product $\prod_{i=1}^n a_{i,\sigma(i)}$ is nonzero, fixes the set $K:=\{i \leq n-j\}$.

More precisely, if the product associated to $\sigma$ is nonzero, then for all $i \leq n-j$, one has $\sigma(i) \geq i$. You want to prove that these inequalities are actually all equalities. Assume that it's not the case and take the minimal $i$ which fails it, calling it $i_0$. Then there is no $i \leq n-j$ such that $\sigma(i) = i_0$. which means that taking $k:=\sigma^{-1}(i_0)$, one has $k>n-j$ and therefore $a_{k,\sigma(k)} = 0$.

As a consequence the determinant is taken on permutations $\sigma$ such that $\sigma(i)=i$ for all $i \leq n-j$. Those permutations are then permutations of $L:=\left\{i>n−j\right\}$. The sum over those permutations is exactly the determinant of $A$.


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