# Determinant of Block Matrix using Leibniz Formula

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.

• 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? Nov 22, 2016 at 18:01
• @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. Nov 22, 2016 at 18:04

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 $$S_j.$$
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 $$A$$.