# 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? – imranfat Nov 22 '16 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. – Mathemphetamine Nov 22 '16 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 $$A$$.