A determinant made of $n \times n$ determinants.

Question

I came across this problem, in a recent exam.

So I was given three matrices $$ A, B, C \in M_{n} (\mathbb{R}) $$ and that $$ 0 \in M_{n}(\mathbb{R}) $$ is the zero matrix.

Then I was also given the matrix $$ \Lambda = \begin{bmatrix} A & B \\ 0 & C\end{bmatrix} \in M_{2n} (\mathbb{R}) $$

I had to determine the formula of determinant of matrix $ \Lambda$ with determinants $ A$, $B$ and $C $ and then prove the latter.

I had the idea to use Gauss method in general and get a formula with that, but I had no time, and I think I wouldn't get anywhere with that. Any help would be appreciated.

Answer 1

There are elementary matrices $E_1$ to $E_n$ such that $A$ = $E_1 \dotsb E_n R$ where $R$ is in RRE form.
There are elementary matrices $\varepsilon_1$ to $\varepsilon_n$ such that $C = \varepsilon_1 \dotsb \varepsilon_n r$ where $r$ is in RRE form.
Form the matrices $E'_i$ in the following way: $E'_i = \begin{bmatrix} E_i & 0 \\ 0 & I\end{bmatrix}$ where $I$ is the identity matrix.
Form the matrices $\varepsilon'_i$ in the following way: $\varepsilon'_i = \begin{bmatrix} I & 0 \\ 0 & \varepsilon_i\end{bmatrix}$
Finally, notice that $\Lambda = E'_1 \dotsb E'_n \varepsilon_1 \dotsb \varepsilon_n \begin{bmatrix}R & * \\ 0 & r\end{bmatrix}$.

Now $\det \Lambda = \det(E'_1 \dotsb E'_n \varepsilon_1 \dotsb \varepsilon_n) \begin{vmatrix}R & * \\ 0 & r\end{vmatrix}$. But also notice that $R$ and $r$ are upper triangular matrices, which means that so is $\begin{bmatrix}R & * \\ 0 & r\end{bmatrix}$, and the determinant of an upper triangular matrix is the product of the entries on the diagonal. But all the diagonal entries are part of $R$ or $r$. So $\det \Lambda = \det(E'_1 \dotsb E'_n \varepsilon_1 \dotsb \varepsilon_n) \det(R) \det(r) = \det(E'_1 \dotsb E'_n R) \det(\varepsilon_1 \dotsb \varepsilon_n r) = \det(A)\det(C)$

Answer 2

Hint:

$$\det \begin{pmatrix} A & B \\ C & D\\ \end {pmatrix}=\det(A)\det (D-CA^{-1}B)$$ Or $$\det \begin{pmatrix} A & B \\ C & D\\ \end {pmatrix} =\det(D)\det (A-BD^{-1}C).$$

To show that, suppose $A $ inversible for example and make: $\begin {pmatrix} I & 0 \\ -CA^{-1} & I \\ \end {pmatrix}\begin {pmatrix} A & B \\ C & D \\ \end {pmatrix}.$

Answer 3

Another hint, which works in greater generality (any commutative base ring): The determinant of $\Lambda$ is the sum of the terms $\left(\operatorname{sign}\sigma\right) \lambda_{1, \sigma\left(1\right)} \lambda_{2, \sigma\left(2\right)} \cdots \lambda_{2n, \sigma\left(2n\right)}$ over all permutations $\sigma$ of $\left\{1, 2, \ldots, 2n\right\}$ (where $\lambda_{i, j}$ means the $\left(i, j\right)$-th entry of $\Lambda$). However, the terms corresponding to some of these permutations $\sigma$ vanish; namely, this happens for every $\sigma$ for which some $i > n$ satisfies $\sigma\left(i\right) \leq n$ (because such a $\sigma$ picks out an entry from the $0$ block of the block matrix $\left[\begin{array}{cc} A & B \\ 0 & C\end{array}\right]$). What can you say about the remaining permutations?

Answer 4

We have $\,\det\Lambda=\det A \cdot\det C\,$ as a consequence of the formula for the computation of a determinant by blocks – a generalisation of the Laplace formula for the development of determinant along a row or a column.

Namely, consider the first $k$ columns of an $n\times n$ matrix $A$, and let $\varphi$ an increasing function from $[1\, ..\,k]$ to $[1\, ..\,n]$; $\varphi$ can be uniqueley extended to a permutation of $[1\, ..\,n]$ such that its restriction to $[k+1\, ..\,n]$ is increasing. We'll still call $\varphi$ this permutation of $[1\, ..\,n]$.

Now for any such $\varphi$, call $\,D_\varphi=\det\bigl(a_{\varphi(i)j}\bigr)_{1\le i,j\le k}\,$ the determinant extracted from $A$ by selecting the first $f$ columns and the $k$ rows $\bigl(\varphi(1),\dots,\varphi(k)\bigr)$. Let $\,D^c_\varphi=\det\bigl(a_{\varphi(i)j}\bigr)_{k+1\le i,j\le n}$ the complementary determinant. Then we have the formula: $$\det A=\sum_{\varphi}\varepsilon(\varphi) D_\varphi\cdot D^c_\phi $$

Here only the $k$-th principal minor $\det A$ will be non-zero, since all other $D_\varphi$ will have at least on line of $0$s, and its complementary determinant is $\,\det C$, whence the formula.