Calculating the determinant of a $2n \times 2n$ matrix by means of a decomposition into four $n \times n$ blocks

Question

I want to calculate the determinant of this matrix:

$\left( \begin{array}{cccccccc} 0 & * & * & * & * & 0 & * & 0 \\ * & 0 & 0 & 0 & 0 & 0 & 0 & * \\ *& 0 & 0 & 0 & 0 & 0 & * & * \\ * & 0 & 0 & 0 & 0 & 0 & * & * \\ * & 0 & 0 & 0 & 0 & 0 & * & * \\ 0 & 0 & 0 & 0 & 0 & 0 & * & * \\ * & 0 & * & * & * & * & 0 &* \\ 0 & * & * & * & * & * & * & 0 \end{array}\right)$

using the methode that says that: $ det \left( \begin{array}{cc} A & B\\ C & D \end{array}\right) =det (AD-BC)$ under some conditions, but I do not achieve it. Can some one help?

Answer 1

Let us name $M$ your matrix.

I am going to show that $\det(M)=0$.

You know that changing the order or rows and columns (said in matricial terms : pre- or post multiplying $M$ by permutation matrices) preserves the determinant up to a sign change. We can do such rows and columns permutations in such a way that the $5 \times 5$ block of zeros visible inside the matrix is moved in the lower-left corner.

We are now with the following matrix :

$$M'=\left( \begin{array}{c|c}A&B\\ \hline 0&D\end{array}\right)=$$ $$ =\left( \begin{array}{cccc|cccc} *& * & * & * & * & * & * & * \\ *& * & * & * & * & * & * & * \\ *& * & * & * & * & * & * & * \\0 & 0 & 0 & 0 & 0 & * & * & *\\ \hline 0 & 0 & 0 & 0 & 0 & * & * & *\\ 0 & 0 & 0 & 0 & 0 & * & * & *\\ 0 & 0 & 0 & 0 & 0 & * & * & *\\ 0 & 0 & 0 & 0 & 0 & * & * & * \end{array}\right)$$

with a $2 \times 2$ partition into blocks themselves $4 \times 4$. Due to the fact that the $C$ block is a block of zeros, we can conclude that :

$$\det(M')=\det(A)\det(D) \tag{1}$$

(see Remark below).

But $A$ has a row of zeros ; therefore $\det(A)=0$, and, as a consequence, $\det(M')=0$,... and $\det(M)=0$.

Remark : (1) is a known result, but one can deduce it from "your" rule,

$$\det(M')=\det(AD-BC)$$

being essential to twin this rule with the condition : iff $C$ and $D$ commute (see here) which is the case here because $C=0$. Then $$\det(AD-BC)=\det(AD)=\det(A)\det(D).$$

Answer 2

If you want to calculate determinant of matrix over than 2x2, you can use a basic technique expansion by row. For example, a matrix is

$$ \left( \begin{array}{} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{array} \right) $$

You can divide it to three 2x2 matrices:

$$ 1\det\left( \begin{array}{} 5 & 6\\ 8 & 9\\ \end{array} \right) - 2\det\left( \begin{array}{} 4 & 6\\ 7 & 9\\ \end{array} \right) + 3\det\left( \begin{array}{} 4 & 5\\ 7 & 8\\ \end{array} \right) $$

Now you can calculate them one by one, and add them together.

If matrix is 4x4, you can divide it to 4 3x3 matrices and divide them to 12 2x2 matrices.