Matrix

Block Diagonal Determinants

The Determinant of a Block Diagonal Matrix. Theory and examples with a solution.

Frequently we encounter determinants with many zero elements and the nonzero elements which form square blocks along the diagonal. For example the following fifth-order determinant is a block diagonal determinant:

$$D_5 = |A| = \left[ \begin{array}{rrr} a_{11} & a_{12} & 0 & 0 & 0 \\ a_{21} & a_{22} & 0 & 0 & 0 \\ * & * & a_{33} & a_{34} & a_{35} \\ * & * & a_{43} & a_{44} & a_{45} \\ * & * & a_{53} & a_{54} & a_{55} \end{array}\right]$$

In this section, we will show that

$$D_5 = \left[ \begin{array}{rrr} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] \cdot \left[ \begin{array}{rrr} a_{33} & a_{34} & a_{35} \\ a_{43} & a_{44} & a_{45} \\ a_{53} & a_{54} & a_{55} \end{array}\right]$$

regardless the values the elements ∗ assume.

By definition

$D_5 = \sum_{i_1=1}^5 \cdot \sum_{i_2=1}^5 \cdot \sum_{i_3=1}^5 \cdot \sum_{i_4=1}^5 \cdot \sum_{i_5=1}^5 \cdot$ $\varepsilon_{i_{1}i_{2}i_{3}i_{4}i_{5}} \cdot$ $a_{i1} 1\cdot a_{i2} 2\cdot a_{i3} 3\cdot a_{i4} \cdot a_{i5} 5$

Since $\text a_{13} \!=\! a_{14} \!=\! a_{15} \!=\! a_{23} \!=\! a_{24} \!=\! a_{25} \!=\! 0$, all terms containing these elements can be excluded from the summation. Thus

$D_5 = \sum_{i_1=1}^5 \cdot \sum_{i_2=1}^5 \cdot \sum_{i_3=\textbf{3}}^5 \cdot \sum_{i_4=\textbf{3}}^5 \cdot \sum_{i_5=\textbf{3}}^5 \cdot$ $\varepsilon_{i_{1}i_{2}i_{3}i_{4}i_{5}} \cdot$ $a_{i1} 1\cdot a_{i2} 2\cdot a_{i3} 3\cdot a_{i4} \cdot a_{i5} 5$

Furthermore, the summation over $i_1$ and $i_2$ can be written as from 1 to 2, since 3, 4, and 5 are taken up by $i_3, i_4,$ or $i_5$, and the Levi-Civita symbol is equal to zero if any index is repeated. Hence

$D_5 = \sum_{i_1=1}^\textbf{2} \cdot \sum_{i_2=1}^\textbf{2} \cdot \sum_{i_3=3}^5 \cdot \sum_{i_4=3}^5 \cdot \sum_{i_5=3}^5 \cdot$ $\varepsilon_{i_{1}i_{2}i_{3}i_{4}i_{5}} \cdot$ $a_{i1} 1\cdot a_{i2} 2\cdot a_{i3} 3\cdot a_{i4} \cdot a_{i5} 5$

Under these circumstances, the permutation of i1, i2, i3, i4, i5 can be separated into two permutations as schematically shown later:

$$\left( \begin{array}{ccc} 1 & 2 & 3 & 4 & 5 \\ i_1=1,2 & i_2=1,2 & i_3=3,4,5 & i_4=3,4,5 & i_5=3,4,5 \end{array}\right)$$ $$= \left( \begin{array}{ccc} 1 & 2 \\ i_1 & i_2 \end{array}\right) \cdot \left( \begin{array}{ccc} 3 & 4 & 5\\ i_3 & i_4 & i_5 \end{array}\right)$$

The entire permutation is even if the two separated permutations are both even or both odd. The permutation is odd if one of the separated permutations is even and the other is odd. Therefore

$$\varepsilon_{i_{1}i_{2}i_{3}i_{4}i_{5}} = \varepsilon_{i_{1}i_{2}} \cdot \varepsilon_{i_{3}i_{4}i_{5}}$$

It follows

$D_5 = \sum_{i_1=1}^5 \cdot \sum_{i_2=1}^5 \cdot \sum_{i_3=\textbf{3}}^5 \cdot \sum_{i_4=\textbf{3}}^5 \cdot \sum_{i_5=\textbf{3}}^5 \cdot$ $\varepsilon_{i_{1}i_{2}} \cdot \varepsilon_{i_{3}i_{4}i_{5}}$ $\cdot a_{i1} 1\cdot a_{i2} 2\cdot a_{i3} 3\cdot a_{i4} \cdot a_{i5} 5$ $= \sum_{i_1=1}^2 \cdot \sum_{i_2=1}^2 \varepsilon_{i_{1}i_{2}} \cdot a_{i1} 1\cdot a_{i2} 2$ $\cdot \sum_{i_3=3}^5 \cdot \sum_{i_4=3}^5 \cdot \sum_{i_5=3}^5 \varepsilon_{i_{3}i_{4}i_{5}}$ $\cdot a_{i1} 1\cdot a_{i2} \cdot a_{i3} 3\cdot a_{i4} \cdot a_{i5} 5 2$ $= \left[ \begin{array}{rrr} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] \cdot \left[ \begin{array}{rrr} a_{33} & a_{34} & a_{35} \\ a_{43} & a_{44} & a_{45} \\ a_{53} & a_{54} & a_{55} \end{array}\right]$

When the blocks are along the “antidiagonal” line, we can evaluate the determinant in a similar way, except we should be careful about its sign. For example,

$\left[ \begin{array}{ccc} 0 & 0 & a_{13} & a_{14} \\ 0 & 0 & a_{23} & a_{24} \\ a_{31} & a_{32} & 0 & 0 \\ a_{41} & a_{42} & 0 & 0 \end{array}\right] \cdot$ $\left[ \begin{array}{rrr} a_{31} & a_{32} \\ a_{41} & a_{42} \end{array}\right]$ $\cdot \left[ \begin{array}{rrr} a_{13} & a_{14} \\ a_{23} & a_{24} \end{array}\right]$    (1)

and

$\left[ \begin{array}{ccc} 0 & 0 & 0 & a_{14} & a_{15} & a_{16} \\ 0 & 0 & 0 & a_{24} & a_{25} & a_{26} \\ 0 & 0 & 0 & a_{34} & a_{35} & a_{36} \\ a_{41} & a_{42} & a_{43} & * & * & * \\ a_{51} & a_{52} & a_{53} & * & * & * \\ a_{61} & a_{62} & a_{63} & * & * & * \end{array}\right]$ $= - \left[ \begin{array}{rrr} a_{41} & a_{42} & a_{43} \\ a_{51} & a_{52} & a_{53} \\ a_{61} & a_{62} & a_{63} \end{array}\right]$ $\cdot \left[ \begin{array}{rrr} a_{14} & a_{15} & a_{16} \\ a_{24} & a_{25} & a_{26} \\ a_{34} & a_{35} & a_{36} \end{array}\right]$    (2)

We can establish the result of (1) by changing it to a block diagonal determinant with an even number of interchanges between two rows. However, we need an odd number of interchanges between two rows to change (2) into a block diagonal determinant, therefore a minus sign.

Example 1

Evaluate    $\left[ \begin{array}{ccc} 0 & 2 & 0 & 7 & 1 \\ 1 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 5 & 1 \\ 1 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right]$

Solution:

$\left[ \begin{array}{ccc} 0 & 2 & 0 & 7 & 1 \\ 1 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 5 & 1 \\ 1 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right] \to (Row 4 - Row 2) = \left[ \begin{array}{ccc} 0 & 2 & 0 & 7 & 1 \\ 1 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right]$ $\left[ \begin{array}{ccc} 0 & 2 \\ 1 & 0 \end{array}\right] \cdot \left[ \begin{array}{ccc} 0 & 5 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] = -2 \cdot 1 = -2$

---   end Example 1   ---