Proofs of determinants of block matrices

Question

I know that there are three important results when taking the Determinants of Block matrices

$$\begin{align}\det \begin{bmatrix} A & B \\ 0 & D \end{bmatrix} &= \det(A) \cdot \det(D) \ \ \ \ & (1) \\ \\ \det \begin{bmatrix} A & B \\ C & D \end{bmatrix} &\neq AD - CB & (2) \\ \\ \det \begin{bmatrix} A & B \\ C & D \end{bmatrix} &= \det \begin{bmatrix} A & B \\ 0 & D - CA^{-1}B \end{bmatrix} \\ \\ &= \underbrace{\det(A)\cdot \det\left(D-CA^{-1}B\right)}_\text{if $A^{-1}$ exists} \\ \\ &= \underbrace{\det\left(AD-CB\right)}_\text{if $AC=CA$} & (3) \end{align}$$

Now I understand in result $(3)$, that all that row operations are being performed to bring it into the form we see in $(1)$, but I can't seem to convince myself that result $(1)$ is true in the first place.

Furthermore in result $(3)$, I understand that, $\det(A)\cdot \det\left(D-CA^{-1}B\right) = \det\left(A(D-CA^{-1}B)\right)= \det(AD-CB)$, via the product rule for determinants I also understand that we need $A^{-1}$ to exist, for the initial row operation to reduce the matrix into an upper triangular form $U$, and I understand that we require $AC = CA$, to allow commutativity when we multiply $ACA^{-1}B$ to reduce it to $CB$.

Can someone provide proofs for results $(1)$ and $(2)$, as I can't seem to find proofs for them in any of the textbooks I have at my disposal

Answer 1

To prove $(1)$, it suffices to note that $$ \pmatrix{A &B\\0&D} = \pmatrix{A & 0\\0 & D} \pmatrix{I&A^{-1}B\\0 & I} $$ From here, it suffices to note that the second matrix is upper-triangular, and to compute the determinant of the first matrix. It is easy to see that the determinant of the first matrix should be $\det(A)\det(D)$ if we use the Leibniz expansion.

Alternatively, the determinant of the first matrix can be further decomposed into $$ \det \left[\pmatrix{A & 0\\0 & I}\pmatrix{I & 0\\0 & D} \right] = \det \pmatrix{A & 0\\0 & I} \det \pmatrix{I & 0\\0 & D}. $$

For an example where $(2)$ fails to hold, consider the matrix $$ \pmatrix{ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0 } = \pmatrix{B&B^T\\B^T&B} $$ For an example where the diagonal blocks are invertible, add $I$ to the whole matrix.

Answer 2

Suppose block $A$ has dimension $r$, block $D$ has dimension $s$. Use the definition of the determinant $\lvert c_{i,j}\rvert,\enspace {1\le i,j\le r+s}$: $$\begin{vmatrix} A&C\\0& D\end{vmatrix} =\sum_{\sigma \in \mathfrak S_{r+s}}\prod_{1\le j\le r+s}(-1)^{\text{sgn}\, \sigma}c_{\sigma(j),j}.$$ Now the non-zero terms are those such that, if $1\le j\le r$, $\;1\le \sigma(j)\le r$. Similarly, if $r+1\le j\le r+s$, $\;r+1\le \sigma(j)\le r+s$. So the non-zero terms are those for which the permutation $\sigma\in \mathfrak S_{r+s}$ is the concatenation of a permutation of $\mathfrak S_r$ and a permutation in $\mathfrak S_s$, and clearly the signature of $\sigma$ is the product of the signatures of its factors.

The formula for the first formula for the determinant follows by distributivity.

Answer 3

Proof of the 3rd identity.

It is a consequence of the following "block diagonalization" identity:

$$\pmatrix{ A&B\\ C&D }=\pmatrix{ I&0\\ CA^{-1}&I }\pmatrix{ A&0\\ 0&S }\pmatrix{ I&A^{-1}B\\ 0&I } \ \ \text{with} \ \ S:=D-CA^{-1}B$$

(S = "Schur's complement" (https://en.wikipedia.org/wiki/Schur_complement)),

Then, it suffices to take determinants on both sides.

Remark : For many matrix formulas, take a look at the amazing compendium : "Matrix Mathematics: Theory, Facts, and Formulas" Second Edition by Dennis S. Bernstein (Princeton University Press, 2009).