Editing Block matrix (section)

===Determinant{{anchor|Determinant}}===
The formula for the determinant of a <math>2 \times 2</math>-matrix above continues to hold, under appropriate further assumptions, for a matrix composed of four submatrices <math>A, B, C, D</math> with <math>A</math> and <math>D</math> square. The easiest such formula, which can be proven using either the [[Leibniz formula for determinants|Leibniz formula]] or a factorization involving the [[Schur complement]], is
:<math>\det\begin{bmatrix}A& 0\\ C& D\end{bmatrix} = \det(A) \det(D) = \det\begin{bmatrix}A& B\\ 0& D\end{bmatrix}.</math><ref name=":0" />

Using this formula, we can derive that [[characteristic polynomial]]s of <math>\begin{bmatrix}A& 0\\ C& D\end{bmatrix}</math> and <math>\begin{bmatrix}A& B\\ 0& D\end{bmatrix}</math> are same and equal to the product of characteristic polynomials of <math>A</math> and <math>D</math>. Furthermore, If <math>\begin{bmatrix}A& 0\\ C& D\end{bmatrix}</math> or <math>\begin{bmatrix}A& B\\ 0& D\end{bmatrix}</math> is [[diagonalizable]], then <math>A</math> and <math>D</math> are diagonalizable too. The converse is false; simply check <math>\begin{bmatrix}1& 1\\ 0& 1\end{bmatrix}</math>.

If <math>A</math> is [[Invertible matrix|invertible]], one has

:<math>\det\begin{bmatrix}A& B\\ C& D\end{bmatrix} = \det(A) \det\left(D - C A^{-1} B\right),</math><ref name=":0" />
and if <math>D</math> is invertible, one has

:<math>\det\begin{bmatrix}A& B\\ C& D\end{bmatrix} = \det(D) \det\left(A - B D^{-1} C\right) .</math><ref>Taboga, Marco (2021). "Determinant of a block matrix", Lectures on matrix algebra.</ref><ref name=":0" />

If the blocks are square matrices of the ''same'' size further formulas hold. For example, if <math>C</math> and <math>D</math> [[commutativity|commute]] (i.e., <math>CD=DC</math>), then 
:<math>\det\begin{bmatrix}A& B\\ C& D\end{bmatrix} = \det(AD - BC).</math><ref>{{Cite journal|first= J. R.|last= Silvester|title= Determinants of Block Matrices|journal= Math. Gaz.|volume= 84|issue= 501|year= 2000|pages= 460–467|jstor= 3620776|url= http://www.ee.iisc.ernet.in/new/people/faculty/prasantg/downloads/blocks.pdf|doi= 10.2307/3620776|access-date= 2021-06-25|archive-date= 2015-03-18|archive-url= https://web.archive.org/web/20150318222335/http://www.ee.iisc.ernet.in/new/people/faculty/prasantg/downloads/blocks.pdf|url-status= dead}}</ref>
Similar statements hold when <math>AB=BA</math>, <math>AC=CA</math>, or {{tmath|1=BD=DB}}. Namely, if <math>AC=CA</math>, then 
:<math>\det\begin{bmatrix}A& B\\ C& D\end{bmatrix} = \det(AD - CB).</math>
Note the change in order of <math>C</math> and <math>B</math> (we have <math>CB</math> instead of <math>BC</math>). Similarly, if <math>BD = DB</math>, then <math>AD</math> should be replaced with <math>DA</math> (i.e. we get <math>\det(DA - BC)</math>) and if <math>AB = BA</math>, then we should have <math>\det(DA - CB)</math>. Note for the last two results, you have to use commutativity of the underlying ring, but not for the first two.

This formula has been generalized to matrices composed of more than <math>2 \times 2</math> blocks, again under appropriate commutativity conditions among the individual blocks.<ref>{{cite journal|last1=Sothanaphan|first1=Nat|title=Determinants of block matrices with noncommuting blocks|journal=Linear Algebra and Its Applications|date=January 2017|volume=512|pages=202–218|doi=10.1016/j.laa.2016.10.004|arxiv=1805.06027|s2cid=119272194}}</ref>

For <math>A = D </math> and <math>B=C</math>, the following formula holds (even if <math>A</math> and <math>B</math> do not commute)
:<math>\det\begin{bmatrix}A& B\\ B& A\end{bmatrix} = \det(A - B) \det(A + B).</math><ref name=":0" />