4
$\begingroup$

On Wikipedia, I saw the following formula

$$\det\begin{bmatrix}A & B\\ C & D\end{bmatrix} = \det(AD-BC)$$

if $C$ and $D$ commute. Is this always true?

Or is there a good counter example for each $2 \times 2$ block matrices?

$\endgroup$
  • $\begingroup$ Perhaps you could try proving it by considering them as linear transformations . . . $\endgroup$ – Shaun Nov 1 '13 at 23:28
11
$\begingroup$

Yes, it's always true, but note that there is a premise to fulfill: $C$ and $D$ have to commute, that is, we need $CD=DC$. If this condition is violated, the formula may fail to hold.

More generally, when the entries of $A,B,C,D$ be matices over a field (this includes, but isn't limited to, the cases where $A,B,C,D$ are real or complex matrices), we have $$ \det \pmatrix{A&B\\ C&D}= \begin{cases} \det(AD-BC) & \text{ if } CD=DC,\\ \det(DA-CB) & \text{ if } AB=BA,\\ \det(DA-BC) & \text{ if } BD=DB,\\ \det(AD-CB) & \text{ if } AC=CA. \end{cases} $$ A proof was given by John Silvester (see his paper). Essentially, suppose $AC=CA$ (this is the last case in the above). Then $$ \pmatrix{I&0\\ -C&A+xI}\pmatrix{A+xI&B\\ C&D}=\pmatrix{A+xI&B\\ AC-CA&(A+xI)D-CB}. $$ By assumption, $AC-CA=0$, so the RHS is block-triangular and $$\det(A+xI)\,\det\pmatrix{A+xI&B\\ C&D}=\det(A+xI)\,\det\left((A+xI)D-CB\right).$$ As $\det(A+xI)$ is a nonzero polynomial in $x$, we can divide both sides by it and obtain $\det\pmatrix{A+xI&B\\ C&D}=\det\left((A+xI)D-CB\right)$. Put $x=0$, the assertion follows. The proofs for the other three cases are similar.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.