5
$\begingroup$

$\textbf{Problem}$: Let a $2n \times 2n$ matrix be given in the form $M=\left[ {\begin{array}{cc} A & B \\ C & D \\ \end{array} } \right]$, where each block is an $n \times n$ matrix. Suppose that $A$ is invertible and that $AC=CA$. Use block multiplication to prove that $\det M= \det(AD-CB)$. Give an example to show that this formula need not hold if $AC \neq CA$

$\textbf{Proof}$: Let $A,B,C,D,X \in \textbf{M}_n(K)$ such that $A+BX$ is invertible. For all $Y \in \textbf{M}_n(K)$, we have:

$$\left[ {\begin{array}{cc} I_n & 0 \\ Y & I_n \\ \end{array} } \right] \left[ {\begin{array}{cc} A & B \\ C & D \\ \end{array} } \right] \left[ {\begin{array}{cc} I_n & 0 \\ X & I_n \\ \end{array} } \right]= \left[ {\begin{array}{cc} A+BX & B \\ YA+C+(YB+D)X & YB+D \\ \end{array} } \right].$$

Let $Y=-(C+DX)(A+BX)^{-1}$. Hence:

$$YA+C+(YB+D)X=Y(A+BX)+(C+DX)=0.$$

Since $\det\left[ {\begin{array}{cc} I_n & 0 \\ Y & I_n \\ \end{array} } \right]= \det\left[ {\begin{array}{cc} I_n & 0 \\ X & I_n \\ \end{array} } \right]= (\det(I_n))^2=1$, we can conclude that:

\begin{align*} \det\left[ {\begin{array}{cc} A & B \\ C & D \\ \end{array} } \right]&=\det\left[ {\begin{array}{cc} A+BX & B \\ 0 & YB+D \\ \end{array} } \right]\\ &= \det(A+BX)\det(-(C+DX)(A+BX)^{-1}B+D). \end{align*}

In particular for $X=0$, we have:

\begin{align*} \det\left[ {\begin{array}{cc} A & B \\ C & D \\ \end{array} } \right]&=\det(A)\det(-CA^{-1}B+D)=\det(-ACA^{-1}B+AD) \\ &=\det(-CAA^{-1}B+AD)=\det(AD-CB). \end{align*}

I just wanted someone to verify my proof and help me with the second part of this question.

Thank you in advance

$\endgroup$
  • $\begingroup$ Some typesetting advices: bmatrix environment is better suited for matrices that array (you get nicer code, and need not "predict" the number of columns). Also, align* environment will help you make nice multiline formulas, so they don't "run out" of the post. $\endgroup$ – Vedran Šego Aug 26 '13 at 10:26
  • 1
    $\begingroup$ It appears that you could let $X=0$ from beginning to simplify proof. $\endgroup$ – Maesumi Aug 26 '13 at 11:09
  • $\begingroup$ See reference to block matrices here $\endgroup$ – Maesumi Aug 26 '13 at 11:45
2
$\begingroup$

Your proof seems fine to me.

As for a counterexample, consider

$$A = \begin{bmatrix} 2 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 2 & 1 & 1 & 1 \\ 1 & 1 & 1 & 2 \end{bmatrix}.$$

In one hand, $\det A = -4$ (check here), and in the other hand, $\det (A_{11} A_{22} - A_{21}A_{12}) = 0$ (check here).

$\endgroup$

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.