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As reference, I use the question How to take the determinant of a partitioned matrix with vectors?

Suppose that

$$ M := \begin{pmatrix} A & B \\ C & D \end{pmatrix} $$

where the block $A$ is $r \times r$ and invertible, the block $B$ is $r \times (p-r)$, the block $C$ is $(p-r) \times r$, the block $D$ is $(p-r) \times (p-r)$ and $p > r$. Further, assume $\operatorname{rank} (M) = \operatorname{rank} (A) = r$. I know from the link above,

$$ \det \left( D - C A^{-1} B \right) = 0 $$

However, I am asked to show $C A^{-1} B = D$. I am not sure how to make the jump from the determinant of $D - C A^{-1} B$ being zero to $D - C A^{-1} B = 0$. Can someone please help me?

Rodrigo de Azevedo
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JJNeedsAssistance
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2 Answers2

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By the rank condition, the columns of the matrix $\pmatrix{B\cr D}$ are linear combinations of the columns of $\pmatrix{A\cr C}$. It follows that there is a matrix $Q$ such that \begin{equation} \pmatrix{B\cr D} = \pmatrix{A\cr C}Q \quad\Longrightarrow\quad B = AQ, \quad D = C Q \end{equation} Multiplying by $A^{-1}$ we get $Q = A^{-1}B$ and finally $D = C A^{-1} B$

Gribouillis
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Let $$P = \begin{pmatrix} A^{-1} & A^{-1}B \\ 0 & -I_{p-r} \end{pmatrix} $$

One has $$ M\times P=\begin{pmatrix} A & B \\ C & D \end{pmatrix} \times \begin{pmatrix} A^{-1} & A^{-1}B \\ 0 & -I_{p-r} \end{pmatrix} = \begin{pmatrix} I_r & 0 \\ CA^{-1} & CA^{-1}B-D \end{pmatrix}$$

Because the matrix $P$ is invertible, then $M \times P$ has the same rank as $M$, i.e. it has rank $=r$. But you see directly from the above computation that this implies that one must have $$\boxed{CA^{-1}B-D=0}$$

TheSilverDoe
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