Proof explanation about matrix for block.

Question

Prove: $$\det\left[\begin{array}[cc]\\A&C\\ 0&B\end{array}\right]=\det(A)\det(B)$$

Proof:

$$A = Q_A R_A, \quad B = Q_B R_B$$

be QR decompositions of $A$ and $B$. Then

\begin{align*} \det \begin{bmatrix} A & C \\ 0 & B \end{bmatrix} &= \det \begin{bmatrix} Q_A R_A & Q_A Q_A^T C \\ 0 & Q_B R_B \end{bmatrix} = \det \left( \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix} \begin{bmatrix} R_A & Q_A^T C \\ 0 & R_B \end{bmatrix} \right) \\ &= \det \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix} \det \begin{bmatrix} R_A & Q_A^T C \\ 0 & R_B \end{bmatrix} = \det Q \det R, \end{align*}

where

$$Q := \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix}, \quad R := \begin{bmatrix} R_A & Q_A^T C \\ 0 & R_B \end{bmatrix}.$$

Notice that $R$ is (upper) triangular, so its determinant is equal to the product of its diagonal elements, so

$$\det R = \det \begin{bmatrix} R_A & 0 \\ 0 & R_B \end{bmatrix}.$$

Combining what we have,

\begin{align*} \det \begin{bmatrix} A & C \\ 0 & B \end{bmatrix} &= \det Q \det R = \det \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix} \det \begin{bmatrix} R_A \\ & R_B \end{bmatrix} \\ &= \det Q_A \det Q_B \det R_A \det R_B = \det (Q_AR_A) \det (Q_B R_B) \\ &= \det A \det B. \end{align*}

I don't understand this line:$$\det \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix} \det \begin{bmatrix} R_A \\ & R_B \end{bmatrix} \\ = \det Q_A \det Q_B \det R_A \det R_B$$

Can explain me that detail?

I think it's a mistake and should say $R_A$ and $R_B$, not $A$ and $B$ — BallBoy, Jan 26 '18 at 01:10
Oh, yes, sorry. @Y.Forman is that. But i dont see why that happen — rcoder, Jan 26 '18 at 01:11
See Émile Greer's answer at https://math.stackexchange.com/questions/148532/general-expression-for-determinant-of-a-block-diagonal-matrix — BallBoy, Jan 26 '18 at 01:15

score 1 · Answer 1 · answered Jan 26 '18 at 01:22

The effort of proving $$ \det \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix} = \det Q_A \det Q_B $$ is basically the same as the effort to show $$ \det \begin{bmatrix} A&C \\0 & B \end{bmatrix} = \det A \det B. $$ You can either work slowly (and painfully) by working on calculating the determinant by definition, or you can do the following. Use Schur Decomposition (or the Jordan Form) to write $A=SJS^{-1}$, $B=TKT^{-1}$, with $J,K$ upper triangular. Then \begin{align} \begin{bmatrix} A&C\\ 0&B\end{bmatrix} &=\begin{bmatrix} SJS^{-1}&C\\ 0&TKT^{-1}\end{bmatrix} =\begin{bmatrix} S&0\\ 0&T\end{bmatrix} \begin{bmatrix} J&S^{-1}CT\\ 0&K\end{bmatrix} \begin{bmatrix} S^{-1}&0\\ 0&T^{-1}\end{bmatrix} \\ \ \\ &=\begin{bmatrix} S&0\\ 0&T\end{bmatrix} \begin{bmatrix} J&S^{-1}CT\\ 0&K\end{bmatrix} \begin{bmatrix} S&0\\ 0&T\end{bmatrix} ^{-1}. \end{align} So $$ \det\begin{bmatrix} A&C\\ 0&B\end{bmatrix}=\det \begin{bmatrix} J&S^{-1}CT\\ 0&K\end{bmatrix} =\det J\,\det K=\det A\,\det B. $$ The fact the determinant of the block matrix with $J,K$ in the diagonal is $\det J\,\det K$ follows easily from the fact that the matrix is upper triangular, and its diagonal is made up of the diagonals of $J$ and $K$.

score 0 · Answer 2 · answered Jan 26 '18 at 02:10

Determinants multiply.

With $A$ square $m$ by $m,$ and $B$ square $n$ by $n,$ also $C$ $m$ by $n,$ $$ \left( \begin{array}{c|c} I_m & 0 \\ \hline 0 & B \end{array} \right) \left( \begin{array}{c|c} A & C \\ \hline 0 & I_n \end{array} \right) = \left( \begin{array}{c|c} A & C \\ \hline 0 & B \end{array} \right) $$

Proof explanation about matrix for block.

2 Answers2