# Determinant of a block upper triangular matrix [duplicate]

How prove the following equality for a block matrix?

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

I tried to use a proof by induction but I'm stuck. Is there a simpler method? Thanks for help.

• Oct 11, 2013 at 12:07
• There is a deeper truth to this than just the proof itself. While a matrix doesn't have the commutative multiplication of a field, most linear algebra results can be rephrased to not assume multiplicative commutivity. Therefore any matrix can be though of as a 2x2 matrix of matrices, and most results follow inductively from the 2x2 case. Oct 11, 2013 at 12:18

Other answers suggest quite elementary proofs, and I upvoted one of them. However, I want to propose a technically easier, but less elementary proof.

If you're familiar with it, you can use QR decomposition. Let

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

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 A \det B = \det (Q_AR_A) \det (Q_B R_B) \\ &= \det A \det B. \end{align*}

Notice that this is far from elementary proof. It uses the QR decomposition, a formula for the determinant of block diagonal matrices, a formula for the determinant of triangular matrices, and block multiplication of matrices.

• I personally don't like too much a proof that works in a much more restricted setting (real matrices) than the result that is being proved requires (here that is matrices over a commutative ring), even it one should be primarily interested in that restricted setting. That is because uch a proof suggests a relation that just isn't there. Oct 11, 2013 at 11:29
• Gaussian elimination definitely does not work for matrices over arbitrary commutative rings. Oct 11, 2013 at 13:16
• But the QR-decompostion is just a special case of the Iwasawa decomposition, so it works at least in somewhat more generality than is suggested... May 14, 2015 at 14:03
• If you prove it for real-valued matrices, you've proved it for matrices with values in an arbitrary commutative unital ring. LHS minus RHS of the desired identity is an integer-coefficient polynomial $F$ in the entries of $A,B,C$. We've proved that $F$ evaluates to 0 at any point of $\mathbb{R}^N$. Thus, $F$ is 0 as an element of $\mathbb{Z}[x_1,\ldots, x_N]$. So it evaluates to 0 any point of $R^N$ for any commutative unital ring $R$. Aug 24, 2016 at 6:43

Since I have criticised some other answers, let me be fair and give my take on this. The result to use is just the Leibniz formula defining the determinant (for once, use the definition!): $$\def\sg{\operatorname{sgn}} \det(M) = \sum_{\sigma \in S_n} \sg(\sigma) \prod_{i = 1}^n M_{i,\sigma(i)}.$$ Now if $M$ is the matrix of the question, and its block $A$ has size $k\times k$, then by the block form $M_{i,j}=0$ whenever $j\leq k<i$ (lower left hand block). So we can drop all permutations from the sum for which $\sigma(i)\leq k$ for any $i\in\{k+1,\ldots,n\}$. But that means that for $\sigma$ to survive, the $n-k$ values $\sigma(i)$ for are $i\in T=\{k+1,\ldots,n\}$ must all be in the $n-k$-element set $T$, and they must of course also be distinct: they form a permutation of the elements of$~T$. So $\sigma$ permutes them elements of $T$ among each other, and then necessarily also the elements of the complement $\{1,\ldots,k\}$ of$~T$. These permutations of subsets are independent, so the surviving permutations are naturally in bijection with the Cartesian product of the permutations of $\{1,\ldots,k\}$ and those of$~T$. Also the sign of the combination of two such permutations is the product of the signs of the individual permutations. All in all one gets \begin{align} \det(M)&=\sum_{\pi \in S_k} \sum_{\tau\in S_T} \sg(\pi)\sg(\tau) \left(\prod_{i = 1}^k M_{i,\pi(i)}\right)\prod_{i \in T}M_{i,\tau(i)} \\&=\left(\sum_{\pi \in S_k}\sg(\pi)\prod_{i = 1}^k M_{i,\pi(i)}\right)\left(\sum_{\tau\in S_T}\sg(\tau)\prod_{i \in T}M_{i,\tau(i)}\right) \\&=\det (A)\det(B). \end{align}

• How would this work if the matrix of zero's was in the lower right? Oct 25, 2016 at 18:59
• @user342914 For a matrix of the form $(\begin{smallmatrix}A&B\\C&0\end{smallmatrix})$ with $B$ and $C$ square has a determinant which is $\pm\det(B)\det(C)$ where the sign is that of the permutation of columns that transforms the matrix to the form $(\begin{smallmatrix}B&A\\0&C\end{smallmatrix})$ (it is $-1$ if and only if both $B$ and $C$ have an odd size). It might be more natural to have $A$ and $0$ square, but in that case there is no obvious formula. Oct 26, 2016 at 6:48

Hint: Use Laplace expansion to get what you need.

• I don't see how I can use the Laplace expansion and why the matrix $C$ hasn't an effect on the result.
– user66407
Oct 11, 2013 at 9:38
• Select last $k$ rows of the matrix, where $k$ is the number of the rows of the $0$-part. What minors of $k$-th order are not equal to 0? What are their algebraic complements? Oct 11, 2013 at 9:46
• The link you provided only does Laplace expansion for one row at the time. So at least you might want to state what Laplace expansion of $k$ rows at the time is exactly. Oct 11, 2013 at 12:06