Let $A \in \mathbb{R}^{n, n}$, $B \in \mathbb{R}^{n, m}$, $C \in \mathbb{R}^{m, n}$ and $D \in \mathbb{R}^{m, m}$ be matrices.

Now, I have seen on Wikipedia the explanation of why determinant of $\left[\begin{matrix} A & 0 \\ C & D\end{matrix}\right] = \det{A}\cdot \det{D}$, but I still did not get it.

Specifically, the explanation is:

This can be seen ... from a decomposition like:

$\left[\begin{matrix} A & 0 \\ C & D\end{matrix}\right] = \left[\begin{matrix} A & 0 \\ C & I_{m}\end{matrix}\right]\left[\begin{matrix} I_n & 0 \\ 0 & D\end{matrix}\right]$

I understood that the equation is true from the standard rules of matrix-matrix multiplication, but it is still not too clear why this should prove what we want to prove or show.

If $A$, $B$, $C$ and $D$ were regular reals (and $I_{i}$ was $1$), then the equation and the explanation would be obvious, because of the standard rules of calculating determinants...

But in this case, I cannot understand why the equation shows that the final determinant is $$\det{A} \cdot \det{D}$$

Those 2 matrices $\left[\begin{matrix} A & 0 \\ C & I_{m}\end{matrix}\right]$ and $\left[\begin{matrix} I_n & 0 \\ 0 & D\end{matrix}\right]$ basically could not be triangular or diagonal matrices, from my understanding...


marked as duplicate by Guy Fsone, muaddib, Parcly Taxel, Hans Engler, Chris Godsil Jan 28 '18 at 14:05

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  • $\begingroup$ You cna see: math.stackexchange.com/questions/266068/… $\endgroup$ – Emilio Novati Jun 5 '15 at 13:16
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    $\begingroup$ That det $\left[\begin{matrix} I_n & 0 \\ 0 & D\end{matrix}\right] = \det D$ can be seen by expanding along the first row, then you end up with an $( n+m-1) \times (n+m-1)$ matrix, expand again along the first row etc. Is this answer clear? $\endgroup$ – wiskundeliefhebber Jun 5 '15 at 13:23
  • $\begingroup$ Using Leibniz formula the summand for the first factor are non-zero only when the last $m$ terms in the product are 1 so you get $\det(A)$, similarly for the second matrix. $\endgroup$ – simonzack Jun 5 '15 at 13:24
  • $\begingroup$ @wiskundeliefhebber More or less... $\endgroup$ – nbro Jun 5 '15 at 13:31
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    $\begingroup$ for example for $4 \times 4$ case, calculate as follows $\det \left[\begin{matrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&a&b\\ 0&0&c&c\\\end{matrix}\right] = 1 \cdot \det \left[\begin{matrix} 1&0&0\\ 0&a&b\\ 0&c&c\\ \end{matrix}\right]= 1 \cdot \det \left[\begin{matrix} a&b\\ c&c\\ \end{matrix}\right] $ $\endgroup$ – wiskundeliefhebber Jun 5 '15 at 13:48

You have mentioned that if $I_i = 1$, then the formula follows obviously. In fact it is meaningful, as it serves as a base case of an induction (on the matrix size $m$ and $n$ respectively).

The two matrices are called block triangular matrices. They are not triangular matrices themselves, but has a similar triangular pattern with each entry replaced by blocks. You can view this block triangular determinant formula as an extension of $$\det\begin{pmatrix} a & 0\\ c & d \end{pmatrix} = ad$$ This factorization is useful because the $(m+n) \times (m+n)$ determinants can be reduced to smaller ones, namely $m \times m$ and $n \times n$. The easier one is to see that $$\det\begin{pmatrix} I_n & 0\\ 0 & D\end{pmatrix} = \det(D)$$ As $n = 1$, then it follows trivially from the definition of determinant. Assume it is true for some positive integer $n$. Then $$\det\begin{pmatrix} I_{n+1} & 0\\ 0 & D\end{pmatrix} = 1\times \det\begin{pmatrix} I_n & 0\\ 0 & D\end{pmatrix} = \det(D)$$ by the definition of determinant again. By induction, the formula is true for all positive integer $n$.

The other formula is a little more difficult to prove. $$\det\begin{pmatrix} A & 0\\ C & I_m\end{pmatrix} = \det(A)$$ In particular, the following explains how to do the induction step on $m$. The determinant of the transpose of a matrix is the same as that of the original matrix. Therefore $$\det\begin{pmatrix} A & 0\\ C & I_{m+1}\end{pmatrix} = \det\begin{pmatrix} A^T & C^T\\ 0 & I_{m+1}\end{pmatrix}$$ Leibniz's rule allows us to expand the determinant not only on the first row, but any other row. In particular, expanding through the last row tells $$\det\begin{pmatrix} A^T & C^T\\ 0 & I_{m+1}\end{pmatrix} = (-1)^{2(m+n+1)} \times \det\begin{pmatrix} A^T & B^T\\ 0 & I_{m}\end{pmatrix} = 1\times\det\begin{pmatrix} A & 0\\ B & I_{m}\end{pmatrix} = \det(A)$$ where $B$ is the matrix obtained by deleting the last row of $C$. At the second last equal sign of the last equation we use the property of the determinant of the transpose again. The inductive hypothesis is used at the last equal sign.


The first thing to note is of course the fact that $\det (AB)=\det A \cdot\det B$. This is well known - if you search you will find several proofs - in some texts this condition is used as an axiom when defining the determinant. So this allows you to assert \begin{equation} \det \begin{bmatrix} A & 0 \\ C & D\end{bmatrix} = \det \begin{bmatrix} A & 0 \\ C & I_{m}\end{bmatrix} \det \begin{bmatrix} I_n & 0 \\ 0 & D\end{bmatrix}.\end{equation} Now it is very simple - if you just apply Laplace expansion (cofactor expansion - which is the usual way to determine the determinant by hand) the answer follows directly, since you can expand \begin{equation}\det \begin{bmatrix} A & 0 \\ C & I_{m}\end{bmatrix}\end{equation} starting at the last column and then the second last column etc. to see that \begin{equation}\det \begin{bmatrix} A & 0 \\ C & I_{m}\end{bmatrix}=1^m \cdot \det A = \det A.\end{equation} Now similarly you can expand \begin{equation} \det \begin{bmatrix} I_n & 0 \\ 0 & D\end{bmatrix} \end{equation} by the first row, then the second, etc. to see that \begin{equation} \det \begin{bmatrix} I_n & 0 \\ 0 & D\end{bmatrix}=1^n \cdot \det D = \det D.\end{equation} If you have difficulty understanding this last part (cofactor expansion), then I think you should spend some time with a good text book to make sure you understand exactly how it works. Basically it allows you to calculate determinants very easily in some cases by making intelligent choices with regards to the rows and columns you are expanding by. In this case in particular there are rows/columns which contain only a single 1 which effectively means you can reduce immediately to determining the cofactor, or minor for that matter since the 1 is in a diagonal position.

  • $\begingroup$ Isn't it easier (once you have split the determinant into a product of two) to use column operations of the form $c_{i} \to c_{i} - a_{ij}c_{j}$ for $j > n$ and $i \leq n$, (which does not change the determinant) to replace $C$ by the zero matrix? $\endgroup$ – Geoff Robinson Jun 7 '15 at 14:11
  • $\begingroup$ It comes down to exactly the same thing...arguably with more effort. But you can view it as just another way to understand it. Laplace expansion is however such a standard technique, therefore I would see no need to even use elementary operations with its corresponding effects on determinant. One textbook I own use elementary operations to prove the Laplace expansion formula, so ... $\endgroup$ – Christiaan Hattingh Jun 7 '15 at 14:15
  • $\begingroup$ Sure, all is equivalent, and it's really a matter of taste and viewpoint. $\endgroup$ – Geoff Robinson Jun 7 '15 at 15:07

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