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I'm trying to prove the following: Let $A$ be a $k\times k$ matrix, let $D$ have size $n\times n$, and $C$ have size $n\times k$. Then,

$$\det\left(\begin{array}{cc} A&0\\ C&D \end{array}\right) = \det(A)\det(D).$$

Can I just say that $AD - 0C = AD$, and I'm done?

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    No, you cannot just say that; it doesn't even make sense, because $AD$ cannot be computed: $A$ has size $k\times k$, and $B$ has size $n\times n$. Unless $n=k$, $AD$ doesn't make sense.2011-10-24
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    Then, what can I do?2011-10-24
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    Depends on your definition of the determinant. If it is a sum over all permutations of (in this case) $n+k$, then you should figure out which terms you know for sure are equal to $0$; the formula will drop out of that if you are careful enough. If your definition of determinant is via expansion by minors, then I suggest expanding along the first row and using induction on $k$.2011-10-24
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    Can you show how to do it both ways?2011-10-24
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    I thought of breaking the problem to 2 matricies, (A 0 in the first column, and 0 I sub n) in the second colum, and (0 I sub n in the first column, and 0 C in the second column) but how to use induction on this?2011-10-24
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    @BuddyHolly: I said that it depends on what your definition of determinant is, and sketched two possibilities. Rather than bother to even tell us what your definition of determinant is, you replied by asking me to do *two* proofs for you; why should I do a double effort when you seem unwilling to do even the small effort required to tell us what your *definition* is?2011-10-24
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    @user153012 What is missing from the top-voted answer? It looks complete to me.2015-04-12
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    @Qudit That answer does not prove the formula of the determinant of blockdiagonal matrices. It referes to the Laplace expansions, but anyway it is not a full detailed proof.2015-04-12
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    @Qudit What is wrong with the (currently top-voted) answer by joriki is that it expends all its effort in obtaining a reduction (to the case $C=0$) that is completely useless to the problem at hand. I contend that whatever method one uses to show that $\det(\begin{smallmatrix} A&0\\0&D \end{smallmatrix}) = \det A\det D$, the same also applies directly to the matrix of this question (given that the top-right block is $0$, there is just no way the bottom-left block contributes anything to the determinant). In order to make progress, one needs to split $A,D$ into _separate_ factors, see my answer2015-04-13
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    @Marc van Leeuwen Yes, I was thinking it was complete because the derivation of the formula for the determinant of block diagonal matrices using the Leibniz formula is clear to me without writing out the details. As you point out however, the same argument applies equally well to the original problem.2015-04-13
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    I just added an answer that takes Artin's approach, which is nice because its independent of explicit definitions of the determinant function. A lot of heavy hitters on this post so I feel a bit out of my league, but please check it out and leave comments!2017-02-10

6 Answers 6

28

If $A$ is singular, its rows are linearly dependent, hence the rows of the entire matrix are linearly dependent, hence boths sides of the equation vanish.

If $A$ is not singular, we have

$$\pmatrix{I&0\\-CA^{-1}&I}\pmatrix{A&0\\C&D}=\pmatrix{A&0\\0&D}\;.$$

The determinants of the two new matrices are perhaps easier to derive from the Laplace expansion than that of the entire matrix. They are $1$ and $\det A \det D$, respectively, and the result follows.

Another way to express this is that if $A$ is not singular you can get rid of $C$ by Gaussian elimination.

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    How do you know that the first matrix has a determinant of 1?2014-04-01
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    @larry It is triangular. Determinants of triangular matrices are the products of their diagonals. The diagonals of the first matrix are all 1.2014-04-23
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    How second determinant is `det(A)det(D)`2014-08-15
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    It's the THIRD matrix whose determinant is $det(A) det(D)$.2014-10-12
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    This answer falls into the trap of proving an elementary result (one of the first one should know about determinants, immediately falling out of the Leibniz formula) by using statements that are special cases or consequences of the result to be proved. The result about triangular matrices that @Arkamis refers too can be obtained by iterating a decomposition into block-triangular matrices until hitting $1\times1$ blocks. But more fundamentally, the RHS matrix is just a special case of a block triangular matrix, and proving its determinant is $\det A\det D$ is not really any easier than the OP.2015-04-12
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    @MarcvanLeeuwen: I'd written "are *perhaps* easier to derive" -- I find it easier, but others may not. To get the result for the right-hand side, you just have to argue that the non-zero terms of the Laplace expansion involve all possible products of the terms of the Laplace expansions of the components, so this is the product of the Laplace expansions multiplied out. Of course you could make a similar argument about the OP, but it involves a bit more thinking about why no non-zero terms involving elements of $C$ occur.2015-04-28
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As @user153012 is asking for a proof in full detail, here is a brute-force approach using an explicit expression of a determinant of an $n$ by $n$ matrix, say $A = (a[i,j])$, $$\det A = \sum_{\sigma\in S_n}\operatorname{sgn}\sigma \prod_i a[{i,\sigma(i)}],$$ where $S_n$ is the symmetric group on $[n] = \{1,\dots, n\}$ and $\operatorname{sgn}\sigma$ denotes the signature of $\sigma$.

In matrix $$B = \left(\begin{array}{cc} A&0\\ C&D \end{array}\right),$$ we have $$b[i,j] = \begin{cases}a[i,j] & \text{if }i \le k, j \le k;\\ d[i-k, j-k] & \text{if }i > k, j > k; \\ 0 & \text{if }i \le k, j > k; \\ c[i-k,j] & \text{otherwise}.\end{cases}$$ Observe in $$\det B = \sum_{\sigma\in S_{n+k}}\operatorname{sgn}\sigma\prod_i b[i, \sigma(i)],$$ if $\sigma(i) = j$ such that $i\le k$ and $j > k$, then the corresponding summand $\prod_i b[i,\sigma(i)]$ is $0$. Any permutation $\sigma\in S_{n+k}$ for which no such $i$ and $j$ exist can be uniquely "decomposed" into two permutations, $\pi$ and $\tau$, where $\pi\in S_k$ and $\tau\in S_n$ such that $\sigma(i) = \pi(i)$ for $i \le k$ and $\sigma(k+i) = k+\tau(i)$ for $i \le n$. Moreover, we have $\operatorname{sgn}\sigma = \operatorname{sgn}\pi\operatorname{sgn}\tau$. Denote the collection of such permutations by $S_n'$. Therefore, we can write $$\begin{eqnarray}\det B &=& \sum_{\sigma\in S_n'}\operatorname{sgn}\sigma\prod_{i=1}^k b[i,\sigma(i)]\prod_{i=k+1}^{k+n} b[i,\sigma(i)] \\ &=& \sum_{\sigma\in S_n'}\operatorname{sgn}\sigma\prod_{i=1}^k a[i,\sigma(i)]\prod_{i=1}^nd[i,\sigma(i+k)-k] \\ & = & \sum_{\pi\in S_k,\tau\in S_n}\operatorname{sgn}\pi\operatorname{sgn}\tau\prod_{i=1}^k a[i,\pi(i)]\prod_{i=1}^nd[i,\tau(i)] \\ &=& \sum_{\pi\in S_k}\operatorname{sgn}\pi\prod_{i=1}^k a[i,\pi(i)]\sum_{\tau\in S_{n}}\operatorname{sgn}\tau\prod_{i=1}^nd[i,\tau(i)] \\ & = & \det A\det D.\end{eqnarray}$$ QED.

Update As @Marc van Leeuwen mentioned in the comment, a similar formula holds for permanents.The proof is basically the same as the proof for determinant except one has to drop off all those signatures of permutations.

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This is a fundamental result about determinants, and like most of such results it holds for matrices with entries in any commutative (unitary) ring. It is therefore good to have a proof that does not rely on the coefficients being in a field; I will use the Leibniz formula as definition of the determinant rather than a characterisation as alternating $n$-linear form. (And my apologies to those who find that distasteful; for some purposes using the definition is really best.)

Writing for a permutation $\sigma$ and a matrix $M$ the abbreviation $\def\sg{\operatorname{sg}}M[\sigma]=\sg(\sigma)\prod_iM_{i,\sigma(i)}$, the Leibniz formula says, for any $m\times m$ matrix $M$ $$ \det(M)=\sum_{\sigma\in S_m}M[\sigma] $$ The result is based on the following simple fact about symmetric groups

The subgroup of $S_{k+n}$ of permutations permuting the first $k$ elements among each other is canonically isomorphic to $S_k\times S_n$, and if $\sigma\in S_{k+n}$ corresponds to $(\pi,\rho)\in S_k\times S_n$ then $\sg(\sigma)=\sg(\pi)\sg(\rho)$.

In fact this is basically just saying that if the first $k$ elements are permuted among each other, then so are the remaining $n$ elements, and the sign of the whole permutation is the product of the signs of its restrictions to those two subsets.

Now if $M=\bigl(\begin{smallmatrix}A&0\\C&D\end{smallmatrix}\bigr)$ note that $M[\sigma]=0$ unless $\sigma$ permutes first $k$ indices among each other. From this it follows that $$ \det(M)=\sum_{\sigma\in S_{k+n}}M[\sigma] =\sum_{(\pi,\rho)\in S_k\times S_n}A[\pi]D[\rho] =\left(\sum_{\pi\in S_k}A[\pi]\right)\left(\sum_{\rho\in S_n}D[\rho]\right) =\det A\det D. $$


Alternatively, if one is willing to assume the property $\det(MN)=\det(M)\det(N)$ (not really easier than the one to be proved here, but maybe better known), and if one considers the special cases where either $A=I$ or $D=I$ to be clear (because one can obtain the result in these cases by repeated Laplace expansion by rows respectively by columns), then one can employ any decomposition of the form $$ \pmatrix{A&0\\C&D} = \pmatrix{A&0\\C_0&I} \pmatrix{I&0\\C_1&D} \qquad\text{with $C=C_0+C_1$} $$ (for instance with one of $C_0,C_1$ equal to zero) to obtain the desired result.

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Yet another proof, in the case of fields, can be obtained if you are willing to enlarge your field to an algebraically closed one. Then any matrix can be put in (lower) triangular form, with the eigenvalues on the diagonal. In particular, there are invertible matrices $S, T$ of appropriate sizes such that $S^{-1} A S$ and $T^{-1} D T$ are in lower triangular form. Then $$ \begin{bmatrix} S& 0\\ 0 & T\\ \end{bmatrix}^{-1} \begin{bmatrix} A&0\\ C&D \end{bmatrix} \begin{bmatrix} S& 0\\ 0 & T\\ \end{bmatrix} = \begin{bmatrix} S^{-1} A S&0\\ T^{-1} C S&T^{-1} D T \end{bmatrix} $$ is in lower triangular form, and the rest is more or less straightforward. Clearly I rely on the multiplicativity of determinants here, or on the fact that the determinant is invariant under conjugation.

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    Wow! Using algebraically closed fields to prove an elementary property of determinants. Something similar to my comment to the answer by joriki would apply here, although I admit that in principle one _could_ argue that the determinant of a triangular matrix is the product of its diagonal entries _without_ either using the result to be proved here, or an argument (Leibniz formula) that would just as easily prove it directly. Notably repeated Laplace expansion does the trick. But then wouldn't you agree a two-factor decomposition as at the end of my answer does the job more economically?2015-04-13
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    @MarcvanLeeuwen, I had *upvoted* your answer, which is definitely the best of the lot. Still, I find that putting matrices in triangular forms (admittedly over an a.c. field) is a generally useful technique (for instance in the the proof of Cayley-Hamilton as given by - if I remember correctly - Lang in one of his books) and I couldn't resist recording this version (no doubt courting *downvotes*).2015-04-13
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    Such a cool proof...thanks so much, @AndreasCaranti :-)2015-10-02
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Here is a sketch of what I consider a simpler direct proof (assuming complete eigenspaces --- it can be adapted for degenerate eigenspaces). Let's assume there is no zero eigenvalue, otherwise it's easy to find the eigenvector that the full matrix sends to zero, and so the determinant must be zero.

We're going to prove that the eigenvalues are the same. Then the fact that the determinant is the product of the eigenvalues finishes the proof.

Consider the eigenvectors of $D$. Now (pre)pad those with zeros (that is create a longer vector with zeros in the first few entries). These are eigenvectors of the full matrix.

Now consider eigenvectors of $A$. We're going to postpad with something. We have to figure out what it is. Let $\lambda$ be an eigenvalue of $A$ and $v_1$ its eigenvector. We'll pad it with $v_2$. So $v$ is made up of the vectors $v_1$ and $v_2$. The full matrix time $v$ has its first entries being $\lambda v_1$. It's second set of entries is $C v_1 + D v_2$. Set this equal to $\lambda v_1$ and solve for $v_2$. Since $D$ does not have a zero eigenvalue, it's solvable. Now we have an eigenvector for the full matrix that corresponds to $\lambda$.

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Here is an approach that does not rely on any explicit definition of the determinant, nor any concept of the inverse. Instead, we can start with the 3 basic properties that the determinant function should satisfy. These three properties are:

(1) Det(I) = 1

(2) The Det() function is multilinear in each of the rows (columns) individually, assuming all other rows (columns) are held constant

(3) If the matrix M is not full rank, Det(M)=0

Artin showed that these three properties alone uniquely determine the form of the determinant function (I don't prove this here). Property 3 I am using here is slightly more general than that used by Artin, but it is equally intuitive and allows me to skip a step. First, you can show that

$$ Det \begin{pmatrix} A & 0 \\ C & D \end{pmatrix} = Det \begin{pmatrix} A & 0 \\ 0 & D \end{pmatrix} $$

To show this, we can expand the determinant of the original matrix M as $$ (4) Det(M)= Det(M_1)+Det(M_2) $$ where $M_1$ is the same as the original matrix but with the first k entries of the $k+1^{th}$ row set to 0, and $M_2$ is the same as the original matrix but with the last n-k elements of the $k+1^{th}$ row seet to 0. Note that the $k+1^{th}$ row of $M_1$ and $M_2$ sum to the $k+1^{th}$ row of M, and therefore (4) holds according to property (2). Note that $ Det(M_1)=0 $ since the resulting matrix is clearly not full-rank (there are now k+1 rows that have non zero entries in only k columns). Therefore we have $Det(M) = Det(M_2)$. We can then repeat this process for each row to show that the above claim is true.

Now, let us denote $$ M^*=\begin{pmatrix} A & 0 \\ 0 & D \end{pmatrix} , A^*= \begin{pmatrix} A & 0 \\ 0 & I \end{pmatrix} , D^*= \begin{pmatrix} I & 0 \\ 0 & D \end{pmatrix} $$ Note that $ M^*=A^*B^* $ I claim that $ Det(M^*) = Det(A^*)*Det(D^*). $ To show this we can show that the function $$(5) F(D^*)=F(d^*_1,...,d^*_n)=\frac{Det(A^*D^*)}{Det(A^*)}$$ also satisfies properties (1)-(3). Clearly if $D=I$, then the RHS of (5) reduces to 1. To show (ii), We can write the j-th column of $M^*=A^**D^*$ as $A^*d_j$. Since we already know the determinant is multilinear in each column, and $A^*d^*_j$ is a linear function of d^*_j, it follows that $F(d^*_1,...,d^*_n)$ is also multininear in each of the columns ${d^*_1,...,d^*_n}$. Finally, to show (iii), we can note that if $d^*_i=d^*_j$ then $A^*d^*_i=A^*d^*_j$, so the numerator on the RHS would be 0. Therefore, $F(d^*_1,...,d^*_n)=Det(d^*_1,...,d^*_n)$, so $Det(A^**D^*)=Det(D^*)*Det(A^*)$, as desired.

In order to finish the proof, one final step is needed. We need to show that $Det(A^*)=Det(A)$ and $Det(D^*)=Det(D)$. We use the same approach as above, by defining a target function which we show to be equal to the determinant. We do this by showing that the function $C(A)=Det(A^*)$ also satisfies properties (1)-(3) as a function of the columns of A, and therefore must be equal to its determinant. The basic ideas are that if A is the identity then so is $A^*$, so property (1) follows; any inear operation on a row (column) of A is also a linear operation on a row (column) of $A^*$, so (2) holds, and if A is not full rank, then neither is $A^*$, so (3) holds. Therefore, the function C(A), which extends A to $A^*$ by adding n-k standard basis vectors and then takes the determinant of the expanded matrix, is in fact equal to $Det(A)$.

Given all of this, we immediately get the result, since $$Det(M)=Det(M^*)=Det(A^**D^*)=Det(A^*)*Det(D^*)=Det(A)*Det(D) $$