# Determinant of a block lower triangular matrix

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. – Arturo Magidin Oct 24 '11 at 4:30
Then, what can I do? – Buddy Holly Oct 24 '11 at 4:32
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 $k$. – Arturo Magidin Oct 24 '11 at 4:34
Honestly, Buddy, if you are trying to prove something about determinants, it's not too much to expect that you actually know something about determinants, at least one of the methods for computing them, and that you can follow Arturo's sketch and fill in the details. If not, ask your teacher, who is paid to help you through these things, or hire a tutor. Sweet dreams, Arturo; now the rest of us get a chance to answer some questions, heh, heh. – Gerry Myerson Oct 24 '11 at 4:59
@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? – Arturo Magidin Oct 24 '11 at 13:41

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? – larry Apr 1 '14 at 3:51
@larry It is triangular. Determinants of triangular matrices are the products of their diagonals. The diagonals of the first matrix are all 1. – Emily Apr 23 '14 at 20:42
How second determinant is det(A)det(D) – codeomnitrix Aug 15 '14 at 9:57
It's the THIRD matrix whose determinant is $det(A) det(D)$. – John Hughes Oct 12 '14 at 20:06
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. – Marc van Leeuwen Apr 12 '15 at 6:11

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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Congratulations with the bounty. However, apart from the initial equation, you forgot to put the necessary signs of permutations anywhere (so you actually showed that the result is true for the permanent instead of the determinant, which is true!). You might want to put the missing signs in. – Marc van Leeuwen Apr 13 '15 at 8:32
Aha! Yes, I will fix that when I get time. Thanks! – Zilin J. Apr 13 '15 at 15:22

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? – Marc van Leeuwen Apr 13 '15 at 8:22
@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). – Andreas Caranti Apr 13 '15 at 8:57
Such a cool proof...thanks so much, @AndreasCaranti :-) – User001 Oct 2 '15 at 0:31

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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