# 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
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
Can you tell us what your definition is, what you have tried, and why you are trying to do this problem? If it is homework, please use the [homework] tag. And, right now, I'm going off to sleep. Sorry. –  Arturo Magidin Oct 24 '11 at 4:45
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 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. –  Arkamis Apr 23 at 20:42
How second determinant is det(A)det(D) –  codeomnitrix Aug 15 at 9:57
It's the THIRD matrix whose determinant is $det(A) det(D)$. –  John Oct 12 at 20:06