# Property of a determinant proof using LDU decomposition

I don't think that this one has been asked, but I've been told that there is proof for a fundamental property of determinants that hinges on LDU decomposition of a matrix. I just can't figure out what property that is and I don't even know why one would want to do a LDU decomposition. I only know about what LDU decomposition is. Please help me.

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How fundamental is it? I doubt mathematicians would be comfortable with such a proof hinging on a coordinate-dependent algorithmic approach! – wj32 Nov 2 '12 at 12:46
Not sure which one you are talking about. But the determinant of an $LDU$ is certainly more obvious than that of the original matrix :). – gt6989b Nov 2 '12 at 12:46
The definition of LDU that I think is standard does not allow all matrices to be decomposed with LDU, so this seems like a fairly flimsy way to prove something about determinants in general. – rschwieb Nov 2 '12 at 13:08
Well it's described as "one of the basic properties of the determinant". I'm just as befuddled as all of you. However, I don't think that the LDU decomposition isn't the lynchpin, but merely that it can be used. Sorry for the confusion. – Additional Pylons Nov 2 '12 at 13:41

Here is a guess. When I was a freshman, they taught me a proof of the fact that $\det AB = \det A \cdot \det B$ that used elementary matrices. I can imagine a similar proof using LDU decomposition.

I mean, suppose that you already know that each matrix has such a decomposition. Suppose that you've also established that multiplying a column of a matrix by a number multiplies the determinant by the same number. Suppose that it's already known that adding one column or row to another doesn't change the determinant.

Using this knowledge, it is very easy to prove that $\det AB = \det A \cdot \det B$ when $B$ is either a diagonal, a unit upper-triangular or a unit lower-triangular matrix. It is also quite easy to establish that $\det LDU = \det D$.

And knowing all this, we can easily prove the equality in general. If $B=LDU$, then $$\det AB = \det ALDU = \det ALD = \det AL \cdot \det D = \det A \cdot \det D = \det A \cdot \det B.$$

But again, this is just a guess.

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This actually seems rather like what I'm looking for, but I'm wondering if you could explain the steps a little more. Why does that U disappear, and then the L? – Additional Pylons Nov 2 '12 at 14:53
Because the $U$ and $L$ are assumed to have unit diagonals -- each diagonal element is 1. – Emily Nov 2 '12 at 15:12
Well yes, this is a rather informal sketch. What I mean is that if a matrix $L$ is unit triangular and $X$ is an arbitrary matrix, then it is easy to see that one can obtain the matrix $XL$ from $X$ by a series of operations of the kind "add one column to another column". Therefore, $\det XL=\det X$, that is how $L$ disappears. Same goes for $U$. – Dan Shved Nov 2 '12 at 15:13
I see. This is excellent. Thank you. – Additional Pylons Nov 2 '12 at 16:16
Now, this means that $\det XD = \det X \cdot \lambda_1 \lambda_2 \ldots \lambda_n$. Plug in $X=E$ (the identity matrix), and you get $\det D = \lambda_1 \ldots \lambda_n$. Plug in $X=AL$ and you get $\det ALD = \det AL \cdot \lambda_1 \ldots \lambda_n = \det AL \cdot \det D$. – Dan Shved Nov 2 '12 at 17:07