1

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.

Additional Pylons
  • 317
  • 1
    How fundamental is it? I doubt mathematicians would be comfortable with such a proof hinging on a coordinate-dependent algorithmic approach! – wj32 Nov 02 '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 02 '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 02 '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 02 '12 at 13:41

1 Answers1

1

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.

Dan Shved
  • 15,862
  • 39
  • 55
  • 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 02 '12 at 14:53
  • Because the $U$ and $L$ are assumed to have unit diagonals -- each diagonal element is 1. – Emily Nov 02 '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 02 '12 at 15:13
  • I see. This is excellent. Thank you. – Additional Pylons Nov 02 '12 at 16:16
  • Just a small update: as was pointed in the comments to your question, this proof still lacks something, because not every matrix has an LDU decomposition. Some hand-waving about an additional permutation of the columns can probably fix this. – Dan Shved Nov 02 '12 at 16:38
  • Oh, but one last thing, could you tell me how you went from detALD to detAL detD? – Additional Pylons Nov 02 '12 at 16:43
  • Suppose that $D$ has numbers $\lambda_1, \ldots, \lambda_n$ on the diagonal, and $X$ is an arbitrary matrix. Then $XD$ can be obtained from $X$ by multiplying the first column by $\lambda_1$, the second by $\lambda_2$, and so on. – Dan Shved Nov 02 '12 at 17:06
  • Alrighty then. Thanks! – Additional Pylons Nov 02 '12 at 17:07
  • 1
    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 02 '12 at 17:07