# An LDV decomposition of a matrix.

1. Oct 2, 2006

### MathematicalPhysicist

i need to prove that if A is symmetric and invertible (i.e A^-1 exists), and A=LDV, when L is lower triangular matrix with ones on it's diagonal and V is an upper triangualr matrix also with ones on the diagonal and D is a diagonal matrix then V=L^t.

what i did is:
i know that V^t is an LTM and L^t is an UTM, and that D=D^t, but from i havent seucceded in getting V=L^t.

2. Oct 2, 2006

### AKG

Hint 1: prove that the equation $(V^T)^{-1}LD = DL^TV^{-1}$ makes sense, and then that it's true.

Hint 2: If L is lower triangular with 1's on the diagonal, then what of L-1?

Bonus Hint (highlight, only if you're stuck): If X and Y are lower triangular and D is diagonal, what of XYD?

3. Oct 3, 2006

### MathematicalPhysicist

from LDV=V^tDL^t and V^-1D^-1L^-1=(L^T)^-1D^-1(V^T)^-1
(V^T)^-1=DL^TV^-1D^-1L^-1
(V^T)^-1LD=DL^TV^-1.

for the second hint i think that it's also an LTM.
and because V is a UTM V^t is an LTM so we have in the last equation in the lhs that we have product of two LTM which is also an LTM, time D which makes another LTM, which in return equals a product of a UTM, so we must have that V=L^T, is this correct?

4. Oct 3, 2006

### AKG

You need to first show that L and V are invertible, otherwise your equations don't even make sense. Once you've done it, it's somewhat less complicated than what you've done, although what you've done is right:

$$LDV=V^tDL^t$$
$$(V^t)^{-1}LDV = DL^t$$
$$(V^t)^{-1}LD = DL^tV^{-1}$$

You could even do those last two steps in one. V is a UTM, so VT is an LTM. Can you prove that (VT)-1 is also an LTM? L is an LTM, and so is D, so can you prove that the whole left hand side is an LTM? Likewise, can you prove that the whole right hand side is a UTM? This means that both sides are both LTMs and UTMs, so they're diagonals. But D is a diagonal, so you in fact know that (VT)-1L and LTV-1 are both diagonal. What can you say about the entries on the diagonal? Well you know that L and V have 1's on the diagonal, so you can prove that (VT)-1L and LTV-1 have 1's on the diagonal. In short, these matrices are the same, and they are the identity matrix. But if LTV-1 = I, then of course LT = V, as desired.

5. Oct 4, 2006

### MathematicalPhysicist

L and V are row equivalent to I, so if they are nxn matrices then they rank is n, and thus they are invertible.