1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Somewhat challenging linear algebra proof

  1. Oct 11, 2008 #1
    1. The problem statement, all variables and given/known data Let LU and L'U' be two LU decompositions for an invertible matrix. Prove L=L' and U=U', thus the LU decomposition for an invertible matrix is unique.

    2. Relevant equations

    3. The attempt at a solution I honestly do not really know what to do. I suppose I could consider something with the diagonals being equal and try to show that the entries would be the same for the identity matrix, but I'm not sure how to approach any of this.
  2. jcsd
  3. Oct 11, 2008 #2


    Staff: Mentor

    What you're trying to prove isn't true in general. Here's a counterexample.
    Code (Text):

    | 3 -6 |
    | -2 5 |
    Note that A is invertible because its determinant is nonzero.

    Here's one decomposition:
    Code (Text):

    | 3   0|  |1  -2 |
    | -2  1|  |0   1 |
    It's pretty easy to check that LU = A, so the matrices above constitute an LU factorization of A.
    Here's another decomposition of A:
    Code (Text):

    | 1      0|  |3  -6 |
    | -2/3  1|  |0   1 |
    It's also easy to check that L'U' = A, so here is a different LU factorization of A.
    (Sorry this is so clunky. I'm not up on writing matrices in LaTeX.)

    What do these decompositions say about the statement you're trying to prove?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Somewhat challenging linear algebra proof
  1. Linear Algebra Proof (Replies: 8)

  2. Linear Algebra proof (Replies: 21)