Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Something that's been frustrating me

  1. Jun 8, 2005 #1
    Ok, I was reading the proof for Singular Value Decomposition in my Linear Algebra textbook, when the author made an assertion (without proof). Basically, he said that if A is an m x n matrix, then the nullspace of A is equal to the nullspace of transpose(A)*A.

    Now, it's obvious to me that any member of N(A) is in N(transpose(A)*A), since Ax=0 implies that transpose(A)*Ax=0. Nevertheless, I can't prove the converse of this statement to myself. Any tips?
  2. jcsd
  3. Jun 10, 2005 #2
    It seems to be true if the underlying vector spaces are real. I don't know about more general spaces. But here's what I'm thinking. What would happen if there were a column vector [itex] x \neq 0 [/itex] that is in the null space of [itex]A^TA[/itex] but not that of A?

    [tex]A^TAx = 0[/tex] ,

    but [tex]Ax = y \neq 0[/tex]

    Well, premultiply the first equation by [itex]x^T[/itex] to get

    [tex]x^T A^T A x = y^T y = 0[/tex].

    But if y is real vector, this equation implies that y = 0, which contradicts our assumption that it is nonzero.
  4. Jun 10, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

    Multiply both sides by [itex]x^T[/itex] so you get:


    Take it from there.
  5. Jun 10, 2005 #4
    Alright, I thank both of you for the help!
  6. Jun 12, 2005 #5


    User Avatar
    Science Advisor
    Homework Helper

    This is kind of confusing conceptually.

    I.e. abstractly, transpose means "precede by". I.e. a linear map A:V-->W induces a linear map AT:W*-->V*, where W* is linear functioins on W, and if L is such a thing then AT(F) = FoA, a linear function on V.

    So if AT(F) = FoA = 0, it means that F vanishes on the image of A, since preceding F by A, gives zero.

    But now how do we precede AT by A? i.e. it makes no sense abstractly to compose a map into W with a map out of W*. But this is where an inner product comes in, giving us an isomorphism of W with W* and also of V with V*.

    So we compose A:V-->W-->W*-->V*, where the map in the middle takes a vector in W to a functional on W by dotting with that vector.

    so if this composition kills v, then it means that "preceding by A", kills "dotting with Av".

    I.e. that for every x in V, we have Av.A(x) = 0.

    applying this to v gives Av.Av= 0, so Av = 0.

    so the point is: the only way that dotting with Av, can kill everything of form Ax, is if Av=0.

    it is much easier computationally as follows: AT is the unique map such that for all x,y, we have Ax.y =x.ATy.

    hence if ATAv = 0, then Av.A( ) is zero no matter what goes in the blank. putting in v gives Av = 0.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook