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!

Homework Help: Proof that a vector space W is the direct sum of Ker L and Im L

  1. Feb 29, 2012 #1
    Hi there. I'm a long time reader, first time poster. I'm an undergraduate in Math and Economics and I am having trouble in Linear Algebra. This is the first class I have had that focuses solely on proofs, so I am in new territory.

    1. The problem statement, all variables and given/known data

    note Although the question doesn't state it, I think P is supposed to be a projection.

    Let W be a vector space. Let P:W→W be a linear map s.t. P2=P.

    Show that W= KerP + ImP and KerP[itex]\cap[/itex]ImP

    namely, W is the direct sum of KerP and ImP

    Hint: To show W is the sum, write an element of W in the form w=w-P(w)+P(w)

    2. Relevant equations

    I am unsure of any relevant equations.

    3. The attempt at a solution


    I am kind of fuzzy on the meaning of P2=P, and this is where I am stuck.

    Would an example be something like:

    P=[itex]\left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right)[/itex]

    P2=[itex]\left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right) \times \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right) = \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right)[/itex] =P?

    From this example: for KerP=P(z)=0 , we would need to satisfy z+0=0, so z=0

    For some v[itex]\in[/itex]W, the ImP is P(v)=v.

    KerP[itex]\cap[/itex]ImP=0→ P(v)=P(z) iff v=0 so z=v where the two intersect.

    I am unsure where to go from here, or if I'm even doing this correctly. If someone could nudge me along towards and answer without giving the me the full proof, it would be much appreciated. I've been stuck on this problem for 4 hours over two days and I can't seem to figure it out.
  2. jcsd
  3. Feb 29, 2012 #2


    User Avatar
    Science Advisor

    P2 = P is sometimes used as the definition of a projection.

    we can always write w = P(w) + (w - P(w))

    no matter what w is.

    P(w) is clearly in Im(P), so all you have to do is show that

    w - P(w) is in ker(P).

    what is P(w - P(w))?
  4. Feb 29, 2012 #3
    P:V→V and P2=P

    Let me see if I can define this better: so if V is the direct sum of U + W

    let P(v)=u

    let Im P be a complete subset of U; if u is an element of U P(u)=u, so Im P =U

    let Ker P be a complete subset of W; if v is an element of V and v is an element of W, the v=0+w for some element w in W, so Ker P =W

    let v=P(v)+(v-P(v))
    Since P(v) is in the Im P, then (v-P(v)) is in the Ker P

    So P(v-P(v))=P(v)-P2(v)=P(v)-P(u)=u-u=0
    therefore, v-P(v) is in the Ker P

    Is this correct so far? Sorry for not using latex. I have only used it in creating this post and was stretched for time.

    I think I can do the rest if I have set this up correctly.
  5. Feb 29, 2012 #4


    User Avatar
    Science Advisor

    we're not worried about "direct sum" just yet, just the sum (or "join").

    none of this makes much sense. and it's unnecessary.

    you don't need to "let" v = P(v) + (v-P(v)) it ALWAYS is, by the rules of vector addition.

    you don't KNOW this yet, you're trying to prove it. proofs start with what you know. in this case, what you know is:

    v = P(v) + (v-P(v)) (simple algebra, always true, since P(v) + -(P(v)) is the 0-vector) and:
    P(v) is in Im(P) (by the definition of what Im(P) is Im(P)= {u in W: u = P(v) for some v in W}.) and:
    P2 = P (so P(P(v)) = P(v), you'll use this later).

    those are your "knowns".

    why bring "u" into this?

    in my last post, i asked you a question:

    what is P(w - P(w))? you don't need any "extra letters" to answer this, you just need one certain fact about P.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook