Proving that the solution of Ax=0 is a vector space

  1. Here is my attempt to answer this guys, i'd really appreciate any corrections.


    a vector space has the 0 vector
    the vector space is closed under vector addition and scalar multiplication (AKA for every vector u, v in the subspace, there exists a vector u + v in the subspace)

    Here we go.

    The trivial solution is an indication that the 0 vector exists. I don't know how to prove that one, unless i write out the linear combination and set all the scalars to 0

    The trivial solution is not the only solution: free variables and possibly 0 columns or rows exist. (also an indication that the 0 vector exists).

    so if we row reduce the matrix to row echalon form and solve for [x1 ... xn] by moving the free variables to the right side of the homogenius equation then that would be the same as the additive property i mentioned above right? but how would i even know that u or v are in the vector space to begin with ? (remindeR: u and v are vector columns of the matrix A.)
     
  2. jcsd
  3. Dick

    Dick 25,651
    Science Advisor
    Homework Helper

    You don't have to solve anything. If u and v are in the null space N={x:Ax=0} that just means Au=0 and Av=0. To show it's closed under vector addition, for example, you just need to show (u+v) is in N. That is, show A(u+v)=0. Can you do that?
     
  4. how can you show that when you are not given u or v, unless its super simple and just happens to be worth a lot of marks.
     
  5. HallsofIvy

    HallsofIvy 40,229
    Staff Emeritus
    Science Advisor

    No, you can't set the scalars to 0, they are given in "A". What you want to do is the other way around: show that the 0 vector satisfies the equation A0= 0. Since A is a linear transformation, what do you know about linear transformations? In particular, what is A0?
    (Think about A(v+ 0)= Av by definition of "0" and use the distributive law.)

    You want to show that if u and v are in the space, (that is, they satisfy Au= 0, Av= 0) then au+ bv is also in the space, for any numbers a and b (What is A(au+ bv)?).

    DON'T think about "row echelon form" or anything like that. Just use the properties of linear transformations (matrix multiplication) themselves: A(u+ v)= Au+ Av and A(bu)= b(Au).

    Finally, u and v are not "vector columns of the matrix A". They are any vectors of the correct dimension to be multiplied by A
     
  6. thanks mate
     
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