1. Not finding help here? Sign up for a free 30min 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!

Linear algebra. Problem with vector spaces dimension

  1. Nov 6, 2009 #1
    1. The problem statement, all variables and given/known data
    First of all sorry if my terminology sounds a bit weird, i have never studied mathematics in english before.

    So this is the problem: We have the space R^2x2 of all the tables with numbers in R. We also have a subspace V of R^2x2 of all the tables with the following property: "If x,y is the first row and z,w the second row then: 3x+8y+5z+w=0. The main question is what is the dimV? There are more questions but i think i can solve them if I know dimV!!
    //EDIT: I will add a few more questions that i see i have a hard time solving even if I take into account that dimV=3
    a) How to prove that dimV=3(i believe it is 3 because the subspace of the solution set of the linear system I provided is 3dimensonial)
    b) Is there any other subspace of R^2x2 different BUT isomorphic with V
    c) If D a subspace of R^2x2 with dimD=4 then D "contains" V
    d) Is there any other subspace of R^2x2, isoporphic to V, that intersected with V has only one element, the zero element.

    2. Relevant equations



    3. The attempt at a solution

    a)I can see that with the given equation not all of the variables are linear independent. This makes me think that dimV=3 but i cant figure out a way to prove it!
    b)I believe the answer is YES. In a previous homework there was the same question about 2-dimensional subspaces and I replied yes because I can think of 2 planes crossing. I am not sure if 3 dimensional spaces can "cross" though.
    c) Not a clue :P

    Thanks in advance! (the deadline is tommorow Saturday at 24:00 so i would appreciate a quick answer!)
     
    Last edited: Nov 6, 2009
  2. jcsd
  3. Nov 6, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The correct word is "matrices", not "tables". "Arrays" and "tables" do not have the addition and multiplication operations that matrices do and addition and scalar multiplication are necessary for a vector space.

    The given vector space of 2 by 2 matrices is, of course, 4 dimensional and, generally, one equation restricting the components reduces the dimension by 1. More precisely, you can solve the given equation for one of the components, say w= -3x- 8y- 5z. Then you can write the matrices as
    [tex]\begin{bmatrix}x & y \\ z & w\end{bmatrix}= \begin{bmatrix}x & y \\ z & -3x- 8y- 5z\end{bmatrix}[/tex]
    [tex]= \begin{bmatrix}x & 0 \\ 0 & -3x\end{bmatrix}+ \begin{bmatrix}0 & y \\ 0 & -8y\end{bmatrix}+ \begin{bmatrix}0 & 0 \\ z & -5z\end{bmatrix}[/tex]
    [tex]= x\begin{bmatrix}1 & 0 \\ 0 & -3\end{bmatrix}+ y\begin{bmatrix}0 & 1 \\ 0 & -8\end{bmatrix}+ z\begin{bmatrix}0 & 0 \\ 1 & -5\end{bmatrix}[/tex]

    All subspaces of the same dimension are isomorphic. Try just moving the components around.

    Unless I have completely misunderstood what "R^2x2", the only subspace of dimension 4 is R^2x2 itself.

    If V has dimension 3 than any subspace that intersects V only at the zero element must have dimension 1.

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Linear algebra. Problem with vector spaces dimension
  1. Linear algebra: Vectors (Replies: 16)

Loading...