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Subspace of a matrix

  1. Jul 12, 2009 #1

    if anybody can help me

    Let S be the matrix a b
    c d

    with a constrain a = -2d, b= 3c -d

    Prove that S is a subspace of R2?
  2. jcsd
  3. Jul 12, 2009 #2

    a b
    c d
  4. Jul 12, 2009 #3
    It looks like you have misread the problem.

    A matrix cannot be a subspace. Even a set of matrices cannot be a subspace of R^2, because R^2 consists of points, not matrices.
  5. Jul 13, 2009 #4


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    As Tac-Tics said, you must have misread the problem. You can't "prove that S is a subspace of R2", it isn't true. A matrix is NOT a subspace of R2.

    It is possible that you were asked to show that the set of all matrices of the form
    [tex]\begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}[/tex]
    is a subspace of the M(2,2), the vector space of all 2 by 2 matrices. You would do that by showing that the set is closed under addition and scalar multiplication. That is, if
    [tex]M= \begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}[/tex]
    [tex]N= \begin{bmatrix}-2y & 3x- y \\ x & y\end{bmatrix}[/tex]
    are matrices in this set and "a" is a number
    [tex]M+ N= \begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}+ \begin{bmatrix}-2y & 3x- y \\ x & y\end{bmatrix}[/tex]
    [tex]aM= a\begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}[/tex]
    also in that set?
    Last edited by a moderator: Jul 13, 2009
  6. Jul 13, 2009 #5
    Maybe there is a condition on matrix and it is asked to consider the set of (c,d) pairs. Then RxR makes sense seemingly...
  7. Jul 13, 2009 #6
    Yes, you are right HallsofIvy. how i profe that they r close under multiplication and addition. do i have to add and them factorise to go back to the original?
  8. Jul 13, 2009 #7
    thank you HallsofIvy you were right. I read wrong the question. it is if it is a subspace of the M(2,2), the vector space of all 2 by 2 matricess.

    How to i proof that they are close under adittion and multiplication and if i consider the set B = |B1 B2|
    B1 = 0 -6
    -2 0

    B2 = 2 4
    1 -1

    find if B is bases of the first question?
  9. Jul 14, 2009 #8


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    I thought I had responded to this but I don't find it now.

    You have the set of all 2 by 2 matrices of the form
    [tex]\begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}[/tex]
    another such is
    [tex]\begin{bmatrix}-2y & 3x- y \\ x & y\end{bmatrix}[/tex]
    Their sum is
    [tex]\begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}+\begin{bmatrix}-2y & 3x- y \\ x & y\end{bmatrix}[/tex]
    [tex]= \begin{bmatrix}-2d-2y & 3c- d+ 3x- y \\ c+ x & d+ y\end{bmatrix}= \begin{bmatrix}-2(d+y) & 3(c+x)- (c+y)\\ c+ x & d+y \end{bmatrix}[/tex]
    Do you see how that final matrix also satisifies the definition of this set of matrices?

    If a is any number then
    [tex]a\begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}= \begin{bmatrix}-2ad & 3ac- ad \\ ac & ad\end{bmatrix}[/tex]
    Again, do you see how this matrix satisfies the condition to be in the set?

    To show that
    [tex]\begin{bmatrix}0 & -6 \\ -2 & 0\end{bmatrix}[/tex]
    [tex]\begin{bmatrix}2 & 4 \\ 1 & -1\end{bmatrix}[/tex]
    form a basis for the subspace, you must show that they satify the conditions for a basis: that they are independent and that they span the subspace.
    [tex]A\begin{bmatrix}0 & -6 \\ -2 & 0\end{bmatrix}+ B\begin{bmatrix}2 & 4 \\ 1 & -1\end{bmatrix}= \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}[/tex]
    Setting the matrices on the two sides of that equation equal gives you four equations to solve for A and B. If the only solution is A= B= 0, the matrices are independent

    Then, given any numbers, c, d so that
    [tex]\begin{bmatrix}-2d & 3c- d \\ c & d\end{bmatrix}[/tex]
    is in the set, you must show that there exist numbers A and B such that
    [tex]A\begin{bmatrix}0 & -6 \\ -2 & 0\end{bmatrix}+ B\begin{bmatrix}2 & 4 \\ 1 & -1\end{bmatrix}= \begin{bmatrix}-2d & 3c-d \\ c & d\end{bmatrix}[/tex]
    Again, you set the two matrices on either side of the equation equal so you get 4 equations for A and B. Here, you must only show that those equations have a solution no matter what c and d are.
  10. Jul 14, 2009 #9
    thank you very much i got it
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