Results 1 to 4 of 4

Math Help - Vector space.

  1. #1
    Member
    Joined
    Sep 2010
    Posts
    100

    Vector space.

    Let V = M_{22} the set of all 2 x 2 matrices. Let the operation of vector addition in V be the usual matrix addition but let scalar multiplication in V be defined by:

     c \cdot A = cA^{T}

    Show that V is not a vector space by demonstrating with an example that one of the properties in the definition of a vector space fails to hold.

    My attempt:
    property I believe fails to hold (let r be any real number):
     r (A+B) = r(A) + r(B)

    I'll define both A and B as the 2 x 2 matrices:
    A=\begin{bmatrix}a & b\\c & d \end{bmatrix} and B=\begin{bmatrix}e & f\\g & h \end{bmatrix}

    r\begin{bmatrix}a & b\\c & d \end{bmatrix}^{T} + r\begin{bmatrix}e & f\\g & h \end{bmatrix}^{T} => r\begin{bmatrix}a & c\\b & d \end{bmatrix} + r\begin{bmatrix}e & g\\f & h \end{bmatrix}=> \begin{bmatrix}ra & rb\\rc & rd \end{bmatrix} + \begin{bmatrix}re & rf\\rg & rh \end{bmatrix}=> \begin{bmatrix}ra+re & rb+rf\\rc+rg & rd+rh \end{bmatrix} \notin M_{22}

    Hence, V is not a vector space since it does not hold for the distributivity of scalar multiplication with respect to vector addition. Did I go about this proof correctly?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,392
    Thanks
    759

    Re: Vector space.

    no, the matrix you would up with certainly is still a 2x2 matrix.

    and, in fact, it IS true that r.(A+B) = r(A+B)^t = r(A^t + B^t) = r.A + r.B.

    how about checking if r.(s.A) = (rs).A....?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2010
    Posts
    100

    Re: Vector space.

    Quote Originally Posted by Deveno View Post
    no, the matrix you would up with certainly is still a 2x2 matrix.

    and, in fact, it IS true that r.(A+B) = r(A+B)^t = r(A^t + B^t) = r.A + r.B.

    how about checking if r.(s.A) = (rs).A....?
    OK:
    r \cdot (s \cdot A) = r \cdot (s \cdot \begin{bmatrix}a & b\\c & d \end{bmatrix}^{T}) => r \cdot (s \cdot \begin{bmatrix}a & c\\b & d \end{bmatrix}) => r \cdot \begin{bmatrix}sa & sc\\sb & sd \end{bmatrix}) =>\begin{bmatrix}rsa & rsc\\rsb & rsd \end{bmatrix}

    While: (r \cdot s) \cdot A => (r \cdot s)\begin{bmatrix}a & b\\c & d \end{bmatrix}^{T} => (r \cdot s)  \begin{bmatrix}a & c\\b & d \end{bmatrix} =>  \begin{bmatrix}ra & rc\\rb & rd \end{bmatrix} \cdot  \begin{bmatrix}sa & sc\\sb & sd \end{bmatrix} =>  \begin{bmatrix}rsa^{2} & rsc^{2}\\rsb^{2} & rsd^{2} \end{bmatrix}

    Clearly, r \cdot (s \cdot A) \neq (r \cdot s) A and therefore, V is not a vector space. Correct?

    Also, a general question: How would you approach such a question in a circumstance where you are pressed for time? I mean, in an exam, it is quite time consuming to check through all eight properties of a vector space to determine whether a set is a vector space or not. Any general or intuitive tips that you could give? Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,392
    Thanks
    759

    Re: Vector space.

    again, this is incorrect.

    r \cdot(s \cdot A) = r \cdot (sA^T) = r(sA^T)^T = r(s(A^T)^T) = r(sA) = (rs)A

    whereas:

    (rs) \cdot A = (rs)A^T

    if A =

    \begin{bmatrix}0&1\\0&0\end{bmatrix}, for example, these two matrices are NOT the same.

    (r \cdot s)A makes no sense at all, a scalar multiplication is a map F \times V \to V, not F \times F \to F
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Dual Space of a Vector Space Question
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 16th 2011, 03:02 AM
  2. Replies: 2
    Last Post: April 1st 2011, 02:40 AM
  3. Banach space with infinite vector space basis?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 24th 2011, 06:23 PM
  4. Replies: 15
    Last Post: July 23rd 2010, 11:46 AM
  5. Isomorphism beetwenn vector space and sub space
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 30th 2008, 10:05 AM

Search Tags


/mathhelpforum @mathhelpforum