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Math Help - Show that the following matrices form a base for M22

  1. #1
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    Show that the following matrices form a base for M22

    Show that the following matrices form a base for M22.
    [3 6] , [0 -1], [0 -8] ,,, [1 0]
    [3 -6], [-1 0], [-12 -4], [-1 2]

    c1M1+c2M2+c3M3+c4M4=B

    c1M1+c2M2+c3M3+c4M4=
    [0 0]
    [0 0]

    c1M1+c2M2+c3M3+c4M4=
    [a b]
    [c d]



    After addition I get:
    [1 0]
    [0 1]

    [c1 0] = [a b]
    [0 c4] = [c d]

    a=d
    b=c

    Not linearly independent, but it should be.
    Thanks for your help.
    Last edited by algie; March 11th 2011 at 06:15 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by algie View Post
    Show that the following matrices form a base for M22.
    [3 6] , [0 -1], [0 -8] ,,, [1 0]
    [3 -6], [-1 0], [-12 -4], [-1 2]

    You can work with coordinates on the canonical basis B of M_{2\times 2}(\mathbb{R}):

    \left[\begin{pmatrix}{a}&{b}\\{c}&{b}\end{pmatrix}\right]_B =(a,b,c,d)

    As \dim M_{2\times 2}(\mathbb{R})=4 you only need to prove that the rank of A is 4 being A the matrix whose rows are the corresponding coordinates of M_i with respect to B .
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  3. #3
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    Actually I do not know what rank is, I read about it just now and from my understanding the rank of a 2x2 matrix cannot be higher than 2.
    I still do not understand, sorry.

    In my book they have a really simple example that prove The standard basis for M_mn.

    Example)
    Show that
    M1=e1=
    [1 0]
    [0 1]
    M2=e2
    M3=e3
    M4=e4
    form a basis for the vector space M_22.

    Solution:
    1. Prove linear independence. / Show that c1M1+c2M2+c3M3+c3M4=0 has only the trivial solution.
    2. Show that every 2x2 matrix
    B=
    [a b]
    [c d]
    can be expresesd as c1M1+c2M2+c3M3+c4M4=B.

    In the end they have
    [c1 c2]=[0 0]
    [c3 c4] [0 0]
    and
    [c1 c2]=[a b]
    [c3 c4] [c d]

    First eq. has c1=c2=c3=c4=0, has only solution 0, linear independence proven.
    Second eq. c1=a, c2=b, c3=c, c4=d, conclusion; they span M_22, and form a basis of M_22.


    My problem is, after addition and row operations on c1M1+c2M2+c3M3+c3M4 I get
    [1 0]
    [0 1]
    and since I get zeroes, I do not understand if this is still a basis of M_22.

    I mean, if this forms a base for M_22, then I might as well get anything in that matrix and say it form the basis of M_22.
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  4. #4
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    I am sorry I cannot understand everything that is said, I failed this exam and really try to understand now!

    I did it in another way but get another, "wrong?" answer again.

    Create a matrix of c1 c2 c3 c4, they span M_22.
    [3 0 0 1]
    [6 -1 -8 0]
    [3 -1 -12 -1]
    [-6 0 -4 2]

    After row operations I get
    [1 0 0 0]
    [0 0 1 0]
    [0 0 0 1]
    [0 1 0 0]
    c1=c2=c3=c4=0
    Is linear independent.

    But still the same since
    c1[1 0]+c2[0 0]+c3[0 1]+c4[0 0]
    ...[0 0].....[0 1]......[0 0].....[1 0]
    equals
    [c1 0]
    [0 0]

    I really do not know what I try to do, might as well skip this problem
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