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Homework Help: Vector space dimension question

  1. Oct 7, 2011 #1
    Prove that the space of 2×2 real matrices forms a vector space of dimension 4 over
    R. [12]

    Im unsure, anyone any idea?
  2. jcsd
  3. Oct 7, 2011 #2
    If I understood right, we have to find an isomorphism from [itex]M^{R}_{2x2}[/itex] to [itex]R^4[/itex].

    In other word, we need to find a linear transformation, [itex]T:M^{R}_{2x2} \rightarrow R^4[/itex] that [itex]ker(T)=0[/itex].

    HINT: look at T[a,b;c,d]=(a,b,c,d)
  4. Oct 7, 2011 #3


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    What's your book's definition of "dimension 4". I use the definition that says that a vector space is 4-dimensional if it contains a linearly independent set with 4 members, but no linearly independent set with 5 members. So to solve your problem, I would find a set of 4 linearly independent matrices and prove that every matrix is a linear combination of those four.
  5. Oct 7, 2011 #4
    So what does ker(T) stand for? Im still really clueless. So do I find an isomorphism from T[a,b;c,d] to (a,b,c,d)?
  6. Oct 7, 2011 #5


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    ker T is the set of all x such that Tx=0. Finding that isomorphism is one way to do it, but not the only one. Edit: Wait, you said "from T[a,b;c,d]". I don't know what you mean by that. I meant that one way to solve the problem is to find an isomorphism from the set of 2×2 matrices to ℝ4.
  7. Oct 7, 2011 #6
    I guess I wasn't much clear. What I meant is that T takes a matrix [a,b;c,d] and gives a vector (a,b,c,d). Actually, I think that this is what he has meant.
  8. Oct 7, 2011 #7
    A = (a b c d)

    A = a (1 0 0 0) + b (0 1 0 0) + c (0 0 1 0) + d (0 0 0 1)

    Please excuse my lack of ability to lay this out correctly but each of these in brackets have 2 rows and 2 columns i.e 2 x 2. not 1 x4

    a,b... Eℝ is a vector space over ℝ

    Is this the correct approach?
  9. Oct 7, 2011 #8
    it seems like you know the answer but don't know how to do this.
    so first of all, lets state that "in order to prove that the space of 2×2 real matrices forms a vector space of dimension 4 over R. we will define an isomorphism from 2x2 real matrices space to R4". Later, we will say. "lets take a look at the following transformation T:M2x2->R4, T([a,b;c,d])=(a,b,c,d)"
    now, what is left to do is to show that it is linear transformation and that it is a isomorphism.

    regarding the linear transformation, it's not a big deal. i am sure you can handle it by yourself.
    to show that T is isomorphism we need to prove that T is a linear transformation and ker(T)=0 or im(T)=4.

    to do that - we need to know first what im(T) is equal. if we take a base of one space and we transform it using T we will get the base of the other space we are transforming to. so lets take the standard base of M_2x2(R). { [1,0;0,0], [0,1;0,0], [0,0;1,0], [0,0;0,1]}. from what we have defined before:
    T([1,0;0,0])=(1,0,0,0); T([0,1;0,0])=(0,1,0,0) ... T([0,0;0,1])=(0,0,0,1).
    in other terms, imT(M_2x2(R) ) = sp{(1,0,0,0),...,(0,0,0,1)}, but the set {(1,0,0,0)...(0,0,0,1)} is the standard base of R4 thus dim(imT)=4. so in conclusion, T is an isomorphism, and we proved what we have asked for.
    Last edited: Oct 7, 2011
  10. Oct 7, 2011 #9


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    Any 2 by 2 matrix is of the form
    [tex]\begin{bmatrix}a & b \\ c & d \end{bmatrix}= a\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}+ b\begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}+ c\begin{bmatrix}0 & 0 \\ 1 & 0 \end{bmatrix}+ d\begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix}[/tex]
    The most "basic" definition of "dimension" is that it is the number of vectors in any space. It is easy to see that these four matrices are a basis.

    Or, if as Boaz does, you wish to find an isomorphism,
    [tex]f\left(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\right)= (a, b, c, d)[/tex]
    works nicely.
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