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LINEAR ALGEBRA: Consider 2X2 Matrices - What are the subspaces?

  1. Sep 25, 2006 #1
    Consider 2-by-2 matrices [itex]\mathbf{A} =\left( \begin{array}{cc}a & b \\c & d \\\end{array} \right) \in \mathbbm{R}^{2 X 2}[/itex]. Which of the following are subspaces of [itex]\mathbbm{R}^{2 X 2}[/itex]?

    (A) {A | c = 0}

    (B) {A | a + d = 0}

    (C) {A | ad - bc = 0}

    (D) {A | b = c}

    (E) {A | Av = 2v}, where v is some vector in [itex]R^2[/itex].

    I think that all except the last would be subspaces of the 2 X 2 set of matrices. I know that choice (C) is the determinent of the matrix, which is in the subspace of it's own matrix, right?

    Also, how would I make a case for the first four choices using the 3 "rules" that determine if a vector is a subspace of another?

    (1) {0} must be in the subset for it to be a subset.

    (2) if v is in W then a * v is in W for all real numbers a.

    (3) if u and v are in W, then u + v is in W.
    Last edited: Sep 25, 2006
  2. jcsd
  3. Sep 25, 2006 #2
    For (C), you should be able to come up with 2 matrices with zero determinant that add to a matrix with non-zero determinant.

    The set (E) obviously violates one of the rules.
  4. Sep 25, 2006 #3


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    (C) {A | ad - bc = 0}

    [tex]\mathbf{U} =\left( \begin{array}{cc}a & b \\c & d \\\end{array} \right) \in S[/tex] and
    [tex]\mathbf{V} =\left( \begin{array}{cc}e & f \\g & h \\\end{array} \right) \in S[/tex], where S is the subspace of [tex]\mathbbm{R}^{2 X 2}[/tex] described with (C).
    Then, U + V = W has to be in S.
    [tex]\mathbf{W} =\left( \begin{array}{cc}a+e & b+f \\c+g & d+h \\\end{array} \right) \in S[/tex]. Further on, [tex](a+e)(d+h)-(c+g)(b+f)=0[/tex] implies [tex]\frac{ah+ed}{bg-fc}=1[/tex], from which we obtain [tex]bg \neq fc[/tex] (1). So, we have shown that W is in the same subspace as U and V just for the matrices U and V such that (1) is true. But, it should be true for every two matrices U and V satisfying only the property stated in (C). Hence, (C) is no subspace.
    Last edited: Sep 25, 2006
  5. Sep 25, 2006 #4
    So, in order to do this "proving" stuff. A person should assign variables (presumably the set of all real numbers) to two of the types of spaces (or matrix m X n, etc.) being questioned as subspaces? Then apply the three rules to see if they are broken?

    How, exactly, do we know that the U and V matrices above do not satisfy rule number 3(if u and v are in W, then u + v is in W)?
    Last edited: Sep 25, 2006
  6. Sep 25, 2006 #5
    Radou did a proof by contradiction. Assume what you want to prove and show that it leads to a contradiction.

    The one I suggested would be a proof by counterexample.
  7. Sep 25, 2006 #6
    Only (A), (B) and (D) are subspaces?
  8. Sep 25, 2006 #7
    Yeah, you need to show that. Radou basically gave you the template for rule 3: show that for arbitrary U and V in the subset, W = U + V is also in the subset.
  9. Sep 25, 2006 #8
    For example, I can use rule 3 for case (A) like so?:

    [tex]\mathbf{U} =\left( \begin{array}{cc}a & b \\0 & d \\\end{array} \right) \in S[/tex]

    [tex]\mathbf{V} =\left( \begin{array}{cc}e & f \\0 & h \\\end{array} \right) \in S[/tex]

    [tex]\mathbf{W} =\left( \begin{array}{cc}a+e & b+f \\0+0 & d+h \\\end{array} \right) \in S[/tex]

    Does this prove (for rule 3 only) that case (A) is a subspace of [itex]R^{2 X 2}[/itex]?


    For rule 2 -

    [tex]a\,\left( \begin{array}{cc}a & b \\0 & d \\\end{array} \right) \in S[/tex], where the first a is a real number.

    is true?
    Last edited: Sep 25, 2006
  10. Sep 25, 2006 #9
    "If v [itex]\in[/itex] W then av [itex]\in[/itex] W for all a [itex]\in[/itex] R."

    This is rule 2, how would I go about defineing a vector v that is in W?

  11. Sep 26, 2006 #10
    Well, simplify everything that can be simplified, and it should be obvious that you have a new element that's a member of the subset.

    Well, choose a variable name that's not going to confuse the real number multiplier with one of the matrix elements.

    Then...do the matrix arithmetic. That's basically all you do to check rules (2) & (3): do the matrix arithmetic for the particular subspace rule, simplify whatever can be simplified, then check to see if the resulting matrix is a member of the subset according to the definition (which, if it's not obvious by inspection, is just more matrix arithmetic). For rule (1), you just need to check that the 0 matrix is a legal member of the subset.
  12. Sep 26, 2006 #11
    The vectors in this space are 2x2 matrices, so just construct an arbitrary 2x2 matrix like V above.
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