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Linear algebra: Prove that the set is a subspace

  1. Feb 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Let [itex]U[/itex] is the set of all commuting matrices with matrix [itex]A= \begin{bmatrix}
    2 & 0 & 1 \\
    0 & 1 & 1 \\
    3 & 0 & 4 \\
    \end{bmatrix}[/itex]. Prove that [itex]U[/itex] is the subspace of [itex]\mathbb{M_{3\times 3}}[/itex] (space of matrices [itex]3\times 3[/itex]). Check if it contains [itex]span\{I,A,A^2,...\}[/itex]. Find the dimension and a basis for [itex]U[/itex] and [itex]span\{I,A,A^2,...\}[/itex].

    2. Relevant equations
    -Commuting matrices
    -Subspaces
    -Vector space span
    Basis and dimension

    3. The attempt at a solution

    [itex]U[/itex] can be defined as [itex]U=\{B\in\mathbb{M_{3\times 3}}: AB=BA\}[/itex].

    Letting [itex]B=\begin{bmatrix}
    a & b & c \\
    d & e & f \\
    g & h & i \\
    \end{bmatrix}[/itex] and solving the equation [itex]AB=BA[/itex] gives [itex]B=\begin{bmatrix}
    i-\frac{2}{3}g & 0 & \frac{1}{3}g \\
    g-3f & i-3g & f \\
    g & 0 & i \\
    \end{bmatrix}[/itex].

    [itex]U[/itex] is a subspace of [itex]\mathbb{M_{3\times3}}[/itex] if [itex]\forall u_1,u_2\in U\Rightarrow u_1+u_2\in U,\forall t\in\mathbb{R}\Rightarrow tu_1\in U[/itex] which is correct.

    It is easy to check that if [itex]C\in span\{I,A,A^2,...\}\Rightarrow C\in U[/itex]:
    Linear combination for C is
    [itex]C=c_0I+c_1A+c_2A^2+...\Rightarrow CA=AC\Rightarrow C\in U[/itex]

    [itex]U[/itex] has the dimension [itex]3[/itex] and a basis are column vectors of identity matrix [itex]3\times 3[/itex].

    How to find the dimension and a basis for [itex]span\{I,A,A^2,...\}[/itex]?

     
  2. jcsd
  3. Feb 9, 2016 #2
    What does it mean to be a basis?

    I don't agree with your explanation of the dimension and basis of U. Column vectors (in this case) are ##3x1## matrices. You're claiming that you can generate a whole space of 3x3 matrices with linear combinations of 3x1 matrices. The basis that you are claiming is not even a subset of U.

    Also, how are you getting B?
     
    Last edited: Feb 9, 2016
  4. Feb 9, 2016 #3
    I made a mistake in the statement for the dimension and a basis of [itex]U[/itex]. You could look at [itex]9\times 1[/itex] vectors.
    But what is the way of algebraically finding a basis (dimension follows) of [itex]U[/itex] and [itex]span\{I,A,A^2,...\}[/itex]?

    Matrix [itex]B[/itex] can be reduced from [itex]9[/itex] to [itex]3[/itex] variables after solving the equation [itex]AB=BA[/itex] (too long to post).
     
  5. Feb 9, 2016 #4
    Your basis has to be a set of 3x3 matrices.

    That is fair, I'm just wondering the method you used.

    You've (assuming your calculations are correct) found what matrices in the subspace U "look like". Can you write that matrix as a linear combination of other matrices? (I haven't verified, but it seems like the ##span\{I,A,A^2, \dots\}## would not have been given to you just to show that it was contained in U.
     
  6. Feb 9, 2016 #5
    I think that it is not possible to form a linear combination on [itex]span\{I,A,A^2,...\}[/itex] because this set is not finite - defined (you could arbitrarily choose that [itex]span\{I,A,A^2,A^3\}[/itex] - this would be finite set).
     
  7. Feb 9, 2016 #6

    Samy_A

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    What can you say about the dimension of [itex]span\{I,A,A^2,...\}[/itex]?
     
  8. Feb 9, 2016 #7
    I don't follow what you are trying to say here.

    I'm sorry, I must be missing something.
     
  9. Feb 9, 2016 #8

    pasmith

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    Homework Helper

    (a) The space spanned by an infinite set can nevertheless be finite-dimensional; as a trivial example, consider [itex]\mathrm{span}\{ (x,0) : 0 < x \leq 1\}[/itex] which is just the one-dimensional subspace of [itex]\mathbb{R}^2[/itex] spanned by (1,0).

    (b) In this case, apply the Cayley-Hamilton theorem.
     
  10. Feb 10, 2016 #9

    HallsofIvy

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    A linear combination on an infinite set is defined to be a linear combination on a finite subset so your objection in post 5 is not relevant.
     
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