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[Linear Algebra] Subspace Proof

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data

    I have a matrix and need to show that it is a subspace of ℝn using the eigenspace identity of: Ax = λx, where x is the eigenvector.

    2. Relevant equations



    3. The attempt at a solution

    I know that for a subspace, you need to show that it holds under addition, scalar multiplication, and that the zero vector is present. But I don't know how to get started on this... any hints? Thanks!
     
  2. jcsd
  3. Dec 8, 2011 #2

    Mark44

    Staff: Mentor

    I don't believe that your problem description is an accurate depiction of what you're supposed to do. Here's what I think that problem actually is:

    For a given n x n matrix A, with a given eigenvalue λ, show that the solutions of the equation Ax = λx form a subspace of Rn. In other words, show that the eigenvectors associated with the eigenvalue λ form a subspace of Rn. ​
     
  4. Dec 8, 2011 #3

    Mark44

    Staff: Mentor

    "it" = what? Math students should never use the word "it" unless the antecedent is obvious to the most casual observer.
     
  5. Dec 8, 2011 #4
    yes this is correct. . .
     
  6. Dec 8, 2011 #5
    i think what im trying to say is that the set of nxn matricies is a subset. . .
     
  7. Dec 8, 2011 #6

    Mark44

    Staff: Mentor

    See? Even you're confused. This really has almost nothing to do with any set of n x n matrices - just one particular matrix, and the set of vectors that are the eigenvectors associated with one eigenvalue of that matrix.

    Your sentence with "it" removed is:
    I know that to show that this set of vectors is a subspace, you need to show that [STRIKE]it holds under addition[/STRIKE] this set is closed under vector addition and scalar multiplication, and that the zero vector is present.

    So you need to check 3 things.
    1) Is the 0 vector in this set? I.e., is A0 = λ0 a true statement.
    2) Is the set closed under vector addition?
    3) Is the set closed under scalar multiplication?
     
  8. Dec 8, 2011 #7
    i am confused, and this is a lot more clear to me. . .

    when i go to show these, do i pick arbitrary real values and show that they hold?
     
  9. Dec 8, 2011 #8

    Mark44

    Staff: Mentor

    The first item is easy to show, so I'll skip it.
    For #2, take arbitrary vectors that belong to the set and show that their sum is also in the set.
    For #3, take an arbitrary vector u and an arbitrary real number c, and show that cx is in the set.

    You don't get to choose vectors or scalars. What you are showing is that no matter which vectors are chosen, and which scalar, u + v is in the set and cu is in the set.

    "in the set" - what is it that defines every vector in the set you're working with? How can you tell if a vector is in the set or not?
     
  10. Dec 8, 2011 #9
    thanks! this really helped a lot to make sense of things. . .
     
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