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Homework Help: Linear Algebra questions

  1. Feb 9, 2005 #1
    Suppose the last column of AB is zero, but B has no zero columns itself. What do we know about A?

    So, Ab(sub)n = 0, but b(sub)n != 0. I have the answer, but I want to make sure I understand it. Going back to Linear Independence section I see that an indexed set of vectors {v1, ... vp} in Rn is linearly dependent if there exist weights c1..., cp, not all zero such that c1v1 + .. + cpvp = 0

    So, I figure, plug in 'bn' for 'cp' and make the set {v1, ... vp} be the columns of A, and you would have A is Linearly dependent if B has columns, not all zero, that make a1v1 + ... anvn = 0.

    So the proof then is that since we know that B has columns not all zero (in fact, it has no zero columns at all), then A must be linearly dependent.

    Is this the right way to figure this? They give the right answer, but not the logic in the answer key. Furthermore, is there a way to remember this type of thing? Is there some logical way to prove this for myself aside from just knowing the theorum?

    -------------------------------
    another question I don't get :

    Let A = [2,5;-3,1] and B = [4,-5;3,k]. What value of k, if any, will make AB = BA.

    I got as far as figuring that 6-3k = -9 and 5k-10 = 15. So, solving for each, I get k = 5 and k = -1. But, the answer is k-5. Is the wording of the question wrong, or what is my problem here? Is there a simple algebra trick to this?

    Thanks!
    Angela.
     
  2. jcsd
  3. Feb 9, 2005 #2

    Galileo

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    Your first answer is correct. Ab_n is the last column of AB and is linear combination of the column vectors of A. Since b_n is not zero, the columns of A are dependent.

    For the second question. Solving 6-3k=-9 correctly gives k=5.
    Likewise 5k-10=15 also yields k=5.
     
  4. Feb 9, 2005 #3
    doh, nevermind!! k does equal 5 (lord help me, i'm algebra deficient! lol!)

    ---
    I understand that A is dependent if it has non-pivot columns, and I guess that if a column is zero it obviously isn't a pivot column and therefore has a free variable and is dependent. I guess that would be a better way to remember for me, is that right?

    Thanks!! :)
    Angela.
     
    Last edited: Feb 9, 2005
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