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Homework Help: Linear alegbra - transpose

  1. Jul 19, 2012 #1
    1. The problem statement, all variables and given/known data



    here is the answer:


    3. The attempt at a solution

    I can't figure out how the matrix listed above in the answer is supposed to add up to -1. that's the only way that a and b can equal each other, that is, if they both add up to -1.
  2. jcsd
  3. Jul 19, 2012 #2
    What precisely do you mean by this? What matrices add up to (summing the elements?) is not something I would think of as a meaningful quantity.

    Are you disputing that [itex]u_1 \cdot v_1 = -1[/itex]? That seems to be true. [itex]2 - 1 - 2 = -1[/itex], after all. When you transpose [itex]u_1[/itex] and multiply it by [itex]v_1[/itex] on the right, you get the same answer. Some people probably take this method of transposing and multiplying as a definition of the inner product of vectors precisely because it sums and multiplies corresponding elements in this manner.

    Finally, the matrix might be referred to as the outer product of the vectors. Notice that the diagonal elements consist of the same terms you get when you compute the dot product, and they too sum to [itex]-1[/itex].
  4. Jul 19, 2012 #3

    I thought you get

    [2 1 -1]
    [-2 -1 1]
    [4 2 -2]

    Even if you don't, then the book has that listed as an answer. what does it refer to?
  5. Jul 19, 2012 #4
    The book has listed the answers to parts a-c and the extra question at the end of the line all in one statement. That is, what they have may be more clearly written as

    Answer to (a): -1
    Answer to (b): [-1]
    Answer to (c): (the matrix that you see)
    Answer to "Which two are equal?": (a) and (b)

    It may be hard to parse what they wrote, but this is sadly pretty common.

    When you transpose u and multiply by v, you get

    [tex]u_1^T v_1 = \begin{bmatrix} 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ -1\end{bmatrix}[/tex]

    Check your understanding of matrix multiplication to see that the final dimensions must be 1x1, not 3x3, for this case.
  6. Jul 19, 2012 #5
    What does that even refer to? If the final dimensions of uT are 1 x 1, then what does that 3 X 3 matrix even refer to?
  7. Jul 19, 2012 #6
    [itex]u^T[/itex] is not 1x1. [itex]u^T v[/itex] is. [itex]u^T[/itex] is 1 row by 3 columns.

    The 3x3 matrix is the result of [itex]u v^T[/itex], a 3 row, 1 column matrix times a 1 row, 3 column matrix, or [3x1][1x3]. Usually one is taught that the "inner" dimensions must match, and the result has the dimensions of those that are left over. In this case, the 1s match, 3x3 is what's left over.
  8. Jul 19, 2012 #7
    well if [itex]u^T v[/itex] is a 3x3 matrix then how can it add up to -1 so that it equals u*v which also = -1?
  9. Jul 19, 2012 #8
    [itex]u[/itex] and [itex]v[/itex] are both 3x1.

    [itex]u^T v[/itex] is 1x1 and is equal to [-1].

    [itex]uv^T[/itex] is not the same thing. It is the 3x3 matrix you've been seeing. There is no reason for it to add up to -1. It's diagonal elements, however, can, should, and do add up to -1.
  10. Jul 19, 2012 #9
    ok, i got it.
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