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Math Help - matrix multiplication help?

  1. #1
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    matrix multiplication help?

    Let ej denote the jth unit column that contains a 1 in the jth
    position and zeros everywhere else. For a general matrix Ann, describe the following products. (a) Aej (c) eTiAej?
    Theorem:
    Rows and Columns of a Product
    Suppose that A = [aij] is m p and B = [bij] is p n.
    [AB]i∗ = Ai∗B [( ith row of AB)=( ith row of A) B]. (3.5.4)

    [AB]∗j = AB∗j [ (jth col of AB)=A ( jth col of B)]. (3.5.5)

    [AB]i∗ = ai1B1∗ + ai2B2∗ + +aipBp∗aikBk∗. (3.5.6)

    [AB]∗j = A∗1b1j + A∗2b2j + + A∗pbpjA∗kbkj (3.5.7)

    These last two equations show that rows of AB are combinations of
    rows of B, while columns of AB are combinations of columns of A.



    For parts a and c im not even sure what they are even asking for. When its saying ej is a unit column does that mean like this (1 0 0...0) as an example?
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  2. #2
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    Re: matrix multiplication help?

    Hey bonfire09.

    You are correct in saying that e^j is a unit vector where the jth element is a 1 and everything else is a zero. The ^t means transpose of that vector.

    These assume the vectors are in R^n and not some other system.
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  3. #3
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    Re: matrix multiplication help?

    When they are asking to describe the product of Aej what do they mean here? Since e^j is a unit vector that means Aej= a11B*1+a21B*2+a31B*1+..+aipBp∗ from (3.5.6)?
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  4. #4
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    Re: matrix multiplication help?

    It will basically be like a dot product and it should return the (i,j)th element of the matrix (it will return one scalar number).

    But remember that ej has the jth entry equal to 1 so you can have n different ej vectors (and ei vectors as well).
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  5. #5
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    Re: matrix multiplication help?

    the way i remember this stuff is that multiplying A on the RIGHT by ej picks out the j-th column of A, and then multiplying A on the LEFT by:

    (ej)T (that is, a unit row-vector) picks out the j-th row of A.

    now let's look at what happens with a product AB.

    the j-th column of AB only need the j-th column of B to compute (we just take the dot product of the different rows of A with that column of B to get the entries).

    if we expand this (it gets messy, but here goes):

    Bj = b1je1 +...+bnjen.

    so (AB)j = A(Bj) = A(b1je1 +...+bnjen)

    = b1jA(e1) +... + bnjA(en)

    = b1j(a11,a21,...,an1) +....+ bnj(a1n,a2n,...,ann)

    and collecting all the coordinates in the same position together:

    = (a11b1j+...+a1nbnj, a21bij+...+a2nbnj,....,an1b1j+...+annbnj)

    this is what (3.5.5) says.

    in particular, if B = ej (that is, B has only one column, which is the j-th standard unit vector) = (0,...,0,1,0,...0)

    then A(ej) = (a1j,a2j,....,anj) (only the j-th entry in the i-th row survives).

    if we want to know the i-th row of AB, we only need to use the i-th row of A.

    let me work out some examples for you:

    A = \begin{bmatrix}2&1&3\\1&4&5\\-1&0&3 \end{bmatrix};\ B = \begin{bmatrix}-2&0&1\\3&2&0\\6&6&0\end{bmatrix}.

    then AB = \begin{bmatrix}17&20&2\\40&38&1\\20&18&-1 \end{bmatrix}.

    see how the third column of AB is just the first column of A? also the second column of AB is just twice the second column of A plus 6 times the 3rd column of A:

    (20,38,18) = 2(1,4,0) + 6(3,5,3).

    in the same fashion, the third row of AB is just 3 times the third row of B minus the first row of B:

    (20,18,-1) = 3(6,6,0) - (-2,0,1).

    it should be clear that, for example:

    \begin{bmatrix}2&1&3\\1&4&5\\-1&0&3 \end{bmatrix} \begin{bmatrix}0\\1\\0 \end{bmatrix} = \begin{bmatrix}1\\4\\0 \end{bmatrix}

    -we just write down the second column of A.

    can you think of a shortcut for writing down:

    \begin{bmatrix}0&1&0 \end{bmatrix} \begin{bmatrix}2&1&3\\1&4&5\\-1&0&3 \end{bmatrix}?
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  6. #6
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    Re: matrix multiplication help?

    it be [1 4 5]. So e^TA is represented by Ai*.
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  7. #7
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    Re: matrix multiplication help?

    so, continuing this line of reasoning, what do you get when you take the j-th column of a matrix that is the i-th row of A (a row is a 1xn matrix)?

    is this the same as the i-th row of the (nx1 matrix) the j-th column of A?
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  8. #8
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    Re: matrix multiplication help?

    I think your trying to say [0 1 0]A^T then it be [1 4 0] which be one of the rows of A^T. Well it does form the second column of Matrix A. Is there also a way to prove that Aej is equal to A*j?
    Last edited by bonfire09; November 20th 2012 at 08:13 AM.
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  9. #9
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    Re: matrix multiplication help?

    what i am asking is, what is (ei)TAej?

    we can find this two ways:

    e_i^TA = \begin{bmatrix}a_{i1}&a_{i2}&\dots&a_{in} \end{bmatrix}

    what is the j-th column of that matrix (the columns are very short)?

    or:

    Ae_j = \begin{bmatrix}a_{1j}\\a_{2j}\\ \vdots\\a_{nj}\end{bmatrix}

    what is the (rather narrow) i-th row of that?
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  10. #10
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    Re: matrix multiplication help?

    Oh I think I see it now. It just be an entry of Aej represented by anj. So (ei)TAej is just anj
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  11. #11
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    Re: matrix multiplication help?

    no, it is aij.
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  12. #12
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    Re: matrix multiplication help?

    oh ok. i was going off your notation but it makes sense.
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