matrix multiplication help?

Let **e**_{j} denote the jth unit column that contains a 1 in the jth

position and zeros everywhere else. For a general matrix A_{n×n}, describe the following products. (a) A**e**_{j }(c) **e**^{T}_{i}A**e**_{j}?

Theorem:

**Rows and Columns of a Product**

Suppose that **A = [a**_{ij}] is m × p and **B = [b**_{ij}] is p × n.

• [**AB]**_{i∗} = **A**_{i∗}**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∗} = **a**_{i1}**B**_{1∗} + **a**_{i2}**B**_{2∗} + · · · +**a**_{ip}**B**_{p∗} =Ʃ **a**_{ik}**B**_{k∗}. (3.5.6)

• [**AB]**_{∗j }= **A**_{∗1}**b**_{1j} + **A**_{∗2}**b**_{2j} + · · · + **A**_{∗p}**b**_{pj}=Ʃ **A**_{∗k}**b**_{kj} (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 **e**_{j} is a unit column does that mean like this (1 0 0...0) as an example?

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.

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)?

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).

Re: matrix multiplication help?

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

(**e**_{j})^{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):

B_{j} = b_{1j}**e**_{1} +...+b_{nj}**e**_{n}.

so (AB)_{j} = A(B_{j}) = A(b_{1j}**e**_{1} +...+b_{nj}**e**_{n})

= b_{1j}A(**e**_{1}) +... + b_{nj}A(**e**_{n})

= b_{1j}(a_{11},a_{21},...,a_{n1}) +....+ b_{nj}(a_{1n},a_{2n},...,a_{nn})

and collecting all the coordinates in the same position together:

= (a_{11}b_{1j}+...+a_{1n}b_{nj}, a_{21}b_{ij}+...+a_{2n}b_{nj},....,a_{n1}b_{1j}+...+a_{nn}b_{nj})

this is what (3.5.5) says.

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

then A(**e**_{j}) = (a_{1j},a_{2j},....,a_{nj}) (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:

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

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

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:

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

-we just write down the second column of A.

can you think of a shortcut for writing down:

$\displaystyle \begin{bmatrix}0&1&0 \end{bmatrix} \begin{bmatrix}2&1&3\\1&4&5\\-1&0&3 \end{bmatrix}$?

Re: matrix multiplication help?

it be [1 4 5]. So e^TA is represented by Ai*.

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?

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 Ae_{j is equal to A*j?}

Re: matrix multiplication help?

what i am asking is, what is (e_{i})^{T}Ae_{j}?

we can find this two ways:

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

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

or:

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

what is the (rather narrow) i-th row of that?

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

Re: matrix multiplication help?

Re: matrix multiplication help?

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