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.
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) Ae_{j }(c) e^{T}_{i}Ae_{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?
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.
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).
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:
.
then .
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:
-we just write down the second column of A.
can you think of a shortcut for writing down:
?
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?
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?}