Computing Matrix Representation of T in the Ordered Bases B and Y

Please see attached image for problem statement (it is letter "e").

So far, I know that to compute the Matrix Representation of T in the ordered Bases B and Y you do the following:

1. Compute T(xi) where xi denotes the standard basis vectors for V.

2. Write the vectors downwards in columns in the matrix.

However, for letter e, I'm not sure what the standard basis vectors are. I googled, and found out that a common representation might be {e1,e2,...,en}, and using this I got the answer to be a 2 by infinity matrix with {e1,e2,...,en} as both column vectors.

Can somebody clarify this for me? Thank you! :)

the i-th standard basis vector, ei is the vector (0,0,...,1,...,0) where the 1 is in the i-th place (all other entries are 0). for example, in R^3:

e1 = (1,0,0) (sometimes called the "x unit vector" or i)
e2 = (0,1,0) (somtimes called the "y unit vector" or j).
e3 = (0,0,1) (sometimes called the "z unit vector" or k), notations vary from text to text.

in the "standard" basis, the vector (a1,a2,...,an) has coordinates (a1,a2,...,an) (it really is a convenient basis to use, because of this).

so in problem (e) T(a1e1+a2e2+...+anen) = a1e1 + a1e2 +...+ a1en,

because (a1,a2,...,an) = a1(1,0,...,0) + a2(0,1,0,...,0) +....+ an(0,...,0,1) =a1e1 + a2e2 +...+ anen.

since every basis vector ej except e1 has 0 in the first coordinate, what will the 2nd through n-th columns be?

what is T(e1) = T(1,0,...,0)?

Quote:

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Deveno;683879]the i-th standard basis vector, ei is the vector (0,0,...,1,...,0) where the 1 is in the i-th place (all other entries are 0). for example, in R^3:

e1 = (1,0,0) (sometimes called the "x unit vector" or i)
e2 = (0,1,0) (somtimes called the "y unit vector" or j).
e3 = (0,0,1) (sometimes called the "z unit vector" or k), notations vary from text to text.

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I understand this now, thank you!

Quote:

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in the "standard" basis, the vector (a1,a2,...,an) has coordinates (a1,a2,...,an) (it really is a convenient basis to use, because of this).

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I'm not sure what you mean by this, could you explain a little more?

Quote:

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so in problem (e) T(a1e1+a2e2+...+anen) = a1e1 + a1e2 +...+ a1en,

because (a1,a2,...,an) = a1(1,0,...,0) + a2(0,1,0,...,0) +....+ an(0,...,0,1) =a1e1 + a2e2 +...+ anen.

since every basis vector ej except e1 has 0 in the first coordinate, what will the 2nd through n-th columns be?

what is T(e1) = T(1,0,...,0)?

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I'm assuming I need to understand the former part to answer this, so I'll wait until you respond. Thanks again for the help! :)

there are two slightly different ways of describing a vector that is an element of a vector space.

1. since a vector space is a set, you can identify a vector as a certain element of that set. for example, if V is the vector space of all real polynomials of degree 2 or less, we might specify an element of V as "the polynomial x^2 + 2".

2. the second way, is to choose a basis for V, and specify a vector as a linear combination of basis elements. so if our basis is B = {v1,v2,...,vn}, we might say the vector

v = a1v1 + a2v2 +....+anvn (here the a's are scalars, and the v's are vectors, but special ones, from our basis).

we might also say, the vector v has B-coordinates (a1,a2,...,an).

with R^n, and the standard basis, the two ways are the same. in the standard basis (i'll call it E), the n-tuple (a1,a2,..,an) has E-coordinates (a1,a2,...,an).

the reason for the distinction, is because not all vector spaces look like R^n. in the polynomial example i gave earlier, in the basis {1,x,x^2}, x^2 + 2 has coordinates (2,0,1):

2(1) + 0(x) + 1(x^2) = 2 + x^2 = x^2 + 2.

but if we choose our basis to be {1, 1-x, 1+x^2}, then we will have very different coordinates. in THIS basis, x^2 + 2 has coordinates (1,0,1):

1(1) + 0(1-x) + 1(1+x^2) = 1 + 1 + x^2 = x^2 + 2.

one often sees things in books (especially in more elementary texts) like "the vector (2,3-5)". this is very bad, what is better is:

the vector that has coordinates (2,3,-5) in the basis (such-and-such), or the vector that has coordinates (2,3-5) in the standard basis for (inset vector space name here).

coordinates reflect an arbitrary choices of axes. there is nothing special about the x,y and z axes (well, ok, there is something special, they are orthogonal, but so are a lot of other ones).

we could have evolved in a culture that used the mirror-image of these axes, or decided it was preferable to have a 2nd axis at 45 degrees to the first.

coordinates, and coordinate systems, are not unique, there is more than one way to label a particular point in space.

in the problems you are trying to do, you are in "the simplest situation", as you are being told to use the basis that is the easiest to use for computing matrices (for R^n, at any rate).

I think I understand now, please follow my reasoning below and let me know if it is correct:


1. The standard basis vectors for Rn are {e1,e2,...,en} where e1={1,0,0...} (meaning the ith place is 1 and all others are 0, although i'm wondering how to integrate the actual dimension into this without just putting "..." to make the notation cleaner and denote the actual dimension)


2. So, in our problem, T(e1)= T(1,0,0...,0)= (1,1,1,...1) and T(en) (where n does not equal 1) is (0,0,0...0) because when n is not equal to 1, the 1 is not in the a1 position, therefore our transformation will make all the elements in the new n-tuple 0.


So, [T]'s first column will be all ones, the second column and on will be all zeros. the number of columns will depend on the dimension of R.


I can write the answer out in words, but how would I write this with correct notation? There seem to be a lot of conditions involved.


Thanks again for your help! :)

it suffices to write out the matrix (using dots to indicate some unspecified number of entries), like:

[1 0 ... 0]
[1 0 ... 0]
...
[1 0 ... 0]

or just as you described in words, T has a matrix whose first column is all 1's, and 0's elsewhere. your answer is fine as it is. you have demonstrated you could (in theory, anyway) explicitly write the matrix for any value of n.

a 3rd alternative is to write:

Thanks for the clarification and your help!