No, you are not.uandvrefer tovectors, not to their components. And it is not necessary for a vector space to be given a specific basis in order to talk about vectors in that space or about a linear transformation from one space to another.

IfI write the linear transformation as a matrix, and write, say,

,thenI am thinking of as representing the vector where and are basis vectors for the two dimensional space U and as representing the vector where and are basis vectors for the two dimensional vector space V.

It is only when we choose to represent vectors and linear transformations as matrices that the components in a specific basis become important.

I’m getting a little confused sinceI don't see how you get that. I could, for example, take A to be theT(u) =vrefers tou,vas vectors and not their coordinates, but thenAu=vwouldn’t make sense in a polynomial space where, for example,u=αx²+βx+γ, unlessu,vrefers to thecoordinatesin column form.derivativeoperator on the space of polynomials of degree 2 or less: . In that u and v are vectors: and .

If I choose to write A in matrix form, taking as basis , in that order,thenI would think of the polynomial as represented by the column vector - notice thatorderof the vectors in the basis is also important- if I had taken the basis to be the same 3 functions but in order , that same polynomial would be represented by .

Using the first order, This linear transformation would be represented as

Well, we don'tIf so, does this mean that given a vector spaceVover a fieldK,we are essentially using elements ofKas entries in a coordinate system with respect to the basis ofVs.t. each distinct element inVhas a unique coordinate representation or equivalently, is a unique L.C. of basis vectors?haveto represent matrices as columns or rows of numbers but wecan.

Yes, this is true! The fact that vectors in any n dimensional vector space over field K can be represented, upon choice of a basis, as ordered n-tuples of elements of K means that all vector spaces of the sameSo does this mean that all spaces over the same field with the same dimension are isomorphic? Since, for example, the row vector (α,β,γ) inK³ and the polynomialαx²+βx+γinK[x]≤2 both have the same coordinates (namely (α,β,γ)) and so a L.T. to coordinates (α’,β’,γ’) would be represented by the same matrix for both spaces since the matrix only transforms onecoordinatesystem to another while applying a given L.T. on the way.

Thanks very much.finitedimension, over the same field, are isomorphic.

You can demonstrate the isomorphism by choosing a specific basis, in a specific order, for each vector space and constructing a function that maps each basis vector into the corresponding basis in the other space, then extending it to other vectors "by linearity" .

Notice the restriction tofinitedimensional spaces. Infinite dimensional spaces (such as the space of all polynomials or the space of all continuous functions) even if their bases have the same cardinality, over the same field, are not necessarily isomorphic. "Linear Algebra" normally restricts itself to finite dimensional vector spaces (Halmos' famous text on Linear Algebra wastitled"Finite Dimensional Vector Spaces") while infinite dimensional vector spaces,typically function spaces, are dealt with in "Functional Analysis".