# Images of Basis and the Representative Matrix

When I have a linear operator $T:V\rightarrow W$ and images for some ordered basis $B=(b_1,\dots,b_n)$ of $V$, what do I have exactly if I put those images in a matrix? Is it by default:

$[T]^B_E = \begin{bmatrix} T(b_1) & \dots & T(b_n) \end{bmatrix}^B_E$

where the output is according to the standard basis of $W$? What are those images if I don't take their coordinate vectors?

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If you have a basis for $W$ (and that this basis is $E$), then the matrix is represented as you have shown, yes. (I am a little tired so I am not sure if the $B$ or the $E$ goes below the other though. If you are sure that they are okay, don't mind me ; I was just telling myself I don't remember which goes where.) Note that "there are no standard basis in $W$". There is no such thing as a "standard basis" in an abstract vector space. You only have such luxury in $K^n$ in general when $K$ is a field (or $\mathbb R^n$, if you like concrete examples better). –  Patrick Da Silva Nov 8 '12 at 9:13
You should get used to thinking about vectors as things that exist without necessarily being expressed numerically in a basis. Just like an object weighs independently of your choosing a unit to measure its weight by. Vectors could be complicated objects like functions satisfying some differential equation, or whatever. You don't know or need to know this; the power of linear algebra is that you can used the same techniques independently of what vectors actually are. –  Marc van Leeuwen Nov 8 '12 at 9:59
What do you mean by "put those images in a matrix"? Those images may not be column vectors. They are just elements of some vector space $W$. They might be polynomials or symmetric matrices or something strange like that; you might not be able to just put them into a matrix. –  littleO Nov 8 '12 at 10:01
@Robert : I'm not sure I explicitly understand your last question, but to answer the other one, when you look at examples of abstract vector spaces, you can understand what it means to have "no standard basis". For instance, take $V$ to be the vector space of functions from $[0,1]$ to $\mathbb R$ that are of the form $ax+b$ on the interval $[0,1/2]$, of the form $cx+d$ on the interval $[1/2,1]$, and are continuous (i.e. the endpoints match at $1/2$). Can you think of a "standard basis" for this vector space?... This is pretty much the point. –  Patrick Da Silva Nov 8 '12 at 10:14
@PatrickDaSilva What I'm talking about is a method for solving problems of this type: math.stackexchange.com/questions/183077/… –  Robert S. Barnes Nov 8 '12 at 11:17

We can define '$n$ dimension vectors' over any set $S$, as the set of all $n$-tuples formed of $S$. $$S^n:=\{[ s_1,...,s_n ]\mid s_i\in S\}$$ Then, what you are asking, $[T(b_1),...,T(b_n)]$ is none other than a '(row) vector over $W$', just an $n$-tuple of elements of $W$. It does make sense. Then, expanding each $T(b_i)$ to coordinate (column-)vectors by the given basis $E$ of $W$, would give the ordinary matrix $[T]^B_E$.