2
$\begingroup$

There is a section in Sheldon Axler's Linear Algebra Done Right (pg.40), where it says:

"Linear maps can be constructed that take on arbitrary values on a basis. Specifically, given a basis $(v_1,...,v_n)$ of V and any choice of vectors $w_1,...,w_n \in W$, we can construct a linear map $T: V \to W$ such that $Tv_j = w_j$ for $j=1,...,n$. There is no choice how to do this - we must define T by $T(a_1v_1 + ...+a_nv_n) = a_1w_1+....+a_nw_n$, where $a_1,...,a_n \in \mathbb{F}$."

I have been confused with this for a VERY LONG time.

1) Specifically, do all linear maps must have an origination vector $v$ that is written with a basis? Meaning, do I have to always use a basis for mapping any $v \in V$?

2) Is the construction of the linear map such that $Tv_j = w_j$ for $j=1,...,n$ just a specific example of a way to construct a linear map or does this condition hold true for ALL linear maps?

3) Why do we need the coefficients $a_1,...,a_n \in \mathbb{F}$?

Overall, I just do not get the general picture of why the author included this description and how it fits into my understanding of linear maps.

Thank you everyone!

$\endgroup$
  • 1
    $\begingroup$ Assigning arbitrary values to the basis vectors is just one way to specify a linear map. It uniquely determines the map on the entire space by linearity as noted by the author. Also, every linear map assigns the basis vectors some value, so yes, every linear map is determined in this way. But you don't necessarily have to give its definition by this procedure. $\endgroup$ – J.R. Feb 10 '14 at 12:15
  • $\begingroup$ You know, when an answer is what you are looking for, it's fair to accept the answer to acknowledge it. $\endgroup$ – 5xum Feb 14 '14 at 12:53
3
$\begingroup$

Let $V$ be a finite-dimensional vector space of dimension $n$ over the field $\Bbb F$ and $\{\,v_1,v_2,\dots,v_n\,\}$ a basis of $V$; let $W$ be another finite-dimensional vector space over the same field. Then, what the author is saying may be the following: given $n$ vectors $\{\,w_1,w_2,\dots,w_n\,\}$ in $W$ there exist one and only one linear application $L:V\to W$ such that $L(v_1)=w_1, L(v_2)=w_2, \dots,L(v_n)=w_n$.

There exist one: in fact, if we define $L(\alpha_1v_1+\dots+\alpha_nv_n)$ to be $\alpha_1w_1+\dots+\alpha_nw_n$, then we have that L is linear and $L(v_1)=w_1, L(v_2)=w_2, \dots,L(v_n)=w_n$.

There exist only one: in fact, if we have a linear map $L$ such that $L(v_1)=w_1, L(v_2)=w_2, \dots,L(v_n)=w_n$, then we must have (by linearity) $L(\alpha_1v_1+\dots+\alpha_nv_n)=\alpha_1w_1+\dots+\alpha_nw_n$, so $L$ is completely already defined.

$\endgroup$
2
$\begingroup$

1) No, you do not need a basis. Consider the natural homomorphism $V \rightarrow V^{**}$.

2) One can always construct a map like this. Also, every map can be written by saying where a basis maps to. This is in general not unique.

3) Because the map is linear.

$\endgroup$
1
$\begingroup$

This characterizes linear maps between vector spaces $V$ and $W$ over a field $F$. Think about it. If you have a linear map $T: V \rightarrow W$ defined on a basis $\{v_i\}$, and a point $x\in V$, the fact that this is a basis means there is a unique set of coefficients $a_i$ with $x=a_1 v_1 +\ldots + a_n x_n$. But then linearity forces (using definition of linearity) $Tx = a_1 Tv_1+\ldots a_n Tv_n$. In other words, the map is already completely defined everywhere from the values on the basis.

For your other questions, note that there is always a basis lurking in the background with such spaces. You can always pick a basis, and no matter how you do it, the same result holds.

The coefficients $a_i$ are artificial in the sense that they depend completely on what basis you have specified. There is no natural basis. But you can always pick one, and once you have, it determines the coefficients ("coordinates") for any point in the space. The idea is that every point in the space can be written as such a linear combination of basis elements, and every such combination is a point in the space. You haven't left anything out in doing this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.