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On p.796 of Anton's Elementary Linear Algebra, there is a question. Prove: If $(v_1, v_2, ...,v_n)$ is a basis for V and $(w_1,w_2, ...w_n)$ are vectors in W, not necessarily distinct, then there exists a linear transformation $T:V \rightarrow W$ such that $T(v_1) = w_1, T(v_2) = w_2, ..., T(v_n) = w_n$

I know that the set $(v_1, ..., v_n)$ is linearly independent and spans V because it is a basis. The set $(w_1, ..., w_n)$ could be linearly dependent.
I let $v=(c_1v_1 +...+c_nv_n)$ be an arbitrary vector in $V$.
Then the transformation of $v$ is $T(c_1v_1 +...+c_nv_n)=w_i$. Let $w_i=c_1w_1+...+c_nw_n$ so $(w_1,...,w_n)$ spans $W$.
To show that $(w_1, ...., w_n)$ is linearly independent,
let $c_1w_1 +...+c_nw_n =0$ only have the trivial solution of $c_1, ...,c_n =0$ meaning $(w_1, ...,w_n)$ are distinct vectors.

Edit: Now show that $T$ is closed under addition and scalar multiplication.

Let $u=k_1v_1+...+k_nv_n$ be an arbitrary vector in $V$.

$T(u+v)=(k_1w_1+...+k_nw_n)+(c_1w_1+...+c_nw_n)=T(u) + T(v)$

Let $d$ be a scalar where $T(du)=dk_1w_1+...+dk_nw_n=d(k_1w_1+...+k_nw_n)=dT(u)$.

Edit 2:

$T(v_i) = w_i$ where v_i is in the basis of $V$ and $w_i$ is in $W$. Let $v_j$ be a vector in basis of $V$. $T(v_i+v_j)=w_i + w_j=T(v_i) + T(v_j)$ Let k be a scalar (real number). $T(kv_i)=kw_i=kT(v_i)$

The transformation is linear because it satisfies the requirements for closure under addition and scalar multiplication.

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  • $\begingroup$ the list $w_1,\ldots,w_n$ doesnt necessarily span $W$, that is, you are assuming that $T(V)=W$ but all that you knows is that $T(V)\subset W$. And observe that $$T(c_1v_1+\ldots+c_nv_n)=c_1w_1+c_2w_2+\ldots+c_nw_n$$ if $T$ is linear. $\endgroup$
    – Masacroso
    Commented Jul 29, 2017 at 2:53
  • $\begingroup$ @Masacroso if I use the word "let..." can I assume it spans $W$? $\endgroup$ Commented Jul 29, 2017 at 3:00
  • $\begingroup$ no, you cant assume that a random list of vectors span the whole space. Anyway the text is not clear, you are mixing different exercises in the same question? $\endgroup$
    – Masacroso
    Commented Jul 29, 2017 at 3:02
  • $\begingroup$ @Masacroso No, it's just one question. Why can't I assume that set $(w_1,...,w_n)$ spans W? $\endgroup$ Commented Jul 29, 2017 at 4:22
  • $\begingroup$ because it is not stated, the exercise doesnt says that $w_1,\ldots,w_n$ spans $W$ or is a basis of $W$, just that $w_1,\ldots,w_n$ is a list of vectors in $W$. To prove that $T$ is linear you only need to prove that it have linear properties, that is: $$T(\lambda v+\mu w)=\lambda Tv+\mu Tw,\quad \forall v,w\in V,\forall\lambda,\mu\in\Bbb F$$ where $\Bbb F$ is the field of $V$, and that the domain of $T$ is $V$ and $T(V)\subset W$. $\endgroup$
    – Masacroso
    Commented Jul 29, 2017 at 10:39

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since the $\bar v_i$ form a basis, any vector $\bar u$ may be written in a unique way as: $$ \bar u=\sum_{i=1}^n u_i\bar v_i $$ define $$ T\bar u = \sum_{i=1}^n u_i\bar w_i $$ and show $T$ has the required properties

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  • $\begingroup$ Well noted. I am going to try to show closure under addition and scalar multiplication. $\endgroup$ Commented Jul 29, 2017 at 2:58
  • $\begingroup$ Just learned from a student at ENS I met that it is wrong to assume that T is one-to-one. He is using different terminology but sounds like I have to work with the matrix of T relative to the bases B and B'. $\endgroup$ Commented Jul 29, 2017 at 6:43
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    $\begingroup$ Yes, $T$ need not be 1-1. and it need not be surjective. the essential points are (1) a linear map is defined completely by the values it takes on the elements of a basis for the domain (2) those values can be chosen with complete freedom. $\endgroup$ Commented Jul 30, 2017 at 1:25

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