Let $V, W$ be nonzero spaces over a field $F$ and suppose that a set $B=\{v_1,...,v_n\}$ subset of $V$ has the following property: For any vectors $w_1,....,w_n$ as the element of $W$, there exists a unique linear transformation $T:V\to W$ such that $T(v_i)=w_i$ for all $i=1,...,n$. Prove that $B$ is a basis of the space $V$.

I'm not very sure how to do it but here is my attempt to the question. First I prove that $B$ is linearly independent. Let $c_1,...,c_k$ be constants such that $c_1v_1+c_2v_2+....+c_kv_k=0$. Apply the linear transformation of $T$ to both sides: $c_1(T(v_1))+....+c_k(T(v_k))=0$, $c_1w_1+.....+c_kw_k=0$. Since $w_1,...,w_n$ are basis for $W$, so it is linearly independent. So we deduce that $c_1=c_2=....=c_k=0$.

But I'm not very sure how to prove that $B$ spans $V$. Thanks all for the help:)

  • 1
    $\begingroup$ How are you sure that you can choose $\{w_1, \dots, w_n\}$ to be a basis? There may not be any $n$-element basis in W. $\endgroup$ – knsam Mar 7 '13 at 6:56
  • $\begingroup$ Oh yeah, then what would you advise me to do?? $\endgroup$ – Chloe Mar 7 '13 at 7:06
  • $\begingroup$ I have just written up an answer that should guide you to the proof. $\endgroup$ – knsam Mar 7 '13 at 7:38

Linear Independence.

You'd have to manipulate the "any" part of the hypothesis: given any set of $n$ vectors in $W$, you get a unique linear map $T: V \to W$ such that $T(v_i) = w_i$.

The case $n=1$ is easy to see: if $\{v_1\}$ was linearly dependent, then, $v_1 = 0$. But, then, there cannot be a linear transformation from $V$ to $W$, if we chose $w_1 \neq 0$ (such a choice is possible, since $W$ is a non-zero vector space).

Assume that $n > 1$, and that there is a linear dependence: without loss of generality, we may assume that $v_n$ is in the linear span of $\{v_1, \dots, v_{n-1}\}$.

$$v_n = \sum_{j=1}^{n-1} c_i v_i$$

Now, deliberately choose, $w_n$ so that $w_n \neq \sum_{i=1}^{n-1} c_i w_i$ (you can always do this, modulo some edge cases!). This should help you finish the proof.

Exercise. Decipher those edge cases and handle them separately. (I'll write this up neatly in a bit.)


To show that $B = \{v_1, \dots, v_n\}$ is a spanning set, we may proceed by contradiction: you may extend this linearly independent subset $B$ to a basis $\mathcal{B}$ of $V$.

Now, if $\mathcal{B} \neq B$, defining $T$ arbitrarily on the set $\mathcal{B} \setminus B$, will give you a lot of linear transformations that agree on $B$. This contradicts the uniqueness of $T$.

You may want to flesh this argument for completing the proof.

  • $\begingroup$ Hi, I still don't get why contradicting the uniqueness of T can prove that B spans V? $\endgroup$ – Chloe Mar 7 '13 at 13:06

To prove linear independence, let $0 \ne w \in W$, and suppose $$ a_1 v_1 + \dots + a_n v_n = 0.\tag{comb} $$ For each $i$, consider the linear map $T_i$ such that $T_i(v_i) = w$, and $T(v_j) = 0$ for $j \ne i$. Applying $T$ to (comb), you get $$ a_i w = 0, $$ so $a_i = 0$ for all $i$, and you have proved the $v_i$ to be linearly independent.

To prove that they span $V$, there's nothing better than the argument of @KannappanSampath. Simply note that if $\langle B \rangle$ is a proper subset of $V$ (so that there is $v_0 \in V \setminus \langle B \rangle$), then any linear map $T$ defined on $v_1, \dots, v_n$ can be extended by choosing an arbitrary value for $T(v_0)$. (And there are at the very least two possible values for it, that is $0$ and $w$.)

  • $\begingroup$ Clever proof for the linear independence. I did not think along your lines. Thank you for writing it up. $\endgroup$ – knsam Mar 8 '13 at 11:18

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