$\{x_1,...,x_n\}$ a basis of $V_1$; SHOW THAT there is a unique linear transformation $f:V_1\to V_2$ such that $f(x_1)=y_i$

Question

The following question is given in the book Linear Algebra by A.R. Rao and P. Bhimasankaram:

Let $V_1$ and $V_2$ be vector spaces over $F$ and let $\{x_1,...,x_n\}$ be a basis of $V_1$. Then, for any given vectors $y_1,...,y_n$ in $V_2$, show that there is a unique linear transformation $f:V_1\to V_2$ such that $f(x_1)=y_i$, $i=1,2,...,n$.

I know that it is actually a theorem because I found related questions on the site, but none explaining the proof of the same(most had questions on consequences of the theorem). I am not yet comfortable with the topic of linear transformations, so I don't know how to proceed with the proof. Please, help!

Answer 1

The fact that $\{x_1,\dots,x_n\}$ is a basis for vector space $V_1$ tells us that every $x\in V_1$ can be written as $x=a_1x_1+\cdots+a_nx_n$ and secondly that this presentation of $x$ is unique.

Now if $f:V_1\to V_2$ is indeed a linear transformation satisfying $f(x_i)=y_i$ for $i=1,\dots n$ then this leads to:$$f(x)=f(a_1x_1+\cdots+a_nx_n)=a_1f(x_1)+\cdots+a_nf(x_n)=a_1y_1+\cdots+a_ny_n$$

showing that $f$ is completely determined by its values on $\{x_1,\dots,x_n\}$.

So proved is actually now that - if such linear transformation indeed exists - then it is unique.

It remains to prove that such $f$ indeed exists.

For that you can just define $f$ by stating that it is prescribed by:$$a_1x_1+\cdots+a_nx_n\mapsto a_1y_1+\cdots+a_ny_n$$ Based on the fact that the presentation $a_1x_1+\cdots+a_nx_n$ is unique we conclude that like this the function is what we call: well-defined.

That is the first step.

It is evident that it is a function $f:V_1\to V_2$ and the second is to check that this function is indeed a linear transformation. For this it is sufficient to prove that it satisfies the two conditions:

• $f(x+x')=f(x)+f(x')$
• $f(ax)=af(x)$

If $x=a_1x_1+\cdots+a_nx_n$ and $x'=a_1'x_1+\cdots+a_n'x_n$ then $x+x'=(a_1+a_1')x_1+\cdots+(a_n+a_n')x_n$ so that:$$f(x+x')=(a_1+a_1')y_1+\cdots+(a_n+a_n')y_n=(a_1y_1+\cdots+a_ny_n)+(a_1'y_1+\cdots+a_n'y_n)=$$$$f(x)+f(x')$$

Further $ax=a(a_1x_1+\cdots+a_nx_n)=aa_1x_1+\cdots+aa_nx_n$ so that $$f(ax)=aa_1y_1+\cdots+aa_ny_n=a(a_1y_1+\cdots+a_ny_n)=af(x)$$

The proof is now complete.

Answer 2

Any vector $v \in V_1$ is of the form $v=\sum_{k=1}^{n} a_ix_i$ where $a_i$'s are scalars. Define $f(v)=\sum_{k=1}^{n} a_iy_i$ and verify that this is linear. If you put $a_1=1$ and $a_i=0$ for $2\leq i \leq n$ we get $f(x_1)=y_1$. Similarly, $f(x_i)=y_i$ for all $i$. This proves existence. For uniqueness just note that $f(\sum_{k=1}^{n} a_iy_i) =\sum_{k=1}^{n} a_if(x_i)=\sum_{k=1}^{n} a_iy_i$ for any linear $f$ with $f(x_i)=y_i$ for all $i$.