We are given that V, W are vector spaces. We are told that $v$1, $v$2 are distinct vectors in V and that $w$1, $w$2 are distinct vectors in W.
The first True/False question states:
1) There is a linear transformation T : V → W such that T($v$1) = $w$1 , T($v$2) = $w$2.
I want to say that it's false because for this to be true, T would have to be onto, so that every $w$i in W was mapped to by a $v$i in V for $i= 1,2,...,n$.
For example, I know this wouldn't be true if $dim(W)$ > $dim(V)$.
However, my lecture notes state the following, and I don't get it:
"False. Let V = W = $\mathbb{R}^2$, let $v_1 = v_2 = e_1$, and let $w_1 = e_1$ and $w_2 = e_2$. Then it is impossible to simultaneously have $T(e_1) = e_1$ and $T(e_1) = e_2$."
Here's what I don't get:
- If $v$1, $v$2 are distinct as the intro states, then how can we say that they are both equal to $e_1$?
- How can we just say that both vector spaces are in $\mathbb{R}^2$?
2) If $v_1,v_2$ are linearly independent, there is a linear transformation T: V → W such that T($v_1$) = $w_1$, T($v_2$) = $w_2$
I wanted to say false because I can't break the thought that $dim(W) > dim(V)$, because then how would it be possible for every $w$i in W to be mapped to by a $v$i in V for $i= 1,2,...,n$ ?
My lecture notes state:
"True. This is a special case of Theorem 16" where Th.16 is the one about how every linear transformation is uniquely determined by its action on a basis.
$\newcommand{\Span}{\operatorname{span}}$Your first bullet point is a legitimate objection, if it refers to the set-up in your question. It could be that your lecture notes refer to a different set of hypotheses. Or, perhaps there's a "typo" in your notes, and your instructor meant to write something like $v_{1} = -v_{2} = e_{1}$ (N.B. the sign on $v_{2}$).
As for your second bullet point, if you're constructing a counterexample, you're free (in this setting) to pick any vector spaces $V$, $W$, and any vectors $v_{i}$, $w_{i}$ you like, so long as $v_{1} \neq v_{2}$ and $w_{1} \neq w_{2}$. (In fact, there's a simpler counterexample: Let $V = W$ be the vector space of real numbers, $v_{1} = 0$, $v_{2} = 1$, $w_{1} = -w_{2} = 1$.)
It seems there's another issue in need of clarification. Carefully compare the following statements:
If $v_{1}$ and $v_{2}$ are non-proportional elements of a vector space $V$, and if $w_{1}$ and $w_{2}$ are distinct elements of a vector space $W$, then there exists a linear transformation $T:V \to W$ such that $T(v_{1}) = w_{1}$ and $T(v_{2}) = w_{2}$.
The linear transformation $T: V \to W$ satisfies the condition: For every $w$ in $W$, there exists a $v$ in $V$ such that $T(v) = w$.
Despite superficial appearances, these conditions are not logically comparable, much less equivalent. The first concerns the existence of a linear transformation satisfying some property (dependent on some choice of vectors $v_{i}$ and $w_{i}$). The second condition makes an assertion about a pre-existing linear transformation (namely, that the transformation is onto).
In the first statement, the vectors $v_{i}$ and $w_{i}$ are specified before $T$ is defined. In the second, the transformation $T$ is specified first, then we ask whether "every vector in $W$ is in the image of $T$".
This distinction may sound like nitpicking, but it's absolutely fundamental. In the context of your post, the answer to your second question is true (statement 1. is true) even though $T$ need not be onto (condition 2. is not always true).
The take-away message is: In your mathematical education, you'll encounter numerous definitions of the general form "For every [blah], there exists [blah blah] such that [blah blah blah]." Many students focus on the [blah blah blah]. Do not fall into this habit. That "preamble" (for every...there exists...) is a crucial part of the concept being defined.
Added to address question in comment: Let $V$ and $W$ be vector spaces, $v_{1} \neq v_{2}$ vectors in $V$, and $w_{1} \neq w_{2}$ vectors in $W$. For simplicity, assume for the moment that $V = \Span\{v_{1}, v_{2}\}$.
When does there exist a linear transformation $T:V \to W$ such that $T(v_{1}) = w_{1}$ and $T(v_{2}) = w_{2}$?
If these conditions hold, and if $c_{1}$ and $c_{2}$ are scalars, then linearity of $T$ implies $$ T(c_{1}v_{1} + c_{2}v_{2}) = c_{1}T(v_{1}) + c_{2}T(v_{2}) = c_{1}w_{1} + c_{2}w_{2}. \tag{1} $$ On general grounds, this formula defines a linear transformation provided the formula is single-valued: $$ \text{If $c_{1}v_{1} + c_{2}v_{2} = c_{1}'v_{1} + c_{2}'v_{2}$, then $c_{1}w_{1} + c_{2}w_{2} = c_{1}'w_{1} + c_{2}'w_{2}$.} \tag{2} $$ More concisely, if $u_{1} = u_{2}$, then $T(u_{1}) = T(u_{2})$. (Each input gives a unique output. This looks very much like "one-one" or "injectivity", but is actually the converse condition.)
Well-definedness of (1) boils down to showing $T(0) = 0$, i.e., if $c_{1}v_{1} + c_{2}v_{2} = 0$ for some scalars $c_{1}$, $c_{2}$, then $c_{1}w_{1} + c_{2}w_{2} = 0$.
If $\{v_{1}, v_{2}\}$ is linearly independent, then $c_{1}v_{1} + c_{2}v_{2} = 0$ implies $c_{1} = c_{2} = 0$, which certainly implies $c_{1}w_{1} + c_{2}w_{2} = 0$; consequently, if $\{v_{1}, v_{2}\}$ is linearly independent, then (1) is well-defined.
We've made these deductions under the additional assumption that $V = \Span\{v_{1}, v_{2}\}$. In general, let $V' = \Span\{v_{1}, v_{2}\}$, and let $V''$ denote an arbitrary complementary subspace in $V$, i.e., a subspace such that $V = V' + V''$ and $V' \cap V'' = \{0\}$. It turns out that such a $V''$ always exists. Once $V''$ is fixed, an arbitrary vector $v$ in $V$ is written uniquely as $v' + v''$, with $v' = c_{1}v_{1} + c_{2}v_{2}$ in $V'$ and $v''$ in $V''$, and we may define $T:V \to W$ by $$ T(v) = T(v' + v'') = T(v') = T(c_{1}v_{1} + c_{2}v_{2}) = c_{1}w_{1} + c_{2}w_{2}. $$
Here are a couple of related exercises:
Extend the preceding discussion to arbitrary finite ordered sets of distinct vectors $(v_{j})_{j=1}^{k}$ in $V$ and $(w_{j})_{j=1}^{k}$ in $W$.
Find necessary and sufficient conditions on the sets $(v_{j})_{j=1}^{k}$ and $(w_{j})_{j=1}^{k}$ under which there exists a linear transformation $T:V \to W$ such that $T(v_{j}) = w_{j}$ for all $j$. (Linear independence of the $v_{j}$ is sufficient, but not necessary; for example, we might have $v_{2} = 2v_{1}$ and $w_{2} = 2w_{1}$.)