Proving that a linear map is well-defined, one to one and onto?

Question

I'm trying to prove the following statement.

Let $V$ be a finite dimensional vector space over $F$. Let $B_1 = \left\{v_1, \ldots v_n\right\}$ be basis for $V$. Let $\alpha_1, \ldots, \alpha_n \in F$. The linear map $x \to [x]_{B_1} = [\alpha_1, \ldots, \alpha_n]^T$, where $x=\sum \alpha_i v_i$, from $V $ to $F^n$ is well-defined, one to one and onto.

I'm not sure if the statement is fully correct itself, but here's my attempt to prove it.

Well-defined: For $a, b \in V$ we have $ a= b \implies [a]_{B_1} = [b]_{B_1}.$

Let $a, b \in V$. We can write $a = \sum{a_i v_i}$ and $b = \sum{b_i v_i}$, where $a_1, \ldots, a_n \in F$ and $b_1, \ldots, b_n \in F.$ Then $a-b = \sum (a_i-b_i)v_i$, so if $a=b$ then $\sum (a_i-b_i)v_i = 0. $ Since elements of $B_1$ are linearly independent, we must have $a_i = b_i$ for every $i \in \overline{1, n}$, so $[a_1, \ldots, a_n]^T = [b_1, \ldots, b_n]^T$, i.e. $[a]_{B_1} = [b]_{B_1}$. So the map is well-defined.

Injective: For $a, b \in V$, we have $[a]_{B_1} = [b]_{B_1} \implies a= b.$

Let $a, b \in V$. Again, we can write $a = \sum{a_i v_i}$ and $b = \sum{b_i v_i}$, where $a_1, \ldots, a_n \in F$ and $b_1, \ldots, b_n \in F. $Let $[a]_{B_1} = [a_1, \ldots, a_n]^T$ and $ [b]_{B_1}= [b_1, \ldots, b_n]^T$. Then $[a]_{B_1} -[b]_{B_1} = [a_1, \ldots, a_n]^T- [b_1, \ldots, b_n]^T= [a_1-b_1, \ldots, a_n-b_n]^T.$ If $[a]_{B_1} = [b]_{B_1}$ then $[a_1-b_1, \ldots, a_n-b_n]^T=0$. Hence $a_i = b_i$ for every $i \in \overline{1, n}$; so if we take $v_i = v_i$ then $a_iv_i = b_iv_i$ by multiplication so $ \sum a_iv_i= \sum b_i v_i$, i.e. $a=b$. Hence the map is injective.

Surjective: for every $a \in V$, there exists $b \in F^n$ s.t. $[a]_{B_1} = b$ .

Let $a \in V$. We can write $a= \sum a_i v_i$ similar to above. Then $b := [a_1, \ldots, a_n]$ is an element of $F^n$ s.t. $[a]_{B_1} = b$.

Questions:

(1) Does the statement and the proof make sense? Any suggestions/insight/critique is appreciated.

(2) For injectivity, does the part where I say if $a_i = b_i$ for every $i \in \overline{1, n}$, then $$v_i = v_i \implies a_iv_i = b_iv_i \implies \sum a_iv_i= \sum b_i v_i$$

rely on any assumptions, like $\left\{v_1, \ldots, v_n \right\}$ being basis -- span/linear independence etc?

I ask this because it seems we only rely on $\left\{v_1, \ldots, v_n \right \}$ spanning $V$ to prove injectivity.

Similarly, to prove surjectivity we haven't used linear independence.

Answer 1

First, you aren’t interpreting “well-defined” quite correctly here.

Note that the definition of the map depends on how you express $v$; what you need to show is either that the expression is unique and so there is no ambiguity, or that if you are given two different ways to express $v$, the result of applying the definition to each of the different ways will be the same. That is essentially what you do, but it’s not “if $a=b$ then the images of $a$ and $b$ are the same”; rather, it’s “if $a=\sum \alpha_i v_i = \sum \beta_iv_i$, then....”. You essentially do this, but the write-up can lead to the misinterpretation that you are verifying you have a function, rather than that its definition does not depend on the representation.

Remember also that it is much easier to prove injectivity of linear maps by checking that only the zero vector maps to $0$.

Of course, for that you first need to show this is a linear map. You never did that!

Your argument for injectivity is slightly off: after proving that $a_i=b_i$, I don’t understand your invocation of $v_i$. Instead, you should note that $$ a = \sum_i a_iv_i = \sum_i b_iv_i = b$$ with the middle equality simply because $a_i=b_i$ for each $i$. You are right that you are only explicitly using that each element can be written as a combination of $B_1$, but remember that you already used the fact that $B_1$ is linearly independent to show the map makes sense in the first place.

Finally, you got surjectivity backwards! You need to show that if you are given $b\in F^n$, then there exists an $a\in V$ such that $[a]_{B_1} = b$. So that part is wrong.

And in any case, again, you are supposed to also show this map is linear. You have not done so.