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From S.L Linear Algebra:

Let $A$ be a non-zero vector in $R^2$. Let $F: mathbb{R}^2
rightarrow W$ be a linear map such that $F(A)=O$. Show that the image
of $F$ is either a straight line or ${0}$.

I've taken following theorems from the book to try and construct the answer (proofs of theorems are omitted):

Theorem 3.2. Let $V$ be a vector space. Let $L: V rightarrow W$ be a linear map of $V$ into another space $W$. Let $n$ be the
dimension of $V$, $q$ the dimension of the kernel of $L$, and $s$ the
dimension of the image of $L$. Then $n = q + s$. In other words,

$$dim V= dim operatorname{Ker} L + dimoperatorname{Im } L$$

Answer that I have constructed assumes all possibilities of dimension that kernel might have (due to cardinality, and Theorem 3.2, $dimoperatorname{Ker}F in {0, 1, 2}$ of a linear map $F$).

Possibility 1) $dimoperatorname{Ker} F = 2$

If the dimension of kernel is $2$, that is, image is zero dimensional according to Theorem 3.2 ($dim operatorname{Im} F = dimmathbb{R}^{2} - dimoperatorname{Ker}F = 2 - 2 = 0$), then considering that kernel is a subspace, $mathbb{R}^2 = operatorname{Ker}F$, and therefore $F$ is a zero map having the image of ${0}$.

Possibility 2) $dimoperatorname{Ker}F = 1$

If kernel is $1$-dimensional, then so is the image according to Theorem 3.2 and thus we have a straight line as the image of $F$, considering that we have one-dimensional image under linear map $F$.

Possibility 3) $dimoperatorname{Ker}F ={0} $ (Presumably impossible)

Kernel can't be zero-dimensional, since it is spanned by two dimensional vectors that are not zero vectors.

Conclusion:

Hence the image of $F$ is either a straight line or ${0}$.

Is this answer sufficient and true? My explanation of Possibility 2) concerns me the most, it might not be specific enough.

Thank you!

Your approach is correct!

P1) $dim(Im F)=0 implies Im(F)={0}$, because the image of a linear function is a subspace and hence $0$ is in it, and can't have anything else because its dimension is zero. So $F(x)=0 forall x$

P2) we have $dim(Ker F)=1$, applying the theorem you get $dim(Im T)=1$ and you can use the fact that two vector spaces are isomorphic (they are "the same space") if their dimension are equal, hence you can say that $Im(T)cong mathbb{R}$ which is a very nice way to justify that "$Im(T)$ is a straight line".

P3) can't be the case that $dim(Ker T)=0$ because this would implie $Ker(T)={0}$, but we know that $Anot=0$ and $Ain Ker(T)$

Your answer is good too! But it seems like it need to be more "direct" in a way... but the question isn't too direct either... I assumed that "being a straight line" is the same that "have dimension one"... but justifying that dimension one implies being isomorphic to the reals is also a good argument (because they are often called THE line).

You're trying to use the Rank-Nullity Theorem.
$$ r + n = text{dim of the domain}$$.

You know 2 $ge$ n $ge$ 1, since A $in$ Ket(T). Thus 0 $le$ r $le$ 1.