Showing that image of a certain linear map is either trivial or a straight line

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,

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

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

Possibility 1) $$\dim\operatorname{Ker} F = 2$$

If the dimension of kernel is $$2$$, that is, image is zero dimensional according to Theorem 3.2 ($$\dim \operatorname{Im} F = \dim\mathbb{R}^{2} - \dim\operatorname{Ker}F = 2 - 2 = 0$$), 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) $$\dim\operatorname{Ker}F = 1$$

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) $$\dim\operatorname{Ker}F =\{0\}$$ (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!

• $\ker F$ can't be $0$ dimensional because $0\neq A\in\ker F$. – Federico Nov 23 '18 at 19:16
• @Federico That is what I tried to imply in the possibility 3, when I mentioned that kernel not being "spanned" by zero vectors makes it non trivial and therefore $> 0$ dimensional. – ShellRox Nov 23 '18 at 22:15

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 $$A\not=0$$ and $$A\in Ker(T)$$

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).

• Thank you! So since two subspaces can be associated with bijective mapping (injective and surjective) they are isomorphic in this case. I have one last question, on my possibility 3 explanation, was I correct when I said that kernel not being "spanned" by only zero vectors makes it non trivial and higher dimensional? – ShellRox Nov 23 '18 at 22:11
• If I got what you said... yes! By definition if a subspace is "spanned" by some vectors, say $W=Span(v_1,\cdots,v_n)$, then $\dim (W)$ is the smallest number of non-zero vectors that we can get from $v_1,\cdots, v_n$ and the Span still be $W$. Knowing that some $v_i$ is not zero you won't have all $W$ if you delet all $v_n$ – Robson Nov 23 '18 at 22:21
• Now I get it! Thank you! – ShellRox Nov 23 '18 at 23:48

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.