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.


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!

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    $\begingroup$ $\ker F$ can't be $0$ dimensional because $0\neq A\in\ker F$. $\endgroup$ – Federico Nov 23 '18 at 19:16
  • $\begingroup$ @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. $\endgroup$ – 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).

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    $\begingroup$ 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? $\endgroup$ – ShellRox Nov 23 '18 at 22:11
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    $\begingroup$ 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 $v_n$ $\endgroup$ – Robson Nov 23 '18 at 22:21
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    $\begingroup$ Now I get it! Thank you! $\endgroup$ – 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.


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