2
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ $\ker F$ 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
1
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Now I get it! Thank you! $\endgroup$ – ShellRox Nov 23 '18 at 23:48
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.