Kernel and image of linear maps

linear algebra

I had these $4$ small questions in a recent exam. I'm pretty sure that I got $3$ & $4$ right. But I need some help with the first two. Thank you.

Let $f,g:\mathbb{R^2}\rightarrow\mathbb{R^2}$ be two linear maps. We suppose that $f$
and $g$ are not identically zero and $f\circ g=0.$

  1. Show that $f$ and $g$ are not isomorphisms.
  2. Show that Ker$(f)$ and Ker$(g)$ are of dimension $1$.
  3. Show that Im$(g)$ is of dimension $1$.
  4. Show that Im$(g)\subset$ Ker$(f)$ and conclude that Im$(g)=$ Ker$(f)$.

My answers:

  1. $f$ is not injective because $f( g(x)) = 0$, but $g$ isn't identically $0.$ (Need some help for $g$).

  2. Not really sure here, but maybe: If $f$ and $g$ were identically $0$, then Ker$(f)=\mathbb{R^2}$. But since this is not the case then their Kernel can't be of dimension $2$. Not sure how to prove they are of dimension $1$.

  3. Obvious by rank-nullity theorem.

  4. Let $x\in$ Im$(g)$. Then $\exists y $ such that $g(y)=x.$ Then we have $f(g(y)) = f(x) = 0.$ So $x\in$ Ker$(f)$. We have Im$(g)\subset$ Ker$(f)$ and dim$($Im$(g)) = 1 = $ dim$($Ker$(f)) \implies $Im(g) = Ker(f).

Best Answer

Ben's answer is perfectly fine. However, I feel it is beneficial to reflect on why such a question might have been asked in an exam.

I think the important take-away is this:

A linear map between vector spaces of equal finite dimension
is an isomorphism if, and only if, it has trivial kernel.

Once you have shown (1), the remainder of the question (parts (2), (3) & (4)) follow almost immediately from the following crucial "theorems":

The Rank-Nullity Theorem (i)
A linear map $T : V \to W$ satisfies $\dim V = \dim \mathrm{im}\ T + \dim \ker T$.

A fact without a name (ii)
A finite-dimension linear subspace $W \subseteq V$ satisfies $\dim W \le \dim V$, with equality only if $W = V$.

A solution to (1) has been covered. For (4), as a matter of style, I would refrain from writing out in excruciating detail why $\mathrm{im}\ g \subseteq \ker f$ since this is precisely what the relation $f \circ g = 0$ says. However, if putting pen to paper aids your understanding here, then do whatever works best for you. Indeed, it seems by the wording of your question that perhaps this was expected in your exam. As you noted, parts (3) & (4) then follow from facts (i) & (ii).

Part (2) is the crux of the problem in my opinion, since this is where the moral of the story comes from. The only linear map with kernel of maximum dimension in any linear space is the zero map. So $\dim \ker f$ must either be $0$ or $1$. But it cannot be $0$ since we showed in (1) that $f$ is not an isomorphism. Thus, it must be of dimension $1$, since there is no other alternative. The same argument applies to $g$, of course.

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