So this is a true or false question (with justification necessary). The question is as follows.

True or false? Let $f: \mathbb{R}^m \rightarrow \mathbb{R}^k$ and $g: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be linear transformations. Then $f ∘ g=0$ implies img $g =$ ker $f$.

So far I have been able to prove that img $g \subset$ ker $f$.

PROOF OF img $g \subset$ ker $f$.

Suppose that $x \in \mathbb{R}^n$. Let $g(x) =y \in$ img $g$. Since $f(y)=0$, any such $y$ is in ker $f$. Thus, img $g \subset$ ker $f$.

From here I am stuck on how to prove ker $f \subset$ img $g$, or even that is true. If this is false, there has to exist a vector $a$ such that $a \in$ ker $f$ and $a \not\in$ img $g$.

The most I have been able to figure out is that because img $g \subset$ ker $f$, that the matrix of $g$ and some vector in $\mathbb{R}^n$ forms a linear combination of a vector in the kernel of $f$. If the columns of $g$ form a basis for the kernal of $f$, then it is possible for ker $f = $ img $g$. However, I am unsure of how to prove such a basis exists.

  • $\begingroup$ Can you think of a counter-example where $\text{Im} g \not = \ker f$, even though $\text{Im} g \subset \ker f$? $\endgroup$ – Olivier Sep 13 '18 at 18:58
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    $\begingroup$ The equality is false. Just consider $g=0$ and $f=0$. $\endgroup$ – amsmath Sep 13 '18 at 18:58
  • $\begingroup$ @amsmath I am face palming myself for not thinking of that. Thank you. $\endgroup$ – user278039 Sep 13 '18 at 19:03

You are correct in thinking that the converse direction is not true. $g$ is not assumed to be onto, so we can imagine that there are some elements of the kernel of $f$ which $g$ does not map to.

For example, the zero map $g=0$ satisfies the hypothesis but is not necessarily the kernel for arbitrary $f$.

  • $\begingroup$ You don't write the zero map as $g(A)\mapsto 0$. Just write $g=0$. $\endgroup$ – amsmath Sep 13 '18 at 19:06

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