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Consider $F: U \to V$ and $g: V \to W$ I have to show that $$rk(g \circ F) \geq rk(F) + rk(g) - \dim(V)$$

Where as usual the rank is the dimension of the image.

Attempts

In general $im(g\circ F) \subseteq im(g)$ whence $rk(g) \geq rk(g\circ F)$. Using the rank theorem:

$$\dim(V) = \dim\ker(g) + \dim(im(g))$$ hence $\dim(V) \geq rk(g)$.

If I consider $im(F) \subseteq V$ then $\dim(im(F)) \leq \dim(V)$ hence $rk(F) \leq \dim(V)$.

So far I got:

  • $rk(g) \geq rk(g \circ F)$
  • $\dim(V) \geq rk(g)$
  • $\dim(V) \geq rk(F)$
  • rk(F) \geq rk(g \circ F)$

Now I thought of looking at the kernel: $\ker(F) \subseteq \ker (g\circ F)$ hence $\dim(\ker(F)) \leq \dim \ker (g\circ F)$. Again using the rank theorem

$$\dim(U) = \dim(im(F)) + \dim\ker(F)$$

$$\dim(U) = \dim(im (g\circ F)) + \dim(\ker(g\circ F))$$

so $\dim(im(F)) + \dim(\ker(F)) = \dim(im(g\circ F)) + \dim(\ker (g\circ F))$$

And here I stopped becasue first of all I don't know whether all of this is right/useful, and I got stuck, not knowing how to go on...

Is there other some "simpler" way (that is: more rigorous or indeed simple) to face the problem?

ADD

I read this answer show that $\operatorname{rank}(g\circ f) \leq \operatorname{rank}(f)+\operatorname{rank}(g)-n$

but I don't understand the following:

  • why is $$\DeclareMathOperator{\rk}{rank}\rk(g\circ f)=\rk\Bigl(g_{\,\bigm\vert_{\,\scriptstyle\Im f}}\Bigr)$$

  • why $$rk(g\circ f)= \dim(\Im f)-\dim(\ker g\cap\Im f)=rk f-\dim(\ker g\cap\Im f). $$

shouldn't the dimension of the composition be the dimension of $W$ minus the dimension of the kernel?

Heidegger
  • 3,229

1 Answers1

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why is $$\DeclareMathOperator{\rk}{rank}\rk(g\circ f)=\rk\Bigl(g_{\,\bigm\vert_{\,\scriptstyle\Im f}}\Bigr)?$$

Note that the rank of a linear transformation is the dimension of its image. The image of $f$ is given by $\Im f = \{f(u): u \in U\}$. With that, we have $$ \Im (g|_{\Im f}) = \{g(x):x \in \Im f\} = \{g(f(u)): u \in U\}. $$ Similarly, $$ \Im (g \circ f) = \{(g \circ f)(u): u \in U\} = \{g(f(u)): u \in U\}. $$ So, $\Im (g \circ f) = \Im (g|_{\Im f})$. It follows that $$ \text{rank} (g \circ f) = \dim Im (g \circ f) = \dim \Im (g|_{\Im f}) = \text{rank}(g|_{\Im f}). $$

why $$\rk(g\circ f)= \dim(\Im f)-\dim(\ker g\cap\Im f)=rk f-\dim(\ker g\cap\Im f)?$$

Again, recall that $\text{rank} (g \circ f) = \text{rank}(g|_{\Im f})$. Applying the rank-nullity theorem to $g|_{\Im f} : \Im f \to W$ yields $$ \rk(g|_{\Im f}) = \dim (\Im f) - \dim \ker (g|_{\Im f}). $$ We know that $\dim(\Im f) = \rk(f)$. The trick from here is to show that $\ker (g|_{\Im f}) = \ker g \cap \Im f$. That just comes from the definition of the kernel and the definition of the restriction $g|_{\Im f}$. We have $$ \begin{align*} \ker g|_{\Im f} &= \{v \in \Im f : g(v) = 0\} \\ & = \{v \in V: v \in \Im f \text{ and } g(v) = 0\} \\ & = \{v \in V: v \in \Im f \} \cap \{v \in V: g(v) = 0\} = \Im f \cap \ker g. \end{align*} $$

Here's an alternative approach you might find helpful. We can rewrite the inequality as $$ \dim(U) - \rk(g \circ F) \leq \dim(U) - [\rk(F) + \rk(g) - \dim(V)] \iff\\ \dim(U) - \rk(g \circ F) \leq (\dim U - \rk F) + (\dim V - \rk g) \iff\\ \dim \ker (g \circ F) \leq \dim \ker F + \dim \ker g. $$ So, it suffices to show that this last inequality holds. One approach to this is to show that $$ \ker(g \circ F) = \{u \in U: F(u) \in \ker(g)\}. $$ From there, one could show that for any subspace $S \subset U$, we have $$ \dim\{u \in U: u \in F(u) \in S\} \leq \dim(S) + \ker(F). $$

Ben Grossmann
  • 225,327