Let $\tau \in \mathcal L(U,V) $ and $\sigma \in \mathcal L(V,W) $ show that $rk(\sigma \tau ) \le \min \{rk( \tau), rk(\sigma) \}$

Question

Let $\tau \in \mathcal L(U,V) $ and $\sigma \in \mathcal L(V,W) $ show that $rk(\sigma \tau ) \le \min \{rk( \tau), rk(\sigma) \}$

I tried to prove that like this:

I have to prove that $rk(\sigma \tau ) \le rk(\sigma)$ and $rk(\sigma \tau ) \le rk( \tau)$

$rk(\sigma \tau) =dim(Im(\sigma \tau ))=dim(\sigma(\tau (U)))$

Since $\tau (U) \subset V \Rightarrow \sigma (\tau(U)) \subset \sigma (V) = Im(\sigma) \Rightarrow rk(\sigma \tau) \le rk(\sigma) $

For proving that $rk(\sigma \tau ) \le rk( \tau)$ :

By The Rank Plus Nullity Theorem :

$rk(\sigma \tau) +dim(ker(\sigma \tau )) =dim(U)=rk(\tau ) +dim(ker(\tau))$

If we take $dim(U) < \infty $ then $dim(ker(\sigma \tau )), dim(ker(\tau)) < \infty $ So we can write :

$ rk(\sigma \tau) = rk(\tau ) +dim(ker(\tau)) -dim(ker(\sigma \tau )) $

Since $ ker(\tau) \subset ker(\sigma \tau )$ we have $dim(ker(\tau)) -dim(ker(\sigma \tau )) \le 0$ then

$rk(\sigma \tau ) \le rk( \tau) $

But in the question there isn't this condition "$dim(U) < \infty $". So I have to prove that is correct for infinity dimension of $U$ but I couldn't do this. Any help?

Answer 1

If $\mathfrak{B} = \{v_i | i \in I\}$ is a basis of $Im(\tau)$ then $\{\sigma(v_i) | i \in I\}$ span $Im(\sigma\tau)$ so $rank(\sigma\tau) \le rank(\tau)$ because a basis has cardinal less than cardinal of a spanning family.