The following is Exercise 3.15 from the German textbook Lineare Algebra by Hans-Joachim Kowalsky and Gerhard O. Michler:
Let $\varphi$ and $\psi$ two linear maps from a finite-dimensional vector space $V$ to a finite-dimensional vector space $W$ with $\dim{\varphi(V)} = m$ and $\dim{\psi(V)} = n$. Show that $$|m-n| \leq \operatorname{rk}(\varphi + \psi) \leq m + n.$$
Progress and Approaches:
- The second inequality can be shown by using the fact that the image of $\varphi + \psi$ is contained in the sum of the images of $\varphi$ and $\psi$ and the dimension formula for subspaces.
- However, the other direction does not turn to work as smoothly. I tried to consider the kernels of $\varphi$, $\psi$ and $\varphi+\psi$ with the hope that I could apply the Rank-Nullity-Theorem and the dimension formula from before. But it turns out that there is no relationship between these.
I also do not know how to acquire the absolute value of $m-n$ in the end. This must mean that one needs to assume that one image is larger than the other. But I do not see how this is supposed to work.
Could you please help me with this exercise? Thank you in advance!
Best Answer
$\newcommand{\rk}{\operatorname{rk}} \newcommand{\im}{\operatorname{im}}$
Note that $\rk \varphi = \rk (- \varphi)$, since $\im \varphi = - \im \varphi$. Now, by the item that you have already solved,
$$ \rk \psi = \rk (\psi + \varphi -\varphi) \leq \rk(\psi+\varphi) + \rk (-\varphi) = \rk(\psi+\varphi) + \rk \varphi. $$
and so $\rk \psi - \rk \varphi \leq \rk (\psi + \varphi)$. Interchanging the roles of $\psi$ and $\varphi$ we get that $\rk \varphi - \rk \psi \leq \rk \psi + \varphi$ and so
$$ |m-n| = |\rk \psi - \rk \varphi| \leq \rk (\psi + \varphi). $$