# Range and kernel of linear transformations over vector spaces

$\newcommand{\Range}{\operatorname{Range}}$

Consider non-zero vector spaces $V_1, V_2, V_3, V_4$ and linear transformations $\phi_1: V_1 \to V_2$, $\phi_2: V_2 \to V_3$, $\phi_3 V_3 \to V_4$ such that $\ker \phi_1 = \{0\}$, $\Range \phi_1 = \ker \phi_2$, $\Range \phi_2 = \ker \phi_3$, and $\Range \phi_3 = V_4$. Then, is it the case that: $$\sum_{i=1}^4 (-1)^i\dim V_i = 0 \tag{1}$$ $$\sum_{i=1}^4 (-1)^i\dim V_i > 0 \tag{2}$$ $$\sum_{i=1}^4 (-1)^i\dim V_i < 0 \tag{3}$$ $$\sum_{i=1}^4 (-1)^i\dim V_i \neq 0 \tag{4}$$

I cannot think of any theorem or general results regarding this. So I tried with random examples and each time got option $1$. But the answer sheet says option $2$ is also true and $3,4$ are wrong. I also observed one space's dimension could restrict the possibilities of other dimensions . But no general results yet

Let $d_1,...,d_4$ be the dimensions of $V_1,...,V_4.$ Since we're given that $\phi_1$ has trivial kernel, it must be injective. Hence the dimension of $\phi_1$'s image must be $d_1.$ In general, we have a relation between the dimensions of the image and kernel of a linear operator - they add to the dimension of the domain. We can use this fact to compute the dimensions of the kernels and images of the other operators.
We know that the dimension of $\phi_2$'s image is $d_1$, so its kernel has dimension $d_2-d_1.$ Similarly we see that $\phi_3$ has an image of dimension $d_3-(d_2-d_1)=d_3-d_2+d_1.$ From the problem, this is also the dimension of $V_4.$