# Kernel of a homomorphism of vector spaces

I am interested in the kernel of $\Phi: U_1 \oplus U_2 \oplus U_3 \rightarrow V$, $(u_1, u_2, u_3) \mapsto u_1+u_2+u_3$ for $V$ a $K$-Vectorspace and $U_1, U_2, U_3$ arbitrary subspaces of $V$.

Is there a nice way to describe the kernel?

So far I can only show $\max(\dim((U_l+U_k)\cap U_j))\leq\dim \ker\Phi \leq \min (\dim((U_{l_1}+U_{k_1})\cap U_{j_1} \oplus (U_{l_2}+U_{k_2})\cap U_{j_2}))$

And of course that $\ker \Phi$ is a subspace of $((U_2+U_3)\cap U_1) \oplus ((U_1+U_3)\cap U_2) \oplus ((U_1+U_2)\cap U_3)$.

Thank you!

Let's think about the simpler case $\Psi:U_1\oplus U_2\to V$ first. Then $\ker\Psi=(u_1,u_2):u_2=-u_1$ is isomorphic to $U_1\cap U_2$, since given any $u_1\in U_1\cap U_2$ there's a unique corresponding kernel element $(u_1,-u_1)$.
So, now we generalize to your $\Phi$. $\Phi=0$ just if $u_1+u_2=-u_3$, so certainly we'll only find kernel elements when $u_3\in U_1+U_2$. In this case there must exist a solution $\Phi(u_1,u_2,u_3)=0$, but it's not unique even for a fixed $u_3$: we can also take $\Phi(u_1+u_1',u_2+u_2',u_3)=0$ whenever $\Phi(u_1',u_2',0)=0.$ This happens, as above, whenever $u_2'=-u_1',$ so we get another solution for each element of $U_1\cap U_2$. Thus the kernel of $\Phi$ is isomorphic to $U_3\cap (U_1+U_2)\oplus U_1\cap U_2$.
• Thanks for the answer. Now that you say it, there's even a more schematic way to see that: $\dim \ker \Phi = \dim(U_1 \oplus U_2 \oplus U_2) - \dim (U_1+U_2+U_3)$ hence the result. Works also for generalization. – user110071 Nov 20 '13 at 6:48