Linear Algebra Done Right -Sheldon Axler, Third Edition, Section 1.C, Page 23.
Condition for a direct sum:
Suppose $U_1,U_2,\ldots,U_m$ are subspaces of $V$. Then $U_1+\cdots+U_m$ is a direct sum if and only if the only way to write $0$ as a sum $u_1+\cdots+u_m$, where each $u_j$ is in $U_j$, is by taking each $u_j=0$.
I don't seem to understand their proof.
They first write $v=u_1+\cdots+u_m$ (where $u_1\in U_1,u_2\in U_2,\ldots,u_m\in U_m$) and also suppose that $v=v_1+\cdots+v_m$ (where $v_1\in U_1,\ldots,v_m\in U_m$).
Next, they subtract the two equations and write:
"Because $u_1-v_1\in U_1,\ldots,u_m-v_m\in U_m$, the equation above implies that each $u_j-v_j=0$. Thus $u_1=v_1,\ldots,u_m=v_m$." I didn't understand how they deduced this. Why should $u_1-v_1\in U_1,\ldots,u_m-v_m\in U_m$ imply that each $u_j-v_j=0$ ?