Isomorphism Between Union of two subspaces and Direct Sum

Question

Prove that for two subspaces 1,2 of a finite dimensional vector space V , 1+2 is isomorphic to 1$\bigoplus$2 if and only if $1 \cap 2 = {0}$.

I know how to show the statement that "$=1 \bigoplus 2$ if and only if $=1+2$ and $1 \cap 2={0}$."(1) But this statement (1) does not seem to be equivalent to the above statement that I need to prove. So to show isomorphism, I need to show bijection between sum and direct sum and dimensions equal.

Since $\dim(U1+U2)=\dim(U1)+\dim(U2)-\dim(U1 \cap U2)=\dim(U1)+\dim(U2)$, we need to show bijection which I am a bit stuck on.

Or is there an easier way to just prove this using (1)?

Answer 1

You can prove this from statement (1): Let $W = U_1 + U_2$. Then $U_1$ and $U_2$ are both subspaces of $W$ (even if $W$ isn't all of $V$), because certainly $U_1$ and $U_2$ are both contained in $W$.

Statement (1), with $W$ playing the role of $V$, says that "$W = U_1 \oplus U_2$ if and only if $W = U_1 + U_2$ and $U_1 \cap U_2 = 0$".

In this case, we know $W = U_1 + U_2$, because we defined it that way. So this statement simplifies to "$U_1 + U_2 = U_1 \oplus U_2$ if and only if $U_1 \cap U_2 = 0$", which was what you wanted to show.

Answer 2

They are essentialy the same, you dont even have to use the statement you cited,

If $U_{1} + U_{2}$ is isomorphic to $U_{1} \bigoplus U_{2}$ of course $U_{1} \cap U_{2} = \left\lbrace 0 \right\rbrace$ since the isomorphism is injective you know that only the null vector lies in the intersection.

For the other implication as you showed since the dim $U_{1} + U_{2}$ = dim $U_{1} \bigoplus U_{2}$ thanks to the non intersecting hypothesis, we know that exists some isomorphism between those two spaces thanks to the complete invariancy of the dimension for finite dimensional spaces.

Answer 3

Hint:

Consider the short exact sequence \begin{alignat}4 0 &\longrightarrow{} & U_1\cap U_2 &\longrightarrow {}& U_1\otimes U_2 &\longrightarrow {} & U_1+U_2 &\longrightarrow 0 \\ && u&\longmapsto &(u,-u) \\ &&&&(u_1,u_2)&\longmapsto &u_1+u_2 \end{alignat}