$V=V_1 \oplus V_2 = V_1 \oplus V_3$ does not mean $V_2 = V_3$ My professor told me that $V=V_1 \oplus V_2 = V_1 \oplus V_3$ does not imply $V_2 = V_3$. I would like to see an example of this claim.
What conditions do I need to have "if $V=V_1 \oplus V_2 = V_1 \oplus V_3$, then $V_2 = V_3$"?
Thank you.
 A: Take $V=\mathbb{R^2}$, $V_1=$ span{$e_1$}, $V_2=$ span{$e_2$}, and $V_3=$ span{(1 , 1)}.
Then $V_2 \neq V_3$ but $V$ is the direct sum of $V_1,V_2$ and $V_1,V_3$.
I can't recall a condition that would force $V_2=V_3$ though.
A: Voldemort's example is very good. It can be generalized to show that, if $V_1\ne V$ and $V_1\ne\{0\}$, then $V_1$ has at least two different complements.
Since $0<\dim V_1<\dim V$ by assumption, we can consider a basis $\{v_1;v_2;\dots;v_k\}$ of $V_1$ and complete it to a basis $\{v_1;\dots;v_k;v_{k+1};\dots;v_n\}$ of $V$.
Then we can take $V_2=\langle v_{k+1};\dots;v_n\rangle$ and $V_1\oplus V_2=V$. But also $\{v_1;\dots;v_k;v_{k+1}+v_1;v_{k+2}\dots;v_n\}$ is a basis of $V$.
Indeed, if
$$
\alpha_1v_1+\dots+\alpha_kv_k+\alpha_{k+1}(v_{k+1}+v_1)+\alpha_{k+2}v_{k+2}
+\alpha_nv_n=0
$$
it follows easily that $\alpha_i=0$ ($i=1,2,\dots,n$).
The subspace $V_3=\langle v_{k+1}+v_1;v_{k+2};\dots;v_n\rangle$ is different from $V_2$, but $V=V_1\oplus V_3$.
Note that, for $n=2$, the example is exactly the same as Voldemort's.
This can be generalized to infinite dimensional spaces (assuming the axiom of choice, of course).
A: Let $V=F^2$ 
$V_1=\{ (z,z):z\in F \}$,
$V_2 = \{(x,0):x\in F\} $ and
$V_3=\{ (0,y):y\in F \}$
Then $V=V_1\bigoplus V_2$ and $V=V_1\bigoplus V_3$ as you can verify but $V_2\neq V_3$
