Direct Sum properties in linear algebra. " If E, F and G are three subspaces of a vector space V such that E⊕F = E⊕G then F = G. "
Is this statement true or false?
(My first intuition to say that this statement is false. But I was wondering, since we have E⊕F and E⊕G does that necessarily mean that E⊕F = E⊕G = V?
Like does the direct sum of any 2 subspaces always sum of to the entirety of the vector space V simply by the very definition that they direct sums?
so basically is this true?
if E ∩ F = {0} and E and F are both subspaces of vector space V, does that necessarily implies that E⊕F=V 
(my intuition tells me no but I am a bit confused.
Thanks in advance...)
 A: The answer to your first question is no. Take the real vector space $\mathbb{R}^2$.
Then $\mathbb{R}^2 = \operatorname{span}\{(1,0)\}\oplus \operatorname{span}\{(0,1)\} = \operatorname{span}\{(1,0)\}\oplus \operatorname{span}\{(1,1)\}$
but $\operatorname{span}\{(0,1)\}\neq \operatorname{span}\{(1,1)\}.$
It is also not true that the direct sum of two spaces is equal to the entire space. Indeed, take any vector space of dimension $d \geq 1$. Then
$$\{0\}= \{0\} \oplus \{0\}$$
is not equal to the entire space.
A less trivial example:
$$\operatorname{span}\{(1,0,0)\}\oplus \operatorname{span}\{(0,1,0)\} = \mathbb{R} \times \mathbb{R} \times \{0\} \neq \mathbb{R}^3$$
A: For a finite dimensional space, if $E\cap F=\{0\}$, $E\oplus F=V$ if and only if $\dim E+\dim F=\dim V$.
On the other hand, a proper subspace has an infinite number of complementary subspaces. For instance, in $K^2$ ($K$ is the base field), the subspace generated by a vector $u$ and the subspace generated by any vector $v$ non-collinear with $u$ is a complementary subspace. In particular, if $v$ and $v'$ are non-collinear such vectors, the generated  subspaces are not identical, but complementary.
