Here's a question that I've been working on from my Linear Algebra textbook (Larson, 8th ed.) and I'd like to check if what I've been doing is correct.
Question: Prove that if $S_{1}$ and $S_{2}$ are subspaces of $\mathbb{R}^{n}$ and if $\mathbb{R}^{n}=S_{1}\bigoplus S_{2}$, then $S_{1}\bigcap S_{2}=\left \{ \mathbf{0}\right \}$.
Edit: $\mathbb{R}^{n}=S_{1}\bigoplus S_{2}$ refers to direct sums, wherein each vector $x$ $\epsilon$ $\mathbb{R}^{n}$ can be uniquely written as a sum of a vector $s_{1}$ from $S_{1}$ and a vector $s_{2}$ from $S_{2}$, such that $x=s_{1}+s_{2}$.
My answer so far:
Let $S_{1}$ and $S_{2}$ be subspaces of $\mathbb{R}^{n}$ s.t. $\mathbb{R}^{n}=S_{1}\bigoplus S_{2}$
Let $\left \{ v_{1},v_{2},...,v_{t} \right \}$ be a basis for $S_{1}$ and $\left \{ v_{t+1},v_{t+2},...,v_{n} \right \}$ be a basis for $S_{2}$.
Since $\mathbb{R}^{n}=S_{1}\bigoplus S_{2}$, then $\left \{ v_{1},v_{2},...,v_{t},v_{t+1},...,v_{n} \right \}$ forms a basis for $\mathbb{R}^{n}$.
After this, I'm kind of stuck.
If I try proving it by contradiction, what can I do to link the two statements together? I figured that any vector that comes from $S_{1}$ and $S_{2}$ should have a dot product of $0$, so I assume that a vector not equal to the zero vector exists and is in the intersection of the two subspaces. Obviously, this wouldn't be the case because any other vector that comes from $S_{1}$ and $S_{2}$ should result in 0 when taken together for the dot product. How do I connect this to the statement above?
Any help would be greatly appreciated. Thank you so much!
It's simpler than that. Let $v \in S_1 \cap S_2$. Then $v = v + 0$ is an expression of $v$ as a sum of an element $v \in S_1$ and an element $0 \in S_2$. But $v = 0 + v$ is another such expression, with $0 \in S_1$ and $v \in S_2$. By unicity, these are the same, so $v + 0 = 0 + v \implies (v = 0 \text{ and } 0 = v)$, in other words $v = 0$.