Prove the following expression in linear algebra

Question

Let $S=\{(x,y,0)\in\mathbb{K}^3,x,y\in\mathbb{K}\}$. Cosider a vector subspace $P\subset\mathbb{K}^3$ such that $K^3=S\bigoplus P$.

1. Prove that their exists $x_0,y_0\in\mathbb{K}$ such that $$(x_0,y_0,1)\in P$$
2. Prove that $P=span\{(x_0,y_0,1)\}$
3. Conversely, prove that if $x_0,y_0\in\mathbb{K}$, then $$S\bigoplus span\{(x_0,y_0,1)\}=\mathbb{K}^3$$ Note $\mathbb{K}=\mathbb{R}\cup\mathbb{C}$

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For my idea that I used to think this problem. I start from $$S\bigoplus P=\mathbb{K}^3\Longleftrightarrow \begin{cases} S\cap P=\{0\}\\S+P=\mathbb{K}^3 \end{cases}$$ Please kindly help me to solve this or give a recommendation idea for me to prove this. Thank in advance.

Answer 1

1. $P$ has an element with $3$rd coordinate nonzero, otherwise $P \subseteq S$ and so $\mathbb{K}^3 \neq S \oplus P$. So, just multiply the element by the multiplicative inverse of the $3$rd coordinate.

2. $S$ is $2$-dimensional, so $P$ is $1$-dimensional: $3 = \dim \mathbb{K}^3 = \dim (S \oplus P) = \dim S + \dim P = 2 + \dim P \implies \dim P = 1$. From (1) we have produced a non-zero vector in $P$ so it spans $P$.

3. Since $S \cap \text{span}\{(x_0, y_0, 1)\} = \{0\}$, the sum is direct. The dimension of the sum is thus $3$ and hence the sum is $\mathbb{K}^3$.