This is an exercise from Hoffman and Kunzes textbook Linear Algebra.
Let $W_{1}$ and $W_{2}$ be subspaces of $V$ such that $W_{1} + W_{2} = V$ and $W_{1} \cap W_{2} =\{0\}$. Prove that for each vector $\alpha \in V$ there are unique vectors $\alpha_{1} \in W_{1}$ and $\alpha_{2} \in W_{2}$ such that $\alpha = \alpha_{1} + \alpha_{2}$.
My proof is as follows:
Let $\alpha \in V$, we know that $\alpha = \alpha_{1} + \alpha_{2}$ where $\alpha_{1} \in W_{1}$ and $\alpha_{2} \in W_{2}$ since $V = W_{1} + W_{2}$. Similarily, let $\beta_{1} \in W_{1}$ and $\beta_{2} \in W_{2}$. Suppose that $\beta_{1} + \beta_{2} = \alpha$ also.
Then $\alpha_{1} + \alpha_{2} = \beta_{1} + \beta_{2} \implies (\alpha_{1} – \beta_{1}) + (\alpha_{2} – \beta_{2}) = 0$. Since $W_{1} \cap W_{2} = \{0\}$, we have that if $w_{i} \in W_{i}$ for example, then $w_{i}$'s inverse must only be in $W_{i}$.
Let $(\alpha_{1} – \beta_{1}) = w_{1}$ and $(\alpha_{2} – \beta_{2}) = w_{2}$. Then:
$w_{1} + w_{2} = 0$, but $w_{1} \neq -w_{2}$ so $\alpha_{1} – \beta_{1} = 0$ and $\alpha_{2} – \beta_{2} = 0$, hence $\alpha_{1} = \beta_{1}$ and $\alpha_{2} = \beta_{2}$. Thus we have uniqueness.
Please let me know what you think of the proof, any critique would be helpful as I'm self-studying so getting some perspective on your work can be quite difficult. The part I'm the most worried about is
Since $W_{1} \cap W_{2} = \{0\}$, we have that if $w_{i} \in W_{i}$ for example, then $w_{i}$'s inverse must only be in $W_{i}$.
Kind of off-topic but doesn't proving this imply that the direct sum consists only of unique linear combinations? I'm asking since the book hasn't yet give a definition of direct sums. Thank you and have a good day/night.
Best Answer
Your proof is basically fine, but could be clearer for uniqueness:
We have $w_i=\alpha_i-\beta_i\in W_i$, and $w_1+w_2=0$, so we do have $w_1=-w_2$, and indeed $-w_2\in W_2$ as well, so $w_1\in W_1\cap W_2$, and thus $w_1=0=w_2$.
Yes, it's true that if $W_1\cap W_2=\{0\}$ then every element of $W_1+W_2$ can be uniquely written in the form $w_1+w_2$.
In such a case the sum of the two subspaces is called 'direct sum', and then we denote $W_1\oplus W_2=W_1+W_2$.