# Show that this set is a basis for $S_1+S_2$

Let $$B_1 = \{v_1, \dots v_n, x_1 \dots x_r\}$$ $$B_2=\{v_1 \dots v_n, y_1 \dots y_s\}$$ $$B_3 = \{v_1 \dots v_n\}$$

be basis for subspaces $$S_1$$ , $$S_2$$ and $$S_1 \cap S_2$$ respectively. Show that the set $$B_4 = \{v_1 \dots v_n , x_1 \dots x_r, y_1 \dots y_s\}$$ is a basis for $$S_1 + S_2$$.

I have managed to prove that the set spans $$S_1 + S_2$$ what I'm having trouble on is showing that the set is linearly independent. So here is what I did..

Let $$\underbrace{\sum_{i=1}^n \alpha_iv_i}_{\alpha} + \underbrace{\sum_{j=1}^r \beta_jx_j}_{\beta} + \underbrace{\sum_{k=1}^s \sigma_k y_k}_{\sigma} = 0$$ and so we wish to show that all the $$\alpha_i$$ , $$\beta_j$$ and $$\sigma_k$$ are zero.

So i consider the case when $$\alpha + \beta =0$$ and so since $$B_1$$ is a basis we get that the each $$\alpha_i$$ and $$\beta_j$$ are zero. and that $$\sigma = 0$$. Now since the set $$\{y_1 \dots y_s\}$$ is a subset of the basis $$B_2$$ they must be linearly independent also so the $$\sigma_k$$ are all zero.

For the other case we have the $$\alpha + \beta \neq 0$$ which implies that $$\sigma = -\beta - \alpha$$. In other words the linearly independent set $$\{y_1 \dots y_s\}$$ spans $$B_1$$ making it a smaller basis which is a contradiction.

My question is mainly is there a more elegant proof of this fact that doesn't rely on case work and if not then is the proof I provided a valid proof. Thanks in advance!

Let

$$\sum_{i = 1}^n \alpha_i v_i + \sum_{j = 1}^r \beta_j x_j + \sum_{k = 1}^s \sigma_k y_k = 0 \tag{A}$$

Then let

$$\tag{B}v := \sum_{i = 1}^n \alpha_i v_i + \sum_{j = 1}^r \beta_j x_j$$

Then we have $$v \in S_1$$ and

$$-v = - \sum_{k = 1}^s \sigma_k y_k \in S_2$$

So $$v \in S_2$$. This implies $$v \in S_1 \cap S_2$$. But then we have unique $$\gamma_1, \cdots, \gamma_n$$ such that

$$v = \sum_{i = 1}^n \gamma_i v_i \tag{C}$$

On the other hand, the linear combination of $$v$$ in equation $$(\mathrm{B})$$ is unique as well, because $$B_1$$ is a basis of $$S_1$$. Making the subsitution $$\alpha_i = \gamma_i$$, it follows immediately that

$$\beta_1 = \beta_2 = \cdots = \beta_r = 0 \tag{D}$$

Because of $$(\mathrm{D})$$, equation $$(\mathrm{A})$$ becomes:

$$\sum_{i = 1}^n \alpha_i v_i + \sum_{k = 1}^s \sigma_k y_k = 0 \tag{E}$$

But because $$B_2$$ is a basis, $$B_2$$ is also linearly independent. And it follows that $$(\mathrm{E})$$ implies

$$\alpha_1 = \alpha_2 = \cdots = \alpha_n = \sigma_1 = \sigma_2 = \cdots = \sigma_s = 0$$

And this proves that the vectors of $$B_4$$ are linearly independent. $$\blacksquare$$

Bonus: As an immediate corollary of this problem we have

$$\mathrm{dim}\; (S_1 + S_2) = \mathrm{dim}\; S_1 + \mathrm{dim}\; S_2 - \mathrm{dim}\; (S_1 \cap S_2)$$

• (D) follows becuase $\{x_1, \dots x_r\}$ is a linearly independent set since its a subset of $B_1$ right? – TAPLON Jan 17 '19 at 0:17
• @JustinStevenson You can write the unique linear combination in equation $(\mathrm{C})$ also as $v = \sum_{i = 1}^n \gamma_i v_i + \sum_{j = 1}^r 0 \cdot x_j$ and then compare with the same unique linear combination in equation $(\mathrm{B})$ – user635162 Jan 17 '19 at 0:20
• Ah yep, that makes sense. Thankyou! – TAPLON Jan 17 '19 at 0:22