3
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
3
  • $\begingroup$ (D) follows becuase $\{x_1, \dots x_r\}$ is a linearly independent set since its a subset of $B_1$ right? $\endgroup$ – TAPLON Jan 17 '19 at 0:17
  • $\begingroup$ @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 $(\mathrm{B})$ $\endgroup$ – user635162 Jan 17 '19 at 0:20
  • 1
    $\begingroup$ Ah yep, that makes sense. Thankyou! $\endgroup$ – TAPLON Jan 17 '19 at 0:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.