Decide whether the specified vectors are a linearly independent system or not. $\frac{1}{3}, \quad \sqrt{2}$ in the $\mathbb{Q}$ vector space $\mathbb{R}$.

I only need to show, that $\lambda_1 \cdot \frac{1}{3} + \lambda_2 \cdot \sqrt{2} = 0$
If $\lambda_1 = \lambda_2 = 0$ is the only solution, then it's linear independent, otherwise dependent, right?

  • $\begingroup$ yes it is all you need to show. $\endgroup$ – Enkidu May 9 '19 at 8:44
  • $\begingroup$ Note that also $\lambda_i\in \mathbb{Q}$. $\endgroup$ – denklo May 9 '19 at 8:45

Let $\lambda_1 \cdot \frac{1}{3} + \lambda_2 \cdot \sqrt{2} = 0$ and $\lambda_1, \lambda_2 \in \mathbb Q$. Assume that $ \lambda_2 \ne 0$. Then we get $ \sqrt{2}=- \frac{\lambda_1}{3 \lambda_2}$.

Since $\frac{\lambda_1}{3 \lambda_2}$ is rational, we have a contradiction. Thus $ \lambda_2 = 0$. It follows that $ \lambda_1 = 0$.

Conclusion: $\frac{1}{3} , \sqrt{2} $ are linearly independent in the $\mathbb Q$- vector space $ \mathbb R.$


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