Linear Independence Proof | Generated by AI
Translation:
- Given that the vectors \( \alpha_1, \alpha_2, \alpha_3 \) are linearly independent, and the vectors \( \beta_1 = \alpha_1 + 2\alpha_2 \), \( \beta_2 = -\alpha_1 + \alpha_2 - 3\alpha_3 \), \( \beta_3 = 3\alpha_1 + 6\alpha_3 \). Prove that the vectors \( \beta_1, \beta_2, \beta_3 \) are linearly related.
Solution:
To prove that the vectors \( \beta_1, \beta_2, \beta_3 \) are linearly related, we need to show that there exist scalars \( c_1, c_2, c_3 \) (not all zero) such that:
\[ c_1 \beta_1 + c_2 \beta_2 + c_3 \beta_3 = 0 \]
Given: \[ \beta_1 = \alpha_1 + 2\alpha_2 \] \[ \beta_2 = -\alpha_1 + \alpha_2 - 3\alpha_3 \] \[ \beta_3 = 3\alpha_1 + 6\alpha_3 \]
We need to find constants \( c_1, c_2, c_3 \) such that:
\[ c_1 (\alpha_1 + 2\alpha_2) + c_2 (-\alpha_1 + \alpha_2 - 3\alpha_3) + c_3 (3\alpha_1 + 6\alpha_3) = 0 \]
This can be rewritten as:
\[ (c_1 - c_2 + 3c_3)\alpha_1 + (2c_1 + c_2)\alpha_2 + (-3c_2 + 6c_3)\alpha_3 = 0 \]
Since \( \alpha_1, \alpha_2, \alpha_3 \) are linearly independent, the coefficients of \( \alpha_1, \alpha_2, \alpha_3 \) must each be zero:
- \( c_1 - c_2 + 3c_3 = 0 \)
- \( 2c_1 + c_2 = 0 \)
- \( -3c_2 + 6c_3 = 0 \)
Let’s solve this system of equations:
From equation 2: \[ 2c_1 + c_2 = 0 \implies c_2 = -2c_1 \]
From equation 3: \[ -3c_2 + 6c_3 = 0 \implies -3(-2c_1) + 6c_3 = 0 \implies 6c_1 + 6c_3 = 0 \implies c_3 = -c_1 \]
Substitute \( c_2 = -2c_1 \) and \( c_3 = -c_1 \) into equation 1: \[ c_1 - (-2c_1) + 3(-c_1) = 0 \implies c_1 + 2c_1 - 3c_1 = 0 \implies 0 = 0 \]
This holds true for any \( c_1 \). Let’s choose \( c_1 = 1 \):
Then: \[ c_2 = -2 \] \[ c_3 = -1 \]
Thus, we have found constants \( c_1 = 1, c_2 = -2, c_3 = -1 \) such that:
\[ 1 \cdot \beta_1 - 2 \cdot \beta_2 - 1 \cdot \beta_3 = 0 \]
This proves that the vectors \( \beta_1, \beta_2, \beta_3 \) are linearly related.