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If $\alpha_1,\alpha_2,\alpha_3$ are linearly independent, and $(\beta_1,\beta_2,\beta_3) = (\alpha_1,\alpha_2,\alpha_3)C$. Prove that if $\mid C\mid \neq 0$, then $\beta_1,\beta_2,\beta_3$ are linearly independent.

This statement seems to be trivial, if $\alpha_1,\alpha_2,\alpha_3$ are linearly independent, then $\det{(\alpha_1,\alpha_2,\alpha_3)} \neq0$, and $\det{(\alpha_1,\alpha_2,\alpha_3) C} = \det{(\alpha_1,\alpha_2,\alpha_3)}*\det{}C\neq 0$, which implies $\det(\beta_1,\beta_2,\beta_3) \neq0$, and hence the columns are linearly independent.

I'm not sure how to prove it rigorously, or if I'm missing any steps here. Would really appreciate any help! Thank you!

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  • $\begingroup$ Just use the definition of linear independence: $\sum_i c_i\alpha_i=0$ holds for scalars $c_i$ not all zero; multiplying the sum by $C\neq 0$ gives $\sum_i c_i\beta_i=0$ (the scalars $c_i$ remain unchanged) which shows $\beta_1,\beta_2,\beta_3$ are linearly independent since not all $c_i$'s are zero. $\endgroup$ Commented Mar 12, 2018 at 0:22

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Let $$( \beta_1, \beta_2, \beta_3) x = 0$$

we want to show that $x=0$.

We have

$$( \alpha_1, \alpha_2, \alpha_3)C x = 0$$

Since $(\alpha_1, \alpha_2, \alpha_3)$ are linearly independent, we have $Cx=0$.

Since $\det(C) \ne 0$, we have $x=0$, hence $\beta_1,\beta_2,\beta_3$ are linearly independent.

Remark: $(\beta_1, \beta_2, \beta_3)$ need not be a square matrix. Hence determinant might not be well defined.

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