Question

If { V1,V2,V3 } is a linearly independent set in R3 , then so is { V1+V2-2V3 , V1-2V2+V3 , -2V1+V2+V3 }.

Answer 1

Answer on Question #40379, Math, Linear Algebra

If $\{V_1, V_2, V_3\}$ is a linearly independent set in $R^3$, then so is $\{V_1 + V_2 - 2V_3, V_1 - 2V_2 + V_3, -2V_1 + V_2 + V_3\}$.

Solution.

It's False.

Since $\{V_1, V_2, V_3\}$ is linearly independent, the only way to write the zero vector as a linear combination of $V_1, V_2, V_3$ is

$$0 \boldsymbol{V}_1 + 0 \boldsymbol{V}_2 + 0 \boldsymbol{V}_3 = \boldsymbol{0}$$

Consider writing the zero vector as a linear combination of $\{V_1 + V_2 - 2V_3, V_1 - 2V_2 + V_3, -2V_1 + V_2 + V_3\}$. That is, what $c_1, c_2, c_3$ satisfy

$$c_1 (V_1 + V_2 - 2V_3) + c_2 (V_1 - 2V_2 + V_3) + c_3 (-2V_1 + V_2 + V_3) = \boldsymbol{0}$$ $$V_1 (c_1 + c_2 - 2c_3) + V_2 (c_1 - 2c_2 + c_3) + V_3 (-2c_1 + c_2 + c_3) = 0$$

Since the set $\{V_1, V_2, V_3\}$ is linearly independent, we know that

$$\left\{ \begin{array}{l} c_1 + c_2 - 2c_3 = 0 \\ c_1 - 2c_2 + c_3 = 0 \\ -2c_1 + c_2 + c_3 = 0 \end{array} \right.$$

Solution of this equation is $c_2 = c_1, c_3 = c_1$

I can put $c_1 = 1$, so that $c_2 = 1$ and $c_3 = 1$. Which means, that exist the value of coefficients which is not equal to 0. But this contradicts our assumption.

Hence, the set is $\{V_1 + V_2 - 2V_3, V_1 - 2V_2 + V_3, -2V_1 + V_2 + V_3\}$ is not linearly independent.

Answer: the set is not linearly independent.