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Assume

  1. $x_1, x_2, x_3, x_4, x_5 \in \mathbb{R}^m$

  2. $(x_1, x_2, x_3)$ is linearly independent

  3. $(x_4, x_5)$ is linearly independent.

  4. $\text{Span}(x_1, x_2, x_3) \cap \text{Span}(x_4, x_5) = \{0_m\}$.

Prove that $(x_1, x_2, x_3, x_4, x_5)$ is linearly independent.

I have difficulty to answer this question, can someone please show me how to do this?

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    $\begingroup$ Suppose $c_1x_1+ c_2x_2+\cdots +c_5x_5=0$, you are going to show that all the $c$s are zero. Just write this equation down, manipulate and use the condition 4, then conditions 2 and 3. $\endgroup$ – Li Chun Min Apr 26 '17 at 2:03
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Let $\lambda_1, \cdots, \lambda_5 \in \mathbb{R}$. Suppose $\lambda_1x_1+\cdots+\lambda_5x_5 = 0$. If you can show that each $\lambda_i$ must be $0$, then $\{x_1,\cdots,x_5\}$ is linearly independent.

First by rearranging, you know that $\lambda_1x_1+\lambda_2x_2+\lambda3x_3=-\lambda_4x_4-\lambda5x_5$. The left side is in $\mathrm{span}\{x_1,x_2,x_3\}$ while the right side is in $\mathrm{span}\{x_4,x_5\}$. As they are both equal, they must be in the intersection, so $-\lambda_4x_4-\lambda5x_5=\lambda_1x_1+\lambda_2x_2+\lambda3x_3\in\mathrm{span}\{x_1,x_2,x_3\}\cap\mathrm{span}\{x_4,x_5\}$. This means they are both equal to $0$.

But you also know that if $\lambda_1x_1+\lambda_2x_2+\lambda3x_3 = 0$, then $\lambda_1 = \lambda_2 = \lambda_3 = 0$ because $\{x_1, x_2, x_3\}$ is linearly independent. Similarly you can show that $\lambda_4 = \lambda_5 = 0$. Together this establishes linear independence of the entire set, $\{x_1,\cdots,x_5\}$.

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