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?

  • 1
    $\begingroup$ 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

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, $\{x_1,\cdots,x_5\}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.