linearly independent

**antman** · March 18th 2009, 05:45 PM

Suppose that S={v1, v2, v3} is a linearly independent set of vectors in a vector space V. How do you show that T={w1, w2, w3} is also linearly independent , where w1=v1+v2+v3, w2=v2+v3, and w3=v3?

**Reckoner** · March 18th 2009, 06:17 PM

Originally Posted by antman

Suppose that S={v1, v2, v3} is a linearly independent set of vectors in a vector space V. How do you show that T={w1, w2, w3} is also linearly independent , where w1=v1+v2+v3, w2=v2+v3, and w3=v3?

Assume the contrary, and suppose that $T$ is linearly dependent. Then there must be some $c_1,\;c_2,\;c_3,$ not all zero, such that

$c_1w_1+c_2w_2+c_3w_3=0$

$\Rightarrow c_1(v_1+v_2+v_3)+c_2(v_2+v_3)+c_3v_3=0$

$\Rightarrow c_1v_1+c_1v_2+c_1v_3+c_2v_2+c_2v_3+c_3v_3=0$

$\Rightarrow c_1v_1+(c_1+c_2)v_2+(c_1+c_2+c_3)v_3=0.$

Since $v_1,\;v_2,\text{ and }v_3$ are linearly independent, we have

$\left\{\begin{array}{ccccccc}<br /> c_1&&&&&=0\\<br /> c_1&+&c_2&&&=0\\<br /> c_1&+&c_2&+&c_3&=0<br /> \end{array}\right.$

$\Rightarrow c_1=c_2=c_3=0.$ But this contradicts the requirement that $c_1,\;c_2,\;c_3$ are not all zero. Hence, our initial assumption is false, and $T$ is linearly independent.

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