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I don't have an idea on how to prove it. What should be my approach?

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closed as off-topic by Namaste, Leucippus, Shailesh, Jyrki Lahtonen Apr 9 '17 at 5:22

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  • $\begingroup$ proof by contradiction $\endgroup$ – Exodd Apr 4 '17 at 17:12
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Here's an idea:

Start with:

$$a_1 v_1 + a_2v_2 + a_3v_3 = 0$$

Then apply $A$ on both sides to get:

$$A (a_1 v_1 + a_2v_2 + a_3v_3) = A(0)$$

Then by linearity of the matrix operation we have:

$$a_1 Av_1 + a_2Av_2 + a_3Av_3 = 0$$

But since $Av_1, Av_2, Av_3$ are linearly independent, we conclude that:

$$a_1 = a_2 = a_3 = 0 $$

as desired

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Hint : Suppose that $c_1v_1+c_2v_2+c_3v_3=0$ and apply $A$ on both side.

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