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Let ${v_1, v_2, v_3}$ be a set of nonzero vectors in $R^m$ such that $v_i^Tv_j=0$ when i≠j. Show that the set is linearly independent. Hint: set $a_1 v_1 + a_2v_2 + a_3v_3=$ and consider $^T$.

According to the hint, I can only get $a_1^2v_1^Tv_1 + a_2^2v_2^Tv_2 + a_3^2v_3^Tv_3 = 0$. But then I hit a dead end. Who can help me continue? Plz..

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    What property does $v^Tv$ have when $v$ is a nonzero vector? What property does $a^2$ have when $a$ is a nonzero real number? – 2017-01-02
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    I'm afraid there is nothing said about that...the exercise consists of the title and the hint I have given. – 2017-01-02
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    Sherwin, I'm not referring to something said in the exercise, I'm referring to what $v^Tv$ and $a^2$ are, encouraging you to notice properties of the things in your computation. What to you get when you take a vector and compute $v^Tv$? Could it be $0$ when $v$ is nonzero, for example? – 2017-01-02
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    Oh, so because $v$ is nonzero, $v^Tv$ is also nonzero, and in turns all $a^2$ must be zero. thank you! – 2017-01-02
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    Not quite. $v^Tv$ isn't just nonzero. If that were all, then the terms could cancel out. – 2017-01-02
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    Hum, do you mean I should say $v^Tv$ is positive? – 2017-01-02
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    You got it! Positivity of $v^Tv$ is key, as is positivity of $a^2$ if $a\neq0$ – 2017-01-02

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For an alternate solution, suppose $$\sum_{i=1}^n a_iv_i=0$$ then $$0 = v_k^T0 = v_k^T\sum_{i=1}^n a_iv_i = \sum_{i=1}^n a_iv_k^Tv_i = a_kv_k^Tv_k$$ because all terms except the $k^{th}$ one vanish. Since $v_k^Tv_k > 0$, it follows that $a_k = 0$. This works for arbitrary $k$, so all coefficients are zero, so the vectors are linearly independent.

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    That's great, thank you! – 2017-01-02