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Let vectors $(a_1, a_2, ..., a_n)$ be linearly independent in some linear subspace over $\mathbb{R}$ (real numbers). Are vectors $(a_1, a_1+2a_2, a_1+2a_2+3a_3, ..., a_1+2a_2+...+na_n)$ linearly independent as well?

Can anyone help prove this? Can't really do it properly. Thanks alot

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    You can do it. Let the second list be $b_1,...,b_n$. Show that if $\lambda_1 b_a + \cdots + \lambda_n b_n = 0$ then you must have $\lambda_k = 0$. – 2017-01-31

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You just need to use the following trick:

Two elements $u,v$ are linearly independent if and only if $v,u-v$ are (prove this yourself).

Now, apply this to your situation. If you subtract one term from the next, you are left with showing that $a_1,2a_2,3a_3, \ldots, na_n$ are linearly independent.

Similarly, $u,v$ are linearly independent if and only if $au,bv$ are, for some (and hence any) non-zero constants $a,b$.

So the constants we choose are : $1,\frac 12, \frac 13 \ldots \frac 1n$. So we get that:$a_1, \frac 12 (2a_2),\frac 13 (3a_3), \ldots \frac 1n (na_n)$ are linearly independent. But then, this set is just $a_1, \ldots, a_n$. So you are done i.e. these are linearly independent.

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    for any *nonzero* constants $a,b$., small nitpick but important none the less. – 2017-01-31
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    @JMoravitz That is very very important.Thank you for pointing it out. I have edited the answer. – 2017-01-31