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First, I'll try not to ramble, although it tends to happen when I type.

I have the following linear algebra problem for my homework.

Prove or give a counterexample: If $v_1, v_2, \ldots , v_m$ are linearly independent vectors in $V$, then $5v_1 − 4v_2, v_2, \ldots , v_m$ is linearly independent.

My first thought was: "It's obviously not linearly independent. $5v_1 - 4v_2$ is a linear combination of $v_1$ and $v_2$."

But then I noticed that $v_1$ wasn't actually in that set, so then I got to thinking that because we're getting a new vector from one vector that's already in the set, and one vector that's outside the set, but that would be independent of the rest if it was in it, then that resulting vector ($5v_1 - 4v_2$) is also independent.

Which one is correct? My gut is telling me that the second one is correct. But I can't think of a way to really prove that $5v_1 - 4v_2$ definitely can't be made by a linear combination of any other vectors in that set.

Any help would be appreciated. Thanks.

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    $\begingroup$ Your reasoning is sound, but that's not quite a proof. You usually prove things are linearly independent by making a linear combination and setting it equal to zero, so give that a try. In this case, it should be rather direct once you set up the form. If you like, I can be more specific in an answer. $\endgroup$
    – walkar
    Commented Jan 30, 2015 at 3:38

1 Answer 1

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In some cases $\def\v{{\bf v}}5\v_1-4\v_2$ might be expressible in terms of $\v_2,\ldots,\v_m$. But the important thing to realise is that in this problem it is given that $\v_1,\ldots,\v_m$ are independent - that makes all the difference.

To see whether $5\v_1-4\v_2,\v_2,\ldots,\v_m$ are independent, make a linear combination equal to the zero vector: $$\lambda_1(5\v_1-4\v_2)+\lambda_2\v_2+\cdots+\lambda_m\v_m={\bf0}\ .\tag{$*$}$$ Rearranging terms, $$(5\lambda_1)\v_1+(\lambda_2-4\lambda_1)\v_2+\lambda_3\v_3+\cdots+\lambda_m\v_m={\bf0}\ .$$ $\color{red}{\hbox{Since $\v_1,\ldots,\v_m$ are independent}}$ (the key fact), all the coefficients on the LHS must be zero: $$5\lambda_1=0\ ,\quad \lambda_2-4\lambda_1=0\ ,\quad \lambda_3=0\ ,\ldots,\quad \lambda_m=0\ .$$ You can easily see from this that $$\lambda_1=0\ ,\quad \lambda_2=0\ ,\quad \lambda_3=0\ ,\ldots,\quad \lambda_m=0\ .$$ So the only way $(*)$ can hold is when all the scalar coefficients are zero; so the vectors are linearly independent.

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