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I have this question that has me pretty stumped. It is as follows:

Assume $A$ is $3 \times 5, v_1, v_2 \in N(A)$ and are not multiples of each other. Assume $(Av_3, Av_4, Av_5)$ is linearly independent. Prove that $(v_1, v_2, v_3, v_4, v_5)$ is a basis of $\mathbb{R}^5.$"

I understand that $v_1$ and $v_2$ are not linear combinations of each other (right?) and $v_3,v_4$ and $v_5$ form a basis, but I don't know how to "word" this very well. Is it right to say that $v_1$ and $v_2$ add to the span of the sequence of vectors? Or that $v_1$ and $v_2$ can also form basis vectors?

I'm still pretty new to Linear and a little lost when it comes to this stuff. So please, please correct me if anything I said doesn't make sense or is incorrect.

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  • $\begingroup$ Hint: suppose they are linearly dependent. Then we have real numbers $\{a_i\}$,not all $0$, with $\sum a_i \vec {v_i}$=0. Now apply $A$ to that equation $\endgroup$ – lulu Feb 24 '17 at 19:35
  • $\begingroup$ What is the definition of $N(A)$? $\endgroup$ – Alexis Olson Feb 24 '17 at 19:36
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You just need to show that $(v_1, v_2, v_3, v_4, v_5)$ are linearly independent.

Since you have information about $Av_i$, the key will be to multiply a linear relation on $v_i$ by $A$.

Now suppose you had a relation $$c_1v_1 + c_2v_2 + c_3v_3 + c_4v_4 + c_5v_5 = 0$$ You want to show all the $c_i$ must be $0$. Multiplying through by $A$ you get a new dependence $$c_3Av_3 + c_4Av_4 + c_5Av_5 = 0$$ since $v_1, v_2$ are in the nullspace of $A$.

From the definition of linear independence, what can you say about $c_3, c_4, c_5$? Can you use this to show all $c_i$ are $0$?

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  • $\begingroup$ So, with my limited knowledge of linear algebra what I'm deducing from this is that if $c_1v_1 + c_2v_2 + c_3v_3 + c_4v_4 + c_5v_5 = 0$, and $c_3Av_3 + c_4Av_4 + c_5Av_5 = 0$ as well, then all $c_i$ must be zero by the definition of linear independence. Is that correct? $\endgroup$ – pstumps Mar 2 '17 at 19:06

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