Show that vectors $v_1$,...,$v_k \in \Bbb R^n$ are linearly dependent if and only if one of these vectors is a linear combination of the others.

I know that $v_1$,...,$v_k$ being linearly dependent means that there exists $c_1$,...,$c_k \in \Bbb R^n$, not all zero, such that $c_1v_1+\dots+c_1v_k = 0$ (vector). Also a linear combination of $v_1,\dots,v_k$ means that a vector $u = c_1v_1+\cdots+c_kv_k$. I am unsure of where to go with this information.


Chose an index $j$ such that $c_j\ne 0$ and then divide everything by $c_j$. What can you deduce from there?

  • $\begingroup$ so then (c1/cj)v1+...+vj+...+(ck/cj)vk = 0, then vj = -((c1/cj)v1+...+(ck/cj)vk), so vj is a linear combination of the others? $\endgroup$ – Tommy Dec 6 '16 at 17:10
  • $\begingroup$ @Tommy yes, exactly. $\endgroup$ – TZakrevskiy Dec 6 '16 at 17:10


  • For the forward direction, since one of the $c_i$'s is nonzero, you can solve for $v_i$ in terms of the other vectors (you can divide by $c_i$ because its nonzero).

  • For the backwards direction, if $v_i$ is a linear combination of the others, then $$ v_i=a_1v_1+a_2v_2+\cdots+a_{i-1}v_{i-1}+a_{i+1}v_{i+1}+\cdots a_nv_n. $$ Observe that $v_i$ is missing from the RHS. Put all your vectors on one side and show that you can write $0$ as a linear combination and some of the coefficients (at least one) is nonzero.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.