We've said in linear algebra lectures that when some vectors are linearly independent then the set containing those vectors is as well, but not the other way around. Why is that so? What's the difference btw those vectors being l.i and the set containing them being l.i.?

  • $\begingroup$ There is no difference. $\endgroup$ – Kavi Rama Murthy Nov 5 '18 at 9:09

The important distinction is between the linear independence of a set of vectors and the linear independence of a list of vectors. For an easy example let's say we're working in $3$-dimensional space with basis vectors $\mathbf i$, $\mathbf j$ and $\mathbf k$. Then the list of vectors $(\mathbf j,\mathbf k,\mathbf i, \mathbf k)$ isn't linearly independent, simply because it contains $\mathbf k$ twice*.

But by definition a set can't contain duplicate elements. So the set $\{\mathbf j,\mathbf k,\mathbf i, \mathbf k\}$ is linearly independent because it is equal to the set $\{\mathbf i,\mathbf j,\mathbf k\}$

*The definition of linear independence of a list $(\mathbf v_i)$ is that there is a list of coefficients $(a_i)$ of the same length such that $\sum_ia_i\mathbf v_i=0$ and such that not all of the $a_i$ are $0$. In this case we are taking our coefficients to be $(0,1,0,-1)$, so $(\mathbf j,\mathbf k,\mathbf i, \mathbf k)$ isn't linearly independent because $0\mathbf j+1\mathbf k+0\mathbf i+(-1) \mathbf k=0$.

  • $\begingroup$ Got it, thanks a lot $\endgroup$ – user10606892 Nov 5 '18 at 12:25

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.