If one vector in a linearly independent list is replaced by a linear combination of itself and one other vector from that list is the resulting list always linearly independent? It seems to me that the answer is yes (as long as some non-zero scalar multiple of that first vector is included in the linear combination).
Consider a linearly independent list $(v_{1},v_{2},v_{3})$
Linear independence means that:
$a_{1}v_{1}+a_{2}v_{2}+a_{3}v_{3}=0$ only if $a_{1}=a_{2}=a_{3}=0$.
If we replace $v_{1}$ with a linear combination of itself and $v_{2}$ our list becomes $(b_{1}v_{1}+b_{2}v_{2}, v_{2}, v_{3})$.
To show that this list is also linearly independent we have to ask if the only linear combination of its elements that equals zero has all-zero coefficients. Which is obviously the case because:
$a_{1}(b_{1}v_{1}+b_{2}v_{2})+a_{2}v_{2}+a_{3}v_{3}=0$ and simply rearranging gets us to ($a_{1}b_{1})v_{1}+(a_{1}b_{2}+a_{2})v_{2}+a_{3}v_{3}=0$ and we already know all those coefficients $=0$. And so as long as $b_{1}\neq0$ then all of our $a$ coefficients must be $0$ so the new list is linearly independent.
Is this correct?
Note: I'm assuming that none of the vectors in the original list are $0$ because we've called that list linearly independent (according to Axler in Linear Algebra Done Right 3rd ed. p33 "Every list of vectors in $V$ containing the $0$ vector is linearly dependent.")
For context, I'm trying to generalize from the exercises Axler provides. More specifically, I'm trying to develop some kind of spatial intuition for all of this instead of just relying on rearranging terms and checking off definitions.
Perhaps there are more fundamental tools for describing what's going on here? A way to describe a vector's contribution to the dimension and span of a list that can't be undone by linear combinations with other vectors in that list? Something analogous to degrees of freedom in statistics?
(I'm studying this stuff on my own, so the ability to post and get feedback here has been invaluable!)
Best Answer
Ironically the spatial intuition you seek can be supplied by determinants, which Axler banishes to the end of his book. I'll just provide an informal sketch over the base field $\mathbb{R}$, but encourage you to read more elsewhere.
The three linearly independent vectors $v_1,v_2,v_3$ span a $3$-dimensional subspace $U$, to which we can restrict attention. A determinant function $D:U^3\to\mathbb{R}$ on $U$ measures the oriented (i.e. signed) volume $D(u_1,u_2,u_3)$ of the parallelepiped formed by the vectors $u_1,u_2,u_3$ in $U$, relative to some fixed chosen unit volume. One fundamental property of $D$ is that it's invariant under shears, for example: $$D(u_1+\alpha_2u_2+\alpha_3u_3,u_2,u_3)=D(u_1,u_2,u_3)\tag{1}$$ for any choice of $\alpha_2,\alpha_3\in\mathbb{R}$. This is because the orientation and volume of the parallelepiped determined by $u_1,u_2,u_3$ are unchanged if we move $u_1$ in any direction along the $2$-dimensional "base" spanned by $u_2,u_3$ -- in this case both the "base" and the "height" over the base are unchanged. Of course nothing is special about $u_1$, and similar shear invariance holds for each $u_i$ relative to the others.
Another fundamental property is scaling, for example: $$D(\alpha_1 u_1,u_2,u_3)=\alpha_1 D(u_1,u_2,u_3)\tag{2}$$ The intution here should hopefully be clear. Note that if $\alpha_1<0$ then the orientation (sign) changes because in this case $\alpha_1u_1$ is on the "other side" of the "base" spanned by $u_2,u_3$ than $u_1$. Again a similar property holds for each $u_i$.
It turns out these properties uniquely characterize $D$ up to the choice of the unit volume (i.e. any other function which also satisfies these properties is just a scalar multiple of $D$). Also, we obtain the important result that $D(u_1,u_2,u_3)\ne 0$ if and only if $u_1,u_2,u_3$ are linearly independent. This intuitively makes sense because $u_1,u_2,u_3$ form a non-degenerate parallelepiped (i.e. are not contained in a plane) if and only if $u_1,u_2,u_3$ are linearly independent.
How does all this relate to your problem? Well, we know that $D(v_1,v_2,v_3)\ne 0$ since $v_1,v_2,v_3$ are linearly independent, and by (1) and (2) $$D(\alpha_1v_1+\alpha_2v_2,v_2,v_3)=\alpha_1 D(v_1,v_2,v_3)$$ If $\alpha_1\ne 0$, then the right-hand side is nonzero, so the left-hand side is nonzero, which by the result above implies that the vectors $\alpha_1v_1+\alpha_2v_2,v_2,v_3$ are linearly independent.