Linear independence of vectors over a subspace I cannot understand the difference between these two statements:


*

*vectors $g_1,...,g_k$ are linearly independent over the subspace $L\subset K$.

*vectors $g_1,...,g_k$ are linearly independent in the subspace $L$.
I have understood that $g_1,...,g_k$ are linearly independent over the subspace $L\subset K$ if the only linear combination of our vectors that belongs to $L$ is trivial. But this seems the common definition of linear independence.
Is there a difference between the above statements? I am studying "Linear Algebra by Shilov".
Thanks.
 A: Vectors $g_1,\dots,g_k$ are linearly independent if $\sum_{i=1}^k a_ig_i=0$ implies $ a_i=0$ for all $i$. As you say, Shilov defines $g_1,\dots,g_k$ to be linearly independent over $L$ if $\sum_{i=1}^ka_ig_i \in L$ implies $a_i=0$ for all $i$. The difference here is that we have replace "$=0$" with "$ \in L$" in our definition. Note that $g_1,\dots,g_k$ are linearly independent in the usual sense if and only if they are linearly independent over the subspace $\{0\}$. Moreover, if $g_1,\dots,g_k$ are linearly independent over any $L$, then they are also linearly independent in the usual sense since $0 \in L$. The converse is false: vectors that are linearly independent need not be linearly independent over a given subspace $L$.
For an illustration, consider $\mathbb R^3$ with the standard basis $\{e_1,e_2,e_3\}$. The vectors $\{e_1,e_2\}$ are linearly independent in the usual sense, and are linearly independent over $\langle e_3 \rangle$, since no linear combination of $e_1$ and $e_2$ with nonzero coefficients can a multiple of $e_3$. However, $e_1$ and $e_2$ are not linearly independent over subspace $\langle e_1+e_2\rangle$, since their sum lies in this subspace.
If I'm not mistaken, to say that $g_1, \dots, g_k$ are linearly independent in $L$ means that they are linearly independent, and that they lie in $L$. This implies that they are not linearly independent over $L$, since any linear combination of the $g_i$'s lies in $L$!
A: Georgi E. Shilov (Linear Algebra, Dover Publications, 1977) defines a set of vectors $\{{\mathbf g}_1, \dots, {\mathbf g}_k\}$ in linear space $K$ to be linearly independent over a subspace $L$ if the only linear combination of the vectors which produces a vector in $L$ has all of its combination coefficients vanish.  
Linear independence over a subspace implies ordinary linear independence.
Proof: Suppose $\sum_{i=1}^k \alpha_i {\mathbf g}_i = {\mathbf 0}$.  Since ${\mathbf 0} \in L$, the definition tells us that $\alpha_1 = \cdots = \alpha_k = 0$.  Conclude $\{{\mathbf g}_1, \dots, {\mathbf g}_k\}$ is a linearly independent set. $\blacksquare$ 
${\mathbf g}_j \notin L$ for each $j \in \{1, \dots, k\}$.
Proof: Suppose ${\mathbf g}_j \in L$.  Since the linear combination $\sum_{i=1}^k \delta_{ij} {\mathbf g}_i = {\mathbf g}_j$ is a vector in $L$, all of the Kronecker deltas must vanish, including $\delta_{jj}$. This contradicts $\delta_{jj} = 1$.  The supposition ${\mathbf g}_j \in L$ must have been false.  $\blacksquare$
We thus see that:
$\quad \bullet$ Each ${\mathbf g}_j$ must lie outside the subspace $L$.
$\quad \bullet$ The collection $\{{\mathbf g}_1, \dots, {\mathbf g}_k\}$ is linearly independent.
Shilov shows that, when $n = \mathsf{dim} K > \mathsf{dim} L = m$, a basis 
$\{{\mathbf f}_1, \dots, {\mathbf f}_m\}$ 
for $L$ can be completed to a basis 
$\{{\mathbf f}_1, \dots, {\mathbf f}_m, {\mathbf g}_1, \dots, {\mathbf g}_{n-m}\}$ 
for $K$ by adding $n - m $ vectors ${\mathbf g}_1, \dots, {\mathbf g}_{n-m}$ linearly independent over $L$.
A: In means ordinary independence, and over is restricted linear independence of vectors for given subspace. To see what this means consider $K\equiv\mathbb{R}^2$ and consider the subspace $L$ consisting zero vector alone. Now let $\left\{e_1,e_2,x\right\}$ where $x=[1\quad1]^T$. Now if you apply the definitions you will see that these vectors are linearly independent over $L$ but not linearly independent in $K$.
