# Linear Independence and "Not in the Span"

I'm studying elementary linear algebra right now, and the current section is on linear independence. As I create matrices from the vectors and row reduce them in a calculator, I get various results.

Some row reduced forms give nonzero integers on the right hand side except the bottom entry, which I understand to be linear dependence. Other forms have all zeros on the right hand side, which I understand to be linear independence. Another instance is when the right hand side is all zeros except the one on the bottom. Normally, I know that this means no solution, but my professor has stated, "The vector is not in the span of the other vectors."

I want to know the mathematical difference between linear independence and "not being in the span." If the vector is not in the span, does that mean it's not linearly dependent? If it cannot be linearly dependent, why not just call it linearly independent since the linear combination will never equal 0 unless every coefficient is 0?

If there are any errors in my understanding or assumptions, please correct me and shed light on my ignorant mind as the final exam approaches.

Another question is, can linearly independent vectors be written as a linear combination? I thought that the coefficients cannot all be 0, but my textbook seems to think they can be. It claims that the zero vector is a linear combination of two vectors, its linear combination that can never be 0 unless every coefficient is 0.

Sorry for another edit. The questions are just piling up. My textbook says that a zero vector is a linear combination of some vectors, but another nonzero vector is not a linear combination. Why can't every coefficient be 0 to make the nonzero vector a linear combination as well?

• You might want to go back and look up the definitions of linear (in)dependence, linear combinations, and span.
– user137731
Mar 31 '15 at 2:04
• @Sam to your last question, I dont see how you are reasoning that a non zero vector can be written as a linear combonations of zeros.. you may need to really study from scratch and review the definitions in general Mar 31 '15 at 2:08
• @PersonaA I apologize. Allow me to clarify. The given vectors are (0, -2, 2) and (1, 3, -1). The questions asks whether (0, 0, 0) is a linear combination of the first two vectors, and the answer is yes. To me, the only way is for both coefficients to be zero. It bugs me that the question did not ask for independence or dependence but whether it's a linear combination. Are linear combinations not allowed to have all zeros? Which would then mean that all vectors can be linear combinations and that the question is flawed? Mar 31 '15 at 2:13
• I see, well yes the coeffecients being zero will always give you 0. There may be other coeffecients that give you zero as well though for example -1(1,0)+1(1,0). Mar 31 '15 at 2:14
• But as you will see as you continue, any set with the zero vector is automatically going to be depedent Mar 31 '15 at 2:15

Consider the set of vectors $\{\mathbf v_1, \mathbf v_2, \dots, \mathbf v_k\}$ in the vector space $V$ over the field $F$ (the vector space might be $\Bbb R^3$ for instance and the field (of scalars) might be $\Bbb R$). Here are some definitions:

• A linear combination of these $k$ vectors is another vector $\mathbf w=a_1\mathbf v_1 + a_2\mathbf v_2 + \cdots + a_k\mathbf v_k$, where $a_1, a_2, \dots, a_k$ are scalars.
• The span of these vectors is the set of all linear combinations of these vectors: i.e. $\operatorname{span}\{\mathbf v_1, \mathbf v_2, \dots, \mathbf v_k\} = \{\mathbf w \in V \mid w=a_1\mathbf v_1 + a_2\mathbf v_2 + \cdots + a_k\mathbf v_k \wedge a_1, a_2, \dots, a_k \in F\}$
• This set of vectors is a linearly independent set if the only linear combination of these vectors that produces the zero vector is the trivial one. I.e. $a_1\mathbf v_1 + a_2\mathbf v_2 + \cdots + a_k\mathbf v_k = 0 \implies a_1=a_2=\cdots=a_k=0$.
• A set of vectors is a linearly dependent set if it is not a linearly independent set.

Now let's show that a linearly dependent set has at least one vector which is a linear combination of the others. Let $\mathbf a,\mathbf b,\mathbf c \in \Bbb R^3$ be a set of linearly dependent vectors. Then by definition, the equation $$x\mathbf a+y\mathbf b+z\mathbf c=\mathbf 0$$ for scalars $x,y,z$ has more than one solution ($x=y=z=0$ is definitely a solution, but it's not the only one). Thus at least one of the scalars $x,y,z$ are nonzero. WLOG let's say it's $x$. Then we can rearrange this equation by subtracting all of the vectors except $x\mathbf a$ on both sides. $$x\mathbf a=-y\mathbf b-z\mathbf c$$ Now because $x \ne 0$, we can divide it on both sides to get $$\mathbf a=-\frac yx\mathbf b -\frac zx\mathbf c$$ Thus $\mathbf a$ is a linear combination of $\mathbf b$ and $\mathbf c$. You can see why this would fail in the linearly independent case -- you wouldn't be able to divide out any of the coefficients because they are all zero.

Now let's consider the set of vectors $\{\mathbf 0\}$. That is the set only containing the zero vector. Is this set linearly independent or linearly dependent? It is linearly dependent because $x\mathbf 0=\mathbf 0$ has infinitely many solutions. Likewise, any set which contains the zero vector will be a linearly dependent set (confirm this for yourself).

I now claim that the zero vector in a vector space $V$ is a linear combination of any non-empty set of of vectors in $V$. Can you see why that must be true?

Does this answer you questions or is there something else I need to hit on?

• Thank you for the definitions. They helped me better understand the concept. I guess the problem at hand is that the question in the book asks for linear combination and not linear dependence. According to your definitions, a linear combination is allowed to have all zero coefficients. The definition for linear combination in my book is as follows: If "w" is a vector in a vector space "V," then "w" is said to be a linear combination of the vectors v1, v2, ..., vr in "V" if "w" can be expressed in the form "w" = k1v1 + k2v2 + . . . + krvr where k1, k2, ..., kr are scalars. Mar 31 '15 at 2:35
• That's exactly the same as my definition.
– user137731
Mar 31 '15 at 2:38
• That's what I thought as well. If you don't mind, could you explain why (0, 4, 5) is not a linear combination of (0, -2, 2) and 1, 3, -1)? Yes, the questions asks for linear combination, not linear dependence. Mar 31 '15 at 2:39
• If it were then you could write $(0,4,5) = k_1(0,-2,2)+k_2(1,3,-1)$. To get the first coordinate correct you can see that $k_2$ would have to equal $0$. Thus we have $(0,4,5)=k_1(0,-2,2)$. Now can you find some $k_1$ such that this is true?
– user137731
Mar 31 '15 at 2:41
• It's not true in the case $k_1=0$ because then you have $(0,4,5)=0(0,-2,2)=(0,0,0)$. But clearly $(0,4,5)\ne(0,0,0)$. In fact there is no $k_1, k_2$ such that this is true. If these vector equations are confusing you just write them as a set of scalar equations. That is $(0,4,5)=k_1(0,-2,2)+k_2(1,3,-1) \iff \begin{cases} 0 = 0k_1+k_2 \\ 4=-2k_1+3k_2 \\ 5=2k_1 -k_2\end{cases}$. The first equation tells us that $k_2=0$. Plugging this into the second and third equations we get $k_1 = -2$ and $k_1=\frac 52$. But $k_1$ can't be two different values, so there's no solution to this system.
– user137731
Mar 31 '15 at 3:07