Could someone explain this statement from my textbook please?
Let U be an independent set, then any vector x that is an element of span(U) can be written uniquely as a linear combination of vectors of U.
I don't get it because, I thought independent sets had the exact characteristics of none of their vectors spanning any other vectors in the set. Wouldn't that mean there is no linear combination of some vectors in the set that will give another vector since no vector spans the other?
Thanks!
Best Answer
Your definition is correct: if $\{v_1,\ldots, v_n\}$ is a linearly independent set, then no $v_i$ can be written as a linear combination of the other vectors in the set. Call this "definition 1."
Note that this is equivalent to saying that the zero vector can be uniquely written as a linear combination of the $v_i$, namely $\vec{0} = 0\cdot v_1+0\cdot v_2+\cdots+0\cdot v_n$. Call this "definition 2."
[To show definitions 1 and 2 are equivalent, just do some rearranging. For example, if $v_1=v_2+2v_3$, this violates linear independence in definition 1. Rearranging gives $\vec{0}=-v_1+v_2+2v_3$, which also violates linear independence in definition 2.]
Now, consider the statement in the book. If $x$ is in the span of $U=\{v_1,\ldots,v_n\}$, then by definition it can be written as a linear combination of elements in $U$. In particular, $x$ may not be in $U$, as pointed out by lulu, so this does not violate definition 1, which I think is your confusion. Now, to show that this linear combination is unique, you need to use linear independence.
Suppose for sake of contradiction $x$ can be expressed as a linear combination in more than one way (not unique): \begin{align} x&=a_1v_1+\cdots+a_n v_n\\ x&=b_1v_1+\cdots +b_n v_n \end{align} Then subtracting the two equations gives $$\vec{0} = (a_1-b_1) v_1 + \cdots + (a_n-b_n) v_n.$$ Why does this violate linear independence? (See definition 2.)