
# 10.3: From Dependent Independent

[ "article:topic", "authortag:waldron", "authorname:waldron" ]

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Now suppose vectors $$v_{1},\ldots, v_{n}$$ are linearly dependent,
$c^{1}v_{1} + c^{2}v_{2}+ \cdots +c^{n}v_{n}=0$
with $$c^{1}\neq 0$$.  Then:
$span \{v_{1},\ldots, v_{n}\} = span \{ v_{2},\ldots, v_{n}\}$
because any $$x\in span \{v_{1},\ldots, v_{n}\}$$ is given by
\begin{eqnarray*}
x &=& a^{1}v_{1} + \cdots a^{n}v_{n} \\
&=& a^{1}\left( -\frac{c^{2}}{c_{1}}v_{2}- \cdots -\frac{c^{n}}{c_{1}}v_{n} \right) + a^{2}v_{2} + \cdots + a^{n}v_{n} \\
&=& \left(a^{2}-a^{1}\frac{c^{2}}{c_{1}}\right)v_{2} + \cdots + \left(a^{n}-a^{1}\frac{c^{n}}{c_{1}}\right)v_{n}.
\end{eqnarray*}
Then $$x$$ is in $$span \{v_{2},\ldots, v_{n}\}$$.

When we write a vector space as the span of a list of vectors, we would like that list to be as short as possible (this idea is explored further in chapter 11).  This can be achieved by iterating the above procedure.

Example 110

In the above example, we found that $$v_{4}=v_{1}+v_{2}$$.  In this case, any expression for a vector as a linear combination involving $$v_{4}$$ can be turned into a combination without $$v_{4}$$ by making the substitution $$v_{4}=v_{1}+v_{2}$$.

Then:
\begin{eqnarray*}
S &=& span \{ 1+t , 1+t^{2}, t+t^{2}, 2+t+t^{2}, 1+t+t^{2} \} \\
&=& span \{ 1+t , 1+t^{2}, t+t^{2}, 1+t+t^{2} \}.
\end{eqnarray*}
Now we notice that $$1+t+t^{2}=\frac{1}{2}(1+t) +\frac{1}{2}(1+t^{2}) + \frac{1}{2}(t+t^{2})$$.  So the vector $$1+t+t^{2}=v_{5}$$ is also extraneous, since it can be expressed as a linear combination of the remaining three vectors, $$v_{1}, v_{2},v_{3}$$.  Therefore
$S = span \{ 1+t , 1+t^{2}, t+t^{2} \}.$

In fact, you can check that there are no (non-zero) solutions to the linear system
$c^{1}(1+t) + c^{2}(1+t^{2}) + c^{3}(t+t^{2})=0.$
Therefore the remaining vectors $$\{ 1+t , 1+t^{2}, t+t^{2} \}$$ are linearly independent, and span the vector space $$S$$.  Then these vectors are a minimal spanning set, in the sense that no more vectors can be removed since the vectors are linearly independent.
Such a set is called a $$\textit{basis}$$ for $$S$$.

Example 111

Let $$\mathbb{Z}_{2}^{3}$$ be the space of $$3\times 1$$ bit-valued matrices (i.e., column vectors).  Is the following subset linearly independent?
$\left\{ \begin{pmatrix}1\\1\\0\end{pmatrix}, \begin{pmatrix}1\\0\\1\end{pmatrix}, \begin{pmatrix}0\\1\\1\end{pmatrix} \right\}$

If the set is linearly dependent, then we can find non-zero solutions to the system:
$c^{1}\begin{pmatrix}1\\1\\0\end{pmatrix}+ c^{2} \begin{pmatrix}1\\0\\1\end{pmatrix} +c^{3} \begin{pmatrix}0\\1\\1\end{pmatrix}=0,$
which becomes the linear system
$\begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ \end{pmatrix}\begin{pmatrix}c^{1}\\c^{2}\\c^{3}\end{pmatrix}=0.$
Solutions exist if and only if the determinant of the matrix is non-zero.  But:
$\det \begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ \end{pmatrix} = 1 \det \begin{pmatrix} 0 & 1 \\ 1 & 1 \\ \end{pmatrix} -1 \det \begin{pmatrix} 1 & 1 \\ 0 & 1 \\ \end{pmatrix} = -1-1=1+1=0$
Therefore non-trivial solutions exist, and the set is not linearly independent.