# Question #13832

Jan 17, 2017

See below.

#### Explanation:

This set of n columns vectors, namely :

$A = \setminus \left\{m a t h b f {v}_{1} , m a t h b f {v}_{2} , \ldots . , m a t h b f {v}_{n} \setminus\right\}$

..where:

$\setminus m a t h b f {v}_{k} = \left(\begin{matrix}{a}_{1 k} \\ \ldots \\ {a}_{n k}\end{matrix}\right)$

....forms this nxn matrix:

$A = \left(\begin{matrix}{a}_{11} \ldots \ldots {a}_{1 n} \\ \ldots \ldots \ldots . \\ {a}_{n 1} \ldots . . {a}_{n \setminus n}\end{matrix}\right)$

.
The set's span is all possible linear combinations of the column vectors, ie

${\sum}_{i = 1}^{n} {c}_{i} {v}_{i}$ where ${c}_{i} \in m a t h c a l R , k \in m a t h c a l Z$

Taking 2 simple examples to illustrate how the import of that turns completely on the singularity or otherwise of matrix A, and also how practically to go about working a span:

• Example 1

$A = \left(\begin{matrix}1 & 2 \\ 3 & 5\end{matrix}\right)$

$S p a n \left(A\right)$ follows from the set of all solutions to:

${c}_{1} \left(\begin{matrix}1 \\ 3\end{matrix}\right) + {c}_{2} \left(\begin{matrix}2 \\ 5\end{matrix}\right) = \left(\begin{matrix}x \\ y\end{matrix}\right)$

Which is the same as solving:

$\left(\begin{matrix}1 & 2 \\ 3 & 5\end{matrix}\right) \left(\begin{matrix}{c}_{1} \\ {c}_{2}\end{matrix}\right) = \left(\begin{matrix}x \\ y\end{matrix}\right)$

And which solves as: ${c}_{1} = - 5 x + 2 y , {c}_{2} = 3 x - y$

In other words, for any $\left\langle x , y\right\rangle$ you specify, there exists a $\left({c}_{1} , {c}_{2}\right)$ pairing that includes that point within the span of A. $S p a n \left(A\right) = m a t h c a l {R}^{2}$.

And:
$\left(\begin{matrix}1 & 2 \\ 3 & 5\end{matrix}\right) \left(\begin{matrix}{c}_{1} \\ {c}_{2}\end{matrix}\right) = \left(\begin{matrix}0 \\ 0\end{matrix}\right) \implies \left({c}_{1} , {c}_{2}\right) = \left(0 , 0\right)$

It is no accident that A is non-singular. $A \setminus m a t h b f c = m a t h b f x$ has only one solution which is $\setminus m a t h b f c = {A}^{- 1} m a t h b f x$. $A \setminus m a t h b f c = m a t h b f 0$ has one solution which is $\setminus m a t h b f c = m a t h b f 0$

Example 2

$A = \left(\begin{matrix}1 & 2 \\ 3 & 6\end{matrix}\right)$

This is clearly singular as column 2 is a multiple of column 1, ie: $A = \left(\begin{matrix}1 & \left(1 \times 2\right) \\ 3 & \left(3 \times 2\right)\end{matrix}\right)$

Without even trying to solve it, we know that $\left\langle x , y\right\rangle$'s will lie only along direction: $\left(\begin{matrix}1 \\ 3\end{matrix}\right)$

Because A is singular: $A \setminus m a t h b f c = m a t h b f x$ will have no solution or infinitely many: $\det \left(A\right) = 0$ so ${A}^{- 1}$ does not exist.

In addition, $A \setminus m a t h b f c = m a t h b f 0$ will have infinitely many solutions:

$\left(\begin{matrix}1 & 2 \\ 3 & 5\end{matrix}\right) \left(\begin{matrix}{c}_{1} \\ {c}_{2}\end{matrix}\right) = \left(\begin{matrix}0 \\ 0\end{matrix}\right) \implies {c}_{1} = - 2 {c}_{2}$

$S p a n \left(A\right) = m a t h c a l R$.