# [SOLVED]T31 vector subtraction is not commutative and not associative.

#### karush

##### Well-known member
Prove, by giving counterexamples, that vector subtraction is not commutative
and not associative.

ok I read all I could on trying to understand this but seem to not see something simple
if we have the example of

 $u=\begin{bmatrix}2\\-3\\4\\2\end{bmatrix} v=\begin{bmatrix}-1\\5\\2\\-7\end{bmatrix} u+v=\begin{bmatrix}2\\-3\\4\\2\end{bmatrix}+\begin{bmatrix}-1\\5\\2\\-7\end{bmatrix} =\begin{bmatrix}2+(-1)\\-3+5\\4+2\\2+(-7)\end{bmatrix} =\begin{bmatrix}1\\2\\6\\-5\end{bmatrix}$ if we replace the + with - does that mean it is not commutative and not associative.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
If vector subtraction would be commutative, we would have $u-v=v-u$.
Suppose we substitute your $u$ and $v$ into it, does the equality hold then?

If vector subtraction would be associative, we would have $u-(v-w)=(u-v)-w$.
So we need example vectors $u,v,w$ such that those are not equal.

#### karush

##### Well-known member
ok, that makes sense

#### Country Boy

##### Well-known member
MHB Math Helper
I don't understand why you added vectors. Why didn't you try subtracting as you were asked to?
Commutative:
$\begin{bmatrix}2\\ -3 \\ 4 \\2 \end{bmatrix}- \begin{bmatrix} -1\\ 5 \\ 2 \\-7\end{bmatrix}= ?$

$\begin{bmatrix} -1\\ 5 \\ 2 \\-7\end{bmatrix}- \begin{bmatrix}2\\ -3 \\ 4 \\2 \end{bmatrix} = ?$

Are they the same?

Associative:
$\left(\begin{bmatrix}2\\ -3 \\ 4 \\2 \end{bmatrix}- \begin{bmatrix} -1\\ 5 \\ 2 \\-7\end{bmatrix}\right)$$- \begin{bmatrix}5\\ -3 \\ 7 \\1 \end{bmatrix}= ?$

$\begin{bmatrix}2\\ -3 \\ 4 \\2 \end{bmatrix}- \left( \begin{bmatrix} -1\\ 5 \\ 2 \\-7\end{bmatrix}- \begin{bmatrix}5\\ -3 \\ 7 \\1 \end{bmatrix}\right)= ?$

Are they the same?

#### karush

##### Well-known member
so if we have a negative scaler like - 2 we write it as $+(-2)$

#### topsquark

##### Well-known member
MHB Math Helper
so if we have a negative scaler like - 2 we write it as $+(-2)$
Honestly any way you write it it is still non-commutative, though it does help to remind you that you are subtracting and not adding because vector addition is commutative.

I realize that the problem is simply making a point about vector arithmetic, but the whole issue is that subtraction is not commutative over complex numbers. $$\displaystyle a_{ij} - b_{ij} \neq b_{ij} - a_{ij}$$ for any complex numbers.

-Dan