T31 vector subtraction is not commutative and not associative.

In summary, the conversation discusses the non-commutativity and non-associativity of vector subtraction, using the example of $u=\begin{bmatrix}2\\-3\\4\\2\end{bmatrix}$ and $v=\begin{bmatrix}-1\\5\\2\\-7\end{bmatrix}$. It is shown that if we replace the + with -, vector subtraction is not commutative and not associative. The conversation also discusses the use of negative scalers and how it does not change the fact that vector subtraction is non-commutative. Additionally, it is mentioned that subtraction is not commutative over complex numbers, which further proves the point that vector subtraction is not commutative.
  • #1
karush
Gold Member
MHB
3,269
5
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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
ok, that makes sense
 
  • #4
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?
 
  • #5
so if we have a negative scaler like - 2 we write it as $+(-2)$
 
  • #6
karush said:
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
 

Related to T31 vector subtraction is not commutative and not associative.

1. Why is T31 vector subtraction not commutative?

T31 vector subtraction is not commutative because the order in which the vectors are subtracted affects the result. In other words, if we have two vectors A and B, A - B may not be equal to B - A.

2. Can you provide an example of T31 vector subtraction not being commutative?

Yes, for example, let A = (1, 2, 3) and B = (4, 5, 6). A - B = (-3, -3, -3), while B - A = (3, 3, 3). Since these two results are not equal, T31 vector subtraction is not commutative.

3. Why is T31 vector subtraction not associative?

T31 vector subtraction is not associative because the grouping of vectors affects the result. In other words, (A - B) - C may not be equal to A - (B - C).

4. Can you provide an example of T31 vector subtraction not being associative?

Yes, for example, let A = (1, 2, 3), B = (4, 5, 6), and C = (7, 8, 9). (A - B) - C = (-3, -3, -3) - C = (-10, -11, -12), while A - (B - C) = (1, 2, 3) - (-3, -3, -3) = (4, 5, 6). Since these two results are not equal, T31 vector subtraction is not associative.

5. How does the non-commutativity and non-associativity of T31 vector subtraction affect scientific calculations?

The non-commutativity and non-associativity of T31 vector subtraction can affect the accuracy of scientific calculations that involve vector operations. It is important for scientists to be aware of these properties and carefully consider the order and grouping of vectors in their calculations to ensure accurate results.

Similar threads

Replies
24
Views
1K
  • Linear and Abstract Algebra
Replies
8
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
889
  • Linear and Abstract Algebra
Replies
6
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
5
Views
2K
  • Linear and Abstract Algebra
Replies
7
Views
919
  • Linear and Abstract Algebra
Replies
5
Views
1K
  • Linear and Abstract Algebra
Replies
5
Views
1K
Back
Top