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Is this multiplication formula in linear algebra true?

Started by June 24, 2014 04:44 PM
11 comments, last by JohnnyCode 10 years, 7 months ago
Can we put an end to this nonsense? Go learn some math. If you have some genuine questions, have some humility and we'll be happy to help you learn. But don't come with this attitude of "I don't believe what everyone says" or "some of theory axioms [sic] are excluding others".

Does anyone else smell troll?

But that is not possible since xT=!=x. What I find strange though (for example, dot product of x and xT is defined, thus their subtraction is defined, but thus subtraction of x and xT can never be a zero vector)

First, you cannot have a dot product of \(x\) and \(x^T\) (unless \(x \in \mathbb R\)). The dot product is for two column vectors \(u, v\) is commonly defined as \(dot(u, v) := u^T \cdot v\). Trying \(dot(x^T, x)\) would once again be impossible because you would have to multiply an element of \(\mathbb R^{n \times 1}\) with and element of \(\mathbb R^{n \times 1}\) which is not defined for \(n \neq 1\).

To every vector (matrix) exists addition neutral vector (matrix) Z
To every vector (matrix) exists multiplication neutral vector (matrix) I such that V*I=V
To every vector (matrix) exists multiplication inverse vector J (matrix) such that V*J=I

You are thinking of the ring axioms but that applies only to square matrices. You cannot even multiply general vectors together like that (as already said above), so talking about multiplicative neutral and multiplicative inverse elements for vectors does not make any sense.
In some fields (like GLSL) it can be convenient to define a component-wise vector multiplication. However, that is something completely orthogonal to standard linear algebra. The presence of this extra operator does not change anything about the theory of matrices nor does it negate associativity of matrix multiplication.

I can only reinforce Álvaro's suggestion to "go learn some math", ideally under supervision. You apparently heard of several concepts but you are lacking the attention to detail and strictness required in the field.

Álvaro: I often smell troll on these forums but with him I'm more inclined to believe he actually believes what he is saying.
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Can we put an end to this nonsense? Go learn some math. If you have some genuine questions, have some humility and we'll be happy to help you learn. But don't come with this attitude of "I don't believe what everyone says" or "some of theory axioms [sic] are excluding others".

Does anyone else smell troll?

I just wanted to talk about linear algebra with more competent people.
I was bringing my concerns to find out what what I missunderstand, not for attitude troll fun or something. For that I thank you

I now undestand that making multiplication relate to dimension shifting as transponing relates is bad idea to exist general, becouse linear algebra is a much wider discipline.

If I will happen to have some concerns again I will post them in a new topic of course

Thank you

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