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$\mathbb{R}^2$, with the real scalar multiplication but addition defined by $$ \begin{bmatrix}x_1 \\ y_1\end{bmatrix}+\begin{bmatrix}x_2 \\ y_2\end{bmatrix}=\begin{bmatrix}x_1+x_2+1 \\ y_1+y_2+1\end{bmatrix}$$

why or why not? I'm confused about how to go about this problem. I know the conditions that they must fulfill, but the specific addition definition is confusing me..

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2 Answers 2

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Hint: Check whether the scalar multiple $0v$ of some vector$~v$ is the neutral element for addition (this has to be true in a vector space for any vector $v$).

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  • $\begingroup$ What does neutral element mean? $\endgroup$
    – Candice
    Nov 12, 2015 at 19:56
  • $\begingroup$ The neutral element for an operation is one such that whenever you apply the operation to it and another value, the result is that other value. Like the zero vector (for the operation of addition) in a vector space. $\endgroup$
    – Marc van Leeuwen
    Nov 13, 2015 at 7:55
  • $\begingroup$ @MarcvanLeeuwen is that just another name for an identity? $\endgroup$
    – Q the Platypus
    Mar 11, 2016 at 7:11
  • $\begingroup$ @QthePlatypus: Probably yes, though it depends on what you mean by an identity; I'd say the term neutral element is more neutral about what the operation could be, and therefore more general. The basic meaning of identity is a map that returns its argument unchanged; this is indeed the neutral element for composition of maps. From there it is a small step to use "identity" for the neutral element for some multiplication (although "unit" is also a term used). However for addition "zero element" seems more appropriate; for instance the neutral element for matrix addition is the zero matrix. $\endgroup$
    – Marc van Leeuwen
    Mar 11, 2016 at 8:20
  • $\begingroup$ @MarcvanLeeuwen I was thinking identity in the sense of group theory. Seems we are in agreement though. $\endgroup$
    – Q the Platypus
    Mar 11, 2016 at 8:35
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You can go about this by checking if there exists a neutral element of the addition, i.e., a vector $z\in\mathbb R^2$ such that $x+z=x$ for all $x\in\mathbb R^2$. Just try any two vectors, e.g., $(1, 0)^T$ and $(1,1)^T$ and see if such $z$ can exist.

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