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From "Linear Algebra DeMystified", David McMahon, 2006, page 10, and page 234.

W is a set of vectors such that:

$$W = \{\ all\ vectors:\ \ [a\ \ b\ \ c]^T\ where\ a=b=c \}$$

Is W a vector space and subset of R^3?

Book says: NO! with out explaining why it not a vector space.

Why is this not a vector space?

considering vector space properties:

  1. closure under addition. (looks like it passes to me)

  2. closure under scalar multiplication (looks like it passes to me)

  3. associative addition

  4. commutative addition

  5. zero vector included. yep. (0,0,0)

  6. additive inverse.

  7. scalar multiply distributive

  8. scalar multiply associative

  9. identity element exists

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  • $\begingroup$ The book is wrong. $\endgroup$
    – Kavi Rama Murthy
    Feb 24, 2019 at 23:35
  • $\begingroup$ that's what i thought.... $\endgroup$
    – DiscreteMath
    Feb 24, 2019 at 23:37

1 Answer 1

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Which book? If, say, you are working over the reals, the set of all vectors of the type $\begin{bmatrix}a&a&a\end{bmatrix}^T$, with $a\in\mathbb R$ (which is your set) is not empty and it is closed with respect to addition and to the product by a scalar. Therefore, it is a vector space.

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    $\begingroup$ what would an example of non-closure under addition look like? $$ [a_1\ \ a_1\ \ a_1]^T + [a_2\ \ a_2\ \ a_2]^T = [a_1+a_2\ \ a_1+a_2\ \ a_1+ a_2]^T \in W$$ looks closed to me under addition. $\endgroup$
    – DiscreteMath
    Feb 24, 2019 at 23:40
  • $\begingroup$ Vectors of the type $\begin{bmatrix}a&b&c\end{bmatrix}^T$ with $a,b,c\in\mathbb R$ such that $a+b+c=1$ are not closed under addition. $\endgroup$
    – José Carlos Santos
    Feb 24, 2019 at 23:44
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    $\begingroup$ i would agree, except it doesn't specify that its equal to 1...unless its implicitly specified somehow... $\endgroup$
    – DiscreteMath
    Feb 24, 2019 at 23:46

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