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

  • $\begingroup$ The book is wrong. $\endgroup$ Feb 24 '19 at 23:35
  • $\begingroup$ that's what i thought.... $\endgroup$ Feb 24 '19 at 23:37

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.

  • 1
    $\begingroup$ 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. $\endgroup$ Feb 24 '19 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. $\endgroup$ Feb 24 '19 at 23:44
  • 1
    $\begingroup$ i would agree, except it doesn't specify that its equal to 1...unless its implicitly specified somehow... $\endgroup$ Feb 24 '19 at 23:46

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