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I was trying to solve this exercise:

Decide whether or not the set $\Bbb R^2$, with addition defined by

<p>$$(x,y) + (a,b) = (x+a+1, y+b)$$ </p>

<p>and with scalar multiplication $$r\cdot(x,y) = (rx+r-1, ry)$$ </p>

<p>is a (real) vector space.</p>

So, I try to prove via its space's properties (associatity, conmutativity, etc) and it works. Even preservation of scale. But when I try to prove for zero $(0\cdot v=0)$ It doesn't work for me, I mean $0\cdot(x, y) = (-1, 0)$ which is clearly not $(0,0)$

But book's answer is: it is an space, so I must be wrong. Now, thinking about it, I'm starting to believe that zero vector of vector spaces isn't necessary $(0,0,\dots,0)$. And then for this example $(-1,0)$ is the zero of the space, and it can be found taking any space element and doing $x - x$ ($x$ minus itself).

Am I right? Or am I omitting something more obvious. Thanks.

Gold
  • 26,547
Randolf Rincón Fadul
  • 762

6 Answers6

5

To find the zero vector, remember that the null vector of a vector space $V$ is a vector $0_V$ such that for all $x\in V$ we have $x + 0_V = x$.

So, let $V = \mathbb{R}^2$ and let $x= (x_1, x_2)$ and $0_V=(a,b)$, hence we must have

$$(x_1,x_2)+(a,b)=(x_1,x_2)$$

However, by the operations defined, the LHS becomes:

$$(x_1+a+1, x_2+b) = (x_1, x_2)$$

And this gives $a+1 = 0$ and $b=0$. So the null vector is really $(-1, 0)$. The point is: the null vector is defined by properties, axioms, things it must satisfy. It's just $(0,0,\dots,0)$ in the case of $\mathbb{R}^n$ with usual operations. You change operations and change the set, you'll get a different zero vector.

Gold
  • 26,547
3

You are right the zero of a vector space is not necessarily the $(0,0,0,0,\dots,0)$

Becareful, the additive neutral element, the so called zero, is in your vector space $(-1,0)$

Dominic Michaelis
  • 19,935
3

Now, thinking about it, I'm starting to believe that zero vector of vector spaces isn't necessary [0,0,...,0].

You are correct; the zero of a vector space is determined by its arithmetic properties, not by some accident of how you decided to write it. If you believe that $[-1,0]$ is the zero, then check that it satisfies the properties that a zero must satisfy!

3

Your computation $0\cdot [x,y] = [-1,0]$ shows that if your space is indeed a vector space, then the "zero vector" in your vector space is $[-1,0]$.

While this looks strange, this is not yet a contradiction. Remember that the zero vector is the neutral element with respect to vector addition.

So let's check if $[-1,0]$ is the neutral element by vector addition. Indeed: $$[x,y] + [-1,0] = [x-1+1, y+0] = [x,y]$$ and $$[-1,0] + [x,y] = [-1+x+1,0+y] = [x,y]$$

azimut
  • 22,696
1

Hint: $\vec{0}$ is the unique element that satisfies $\vec{0}+\vec{x} = \vec{x}$ for all $\vec{x}$ in the vector space. You will also need to check that $\vec{0} \cdot \vec{x} = \vec{0}$.

Note that as you observed $\vec{0}$ does not have to be $[0,0]$ in your case. One way to find it, is indeed to consider $\vec{x} - \vec{x}$...

gt6989b
  • 54,422
0

You do not have to show that $0\cdot \vec{v} = \vec{0}$ for any given $\vec{v}$ in your vector space. You need to show that the set in question is closed with respect to scalar multiplication. Hence it is only required that $0\cdot \vec{v}$ is still in your set (which it is).