# Independence, Inverse and Additive Identity for vector with defined vector addition and scalar multiplication

I am struggling to find my bearings on this question. I am confident that I can do parts a and d. I have no clue how to approach b. I was also wondering if the redefined vector addition and scalar multiplication will effect part c?

If you could help me by explaining how to approach each part that would be lovely.Thanks in advance!

Here is the question.

Let $$V=\left\{(x,y,z) \mid x,y,z \in \mathbb R, x>0 \right\}$$ be a set with vector addition defined by $$(x_1,y_1,z_1) + (x_2,y_2,z_2) = (x_1x_2, (y_1^3 +y_2^3)^{1/3}, z_1+z_2+1)$$ and scalar multiplication defined by

$$\alpha(x,y,z) = (x^\alpha,\alpha^\frac{1}{3}y, \alpha z+ \alpha -1).$$

a) What is the additive identity on $V$?

b) What is the inverse of $(x,y,z)$ in $V$?

c) The list $((1,1,1),(1,1,4),(1,0,2),(1,1,-2))$ is linearly dependent in $V$. Reduce it to a linear independent list in $V$ that spans the same subspace as the original list.

d) Either write $(2,0,1)$ as a linear combination of the the elements of your answer to part c or show it is outside the span of your answer.

• Where did you get that exercise from?
– mvw
Commented Feb 8, 2016 at 5:34

Hints:

a) Find $z \in V$ with $v \oplus z = v$ for all $v \in V$

b) The additive inverse $v_i$ of $v$ fulfills $v \oplus v_i = z$

c) If $S$ is the subspace spawned by the original list $L$ of vectors, remove vectors until you end up with a linear independent list $L'$ of the remaining vectors, which still span $S$. The span of the $L$ vectors is $$S = \{ (\alpha_1 \odot v_1) \oplus (\alpha_2 \odot v_2) \oplus (\alpha_3 \odot v_3) \oplus (\alpha_4 \odot v_4) \mid v_i \in V, \alpha_i \in \mathbb{R} \}$$ but note that the given non-standard addition and scalar multiplication has to be used.

d) This runs down to check if you can come up with the coefficients for a linear combination of vectors from $L'$ that result in the given vector.

Tackling c):

1. The zero vector is $z = (1, 0, -1)$.

2. The one dimensional subspaces are $$\alpha \odot v = (x^\alpha, \alpha^{1/3} y, \alpha (1 + z) - 1)$$

3. The linear combination of two vectors is \begin{align} (\alpha_1 \odot v_1) \oplus (\alpha_2 \odot v_2) &= (x_1^{\alpha_1}, \alpha_1^{1/3} y_1, \alpha_1 z_1 + \alpha_1 - 1) \oplus (x_2^{\alpha_2}, \alpha_2^{1/3} y_2, \alpha_2 z_2 + \alpha_2 - 1) \\ &= ( x_1^{\alpha_1}x_2^{\alpha_2}, (\alpha_1 y_1^3 + \alpha_2 y_2^3)^{1/3}, \alpha_1 (1 + z_1) + \alpha_2 (1 + z_2) - 1 ) \end{align} They are linear independent if $$z = (\alpha_1 \odot v_1) \oplus (\alpha_2 \odot v_2)$$ only for $\alpha_1 = \alpha_2 = 0$. This means we have to examine $$x_1^{\alpha_1}x_2^{\alpha_2} = 1 \\ (\alpha_1 y_1^3 + \alpha_2 y_2^3)^{1/3} = 0 \\ \alpha_1 (1+z_1) + \alpha_2 (1+z_2) - 1 = -1$$ or $$x_1^{\alpha_1}x_2^{\alpha_2} = 1 \\ \alpha_1 y_1^3 + \alpha_2 y_2^3 = 0 \\ \alpha_1 (1+z_1) + \alpha_2 (1+z_2) = 0$$

4. For $v_1 = (1,1,1)$ and $v_2 = (1,1,4)$ this turns into $$1^{\alpha_1} 1^{\alpha_2} = 1 \\ \alpha_1 1^3 + \alpha_2 1^3 = \alpha_1 + \alpha_2 = 0 \\ 2 \alpha_1 + 5 \alpha_2 = 0$$ and has only $\alpha_1 = \alpha_2 = 0$ as solution, so $v_1$ and $v_2$ are linear independent.

5. For $v_1$, $v_2$ and $v_3 = (1,0,2)$ we check the linear combination of three vectors $$x_1^{\alpha_1} x_2^{\alpha_2} x_3^{\alpha_3} = 1 \\ \alpha_1 y_1^3 + \alpha_2 y_2^3 + \alpha_3 y_3^3 = 0 \\ \alpha_1 (1+z_1) + \alpha_2 (1+z_2) + \alpha_3 (1+z_3) = 0$$ which turns into $$1^{\alpha_1} 1^{\alpha_2} 1^{\alpha_3} = 1 \\ \alpha_1 1^3 + \alpha_2 1^3 = \alpha_1 + \alpha_2 = 0 \\ 2 \alpha_1 + 5 \alpha_2 + 3 \alpha_3 = 0$$ which leads to $$\alpha_1 = - \alpha_2 \\ \alpha_3 = - \alpha_2$$ so we have the solutions $(\alpha_1, \alpha_2, \alpha_3) = (t, -t, t)$ for $t\in \mathbb{R}$. Which means the three vectors are linear dependent, because non-zero solutions for the coefficients $\alpha_i$ exist, e.g. $(1,-1,1)$. This means $v_3 = v_2 \oplus (-1 \odot v_1)$.
Test: $v_2 \oplus (-1 \odot v_1) = (1,1,4) \oplus (-1 \odot (1,1,1)) = (1,1,4) \oplus (1,-1,-3) = (1, 0, 2)$.

6. We next check the linear combinations of $v_1$, $v_2$, $v_4 = (1,1,-2)$: $$1^{\alpha_1} 1^{\alpha_2} 1^{\alpha_3} = 1 \\ \alpha_1 1^3 + \alpha_2 1^3 + \alpha_3 1^3 = \alpha_1 + \alpha_2 + \alpha_3 = 0 \\ 2 \alpha_1 + 5 \alpha_2 - \alpha_3 = 0$$ This leads to $$3 \alpha_1 + 6 \alpha_2 = 0 \iff \\ \alpha_1 = -2 \alpha_2$$ and $$\alpha_2 - \alpha_3 = 0$$ So we have the solutions $(\alpha_1, \alpha_2, \alpha_3) = (-2t, t, t)$. So these vectors are linear dependent as well. A non-zero solution is $(-2,1,1)$. We have $v_4 = (2 \odot v_1) \oplus (-1 \odot v_2$).
Test: $(2 \odot (1,1,1)) \oplus (-1 \odot (1,1,4)) = (1, 2^{1/3}, 3) \oplus (1,-1, -6) = (1, 1, -2)$.

7. This means we can reduce $L$ to $L' = (v_1, v_2)$, as we can combine $v_3$ and $v_4$ from $v_1$ and $v_2$ each.

• Then the identity would be $z=(1,0,-1)$ and the inverse $v_i=(\frac{1}{x_1},-y_1,-(z_1+2))$. Thanks for you help! Commented Feb 7, 2016 at 17:39
• I am still confused with how to reduce the dependent list to and independant list. I understand the basics of what I need to do, however I am struggling to implement the vector addition and scalar multiplication. Commented Feb 7, 2016 at 18:36
• @ddbb1994 Can you solve d) with the above?
– mvw
Commented Feb 8, 2016 at 3:43
• Note: The above used those unusual definitions of vector addition and scalar multiplication down to the question of linear dependence. I wonder if I oversaw an easier way.
– mvw
Commented Feb 8, 2016 at 5:32