The problem you provided asks about the linear independence of two sets of vectors formed from three given linearly independent vectors $x, y,$ and $z$ in $\mathbb{R}^n$.

Part (a)

Question: Are the vectors $(x - y), (y - z), (z - x)$ linearly independent?

To determine the linear independence of $(x - y), (y - z), (z - x)$, we check if the only solution to the equation $a(x-y) + b(y-z) + c(z-x) = 0$ is $a = b = c = 0$.

Expanding and rearranging the terms, we get: $a x - a y + b y - b z + c z - c x = 0$ $(a - c)x + (b - a)y + (c - b)z = 0$

Since $x, y,$ and $z$ are linearly independent, the coefficients of $x, y,$ and $z$ must each be zero: $a - c = 0$ $b - a = 0$ $c - b = 0$

These equations imply $a = b = c$. This means the equation can be true for any non-zero scalar $a = b = c \neq 0$. Hence, the vectors $(x-y), (y-z), (z-x)$ are not linearly independent as they are linearly dependent on each other.

Part (b)

Question: Are the vectors $(x + y), (y + z), (z + x)$ linearly independent?

To determine the linear independence of $(x + y), (y + z), (z + x)$, we consider the equation: $a(x+y) + b(y+z) + c(z+x) = 0$

Expanding this, we get: $a x + a y + b y + b z + c z + c x = 0$ $(a + c)x + (a + b)y + (b + c)z = 0$

Again, using the linear independence of $x, y,$ and $z$, we must have: $a + c = 0$ $a + b = 0$ $b + c = 0$

These equations imply $a = -b, b = -c,$ and $c = -a$. Solving these gives $a = b = c = 0$. Hence, the vectors $(x + y), (y + z), (z + x)$ are linearly independent.

Would you like further details on any part or have any other questions about linear independence or vector spaces? Here are some related questions and a tip:

1. How can you determine linear independence in higher dimensions?
2. What other types of vector operations might affect linear independence?
3. Can adding more vectors to a set ever lead to a loss of linear independence?
4. What are the implications of linear independence in solving systems of linear equations?
5. How do properties of linear independence apply to function spaces?

Tip: When solving problems involving vector independence, organizing terms by vector components can simplify the visualization of the problem and solutions.