To find the preimage of the vector $w = \left(-4\sqrt{2}, 0, -12\right)$ for the transformation $T(v_1, v_2)$, let's break down the problem:

The transformation $T(v_1, v_2)$ is given by: $T(v_1, v_2) = \left( \frac{\sqrt{2}}{2} v_1 - \frac{\sqrt{2}}{2} v_2, v_1 + v_2, 2v_1 - v_2 \right)$

We need to find values of $v_1$ and $v_2$ such that: $\left( \frac{\sqrt{2}}{2} v_1 - \frac{\sqrt{2}}{2} v_2, v_1 + v_2, 2v_1 - v_2 \right) = \left(-4\sqrt{2}, 0, -12\right)$

This gives us a system of equations:

1. $\frac{\sqrt{2}}{2} v_1 - \frac{\sqrt{2}}{2} v_2 = -4\sqrt{2}$
2. $v_1 + v_2 = 0$
3. $2v_1 - v_2 = -12$

Solving the system

1. From the second equation $v_1 + v_2 = 0$, we get $v_2 = -v_1$.
2. Substitute $v_2 = -v_1$ into the first equation: $\frac{\sqrt{2}}{2} v_1 - \frac{\sqrt{2}}{2} (-v_1) = -4\sqrt{2}$ Simplifying, we get: $\sqrt{2} v_1 = -4\sqrt{2}$ Dividing by $\sqrt{2}$, we find $v_1 = -4$.
3. Substitute $v_1 = -4$ back into $v_2 = -v_1$: $v_2 = -(-4) = 4$

Solution

The preimage of $w$ is: $(v_1, v_2) = (-4, 4)$

Would you like more details on any part of this solution?

Here are some follow-up questions to deepen your understanding:

1. What is the geometric interpretation of a preimage in a linear transformation?
2. How would you verify that the solution $(-4, 4)$ satisfies the transformation equations?
3. How can we check if the transformation $T(v_1, v_2)$ is invertible?
4. What would happen if the transformation led to a system of equations with no solutions?
5. How would the solution change if the transformation matrix were altered?

Tip: To verify solutions for transformations, substitute your answers back into the original transformation equations to ensure they satisfy all conditions.