To find the minimum and maximum values of the function $Q(x, y, z) = 2x^2 + 2y^2 + z^2$ subject to the constraint $X^T X = I$, we can interpret $X$ as a vector in $\mathbb{R}^3$, i.e., $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$. The constraint $X^T X = I$ implies that the vector $X$ has a unit length, i.e., $x^2 + y^2 + z^2 = 1$.

Given this setup, we need to find the extrema of $Q(x, y, z) = 2x^2 + 2y^2 + z^2$ subject to $x^2 + y^2 + z^2 = 1$. This is a problem that can be solved using the method of Lagrange multipliers.

Step 1: Set up the Lagrangian

The Lagrangian function is given by:

$\mathcal{L}(x, y, z, \lambda) = 2x^2 + 2y^2 + z^2 + \lambda(1 - x^2 - y^2 - z^2)$

Step 2: Find the Partial Derivatives

To find the extrema, we take partial derivatives of $\mathcal{L}$ with respect to $x$, $y$, $z$, and $\lambda$ and set them equal to zero:

1. Partial derivative with respect to $x$: $\frac{\partial \mathcal{L}}{\partial x} = 4x - 2\lambda x = 0 \implies x(4 - 2\lambda) = 0$

2. Partial derivative with respect to $y$: $\frac{\partial \mathcal{L}}{\partial y} = 4y - 2\lambda y = 0 \implies y(4 - 2\lambda) = 0$

3. Partial derivative with respect to $z$: $\frac{\partial \mathcal{L}}{\partial z} = 2z - 2\lambda z = 0 \implies z(2 - 2\lambda) = 0$

4. Partial derivative with respect to $\lambda$: $\frac{\partial \mathcal{L}}{\partial \lambda} = 1 - x^2 - y^2 - z^2 = 0$

Step 3: Solve the Equations

From the first three equations, we get two possibilities for $\lambda$:

1. If $\lambda = 2$:

   • From the equations $x(4 - 2\lambda) = 0$ and $y(4 - 2\lambda) = 0$, both $x$ and $y$ must be $0$.
   • The constraint $x^2 + y^2 + z^2 = 1$ implies $z = \pm 1$.

   Substituting these into $Q$: $Q = 2(0^2) + 2(0^2) + 1^2 = 1$

   So, $Q = 1$ in this case.

2. If $\lambda = 4$:

   • From $z(2 - 2\lambda) = 0$, $z = 0$.
   • Then $x^2 + y^2 = 1$.

   Substituting these into $Q$: $Q = 2x^2 + 2y^2 + 0^2 = 2(x^2 + y^2) = 2 \cdot 1 = 2$

   So, $Q = 2$ in this case.

Step 4: Conclusion

• The minimum value of $Q$ is 1, occurring when $(x, y, z) = (0, 0, \pm 1)$.
• The maximum value of $Q$ is 2, occurring when $(x, y, z) = (\pm 1, 0, 0)$ or $(0, \pm 1, 0)$.

Would you like more details or have any questions about these steps?

Related Questions:

1. How does the method of Lagrange multipliers work for finding extrema of functions?
2. Can this method be used for functions with more than three variables?
3. What are some practical applications of Lagrange multipliers?
4. How would the solution change if the constraint were different, like $x^2 + y^2 + z^2 = c$ for some $c \neq 1$?
5. How would we approach this problem if the objective function had additional terms involving $x, y,$ and $z$?

Tip:

When dealing with optimization problems, always ensure the constraints are correctly interpreted and applied, as they define the feasible region where the extrema need to be found.