To derive the optimal step size $\alpha_k$ for minimizing the function $g(z)$ using gradient descent, we first need to calculate the gradient of $g(z)$ and then find the step size that minimizes $g(z)$ along the direction of the negative gradient at each iteration.

Step 1: Define the function and compute its gradient

The function $g(z)$ is given by:

$g(z) = \frac{1}{2} \|Az - b\|_2^2 + \frac{1}{2} \|z\|_2^2$

This can be expanded as:

$g(z) = \frac{1}{2} \left((Az - b)^T (Az - b)\right) + \frac{1}{2} \left(z^T z\right)$

Expanding the first term:

$(Az - b)^T (Az - b) = z^T A^T A z - 2b^T A z + b^T b$

Thus, $g(z)$ becomes:

$g(z) = \frac{1}{2} \left( z^T A^T A z - 2b^T A z + b^T b \right) + \frac{1}{2} z^T z$

The gradient of $g(z)$ with respect to $z$ is:

$\nabla g(z) = \frac{\partial g(z)}{\partial z} = A^T A z - A^T b + z$

So,

$\nabla g(z) = (A^T A + I)z - A^T b$

Step 2: Gradient descent update rule

In gradient descent, the update rule is given by:

$z_k = z_{k-1} - \alpha_k \nabla g(z_{k-1})$

where $\alpha_k$ is the step size at iteration $k$.

Step 3: Derive the optimal step size $\alpha_k$

The optimal step size $\alpha_k$ is determined by minimizing $g(z)$ along the direction of the negative gradient. This requires minimizing the function:

$\phi(\alpha) = g(z_{k-1} - \alpha \nabla g(z_{k-1}))$

Substituting $z_k = z_{k-1} - \alpha \nabla g(z_{k-1})$ into $g(z)$:

$\phi(\alpha) = g(z_{k-1} - \alpha \nabla g(z_{k-1}))$

Expanding $g(z_{k-1} - \alpha \nabla g(z_{k-1}))$:

$\phi(\alpha) = \frac{1}{2} \|A(z_{k-1} - \alpha \nabla g(z_{k-1})) - b\|_2^2 + \frac{1}{2} \|z_{k-1} - \alpha \nabla g(z_{k-1})\|_2^2$

This expands to:

$\phi(\alpha) = \frac{1}{2} \left\{ \|Az_{k-1} - \alpha A \nabla g(z_{k-1}) - b\|_2^2 + \|z_{k-1} - \alpha \nabla g(z_{k-1})\|_2^2 \right\}$

To find the optimal $\alpha_k$, differentiate $\phi(\alpha)$ with respect to $\alpha$ and set it to zero:

$\frac{d\phi(\alpha)}{d\alpha} = -\nabla g(z_{k-1})^T \left( A^T(Az_{k-1} - b) + z_{k-1} \right) + \alpha \nabla g(z_{k-1})^T (A^T A + I) \nabla g(z_{k-1})$

Setting this equal to zero and solving for $\alpha$:

$0 = -\nabla g(z_{k-1})^T \nabla g(z_{k-1}) + \alpha \nabla g(z_{k-1})^T (A^T A + I) \nabla g(z_{k-1})$

$\alpha_k = \frac{\nabla g(z_{k-1})^T \nabla g(z_{k-1})}{\nabla g(z_{k-1})^T (A^T A + I) \nabla g(z_{k-1})}$

Conclusion

The optimal step size $\alpha_k$ for the gradient descent at iteration $k$ is given by:

$\alpha_k = \frac{\|\nabla g(z_{k-1})\|_2^2}{\nabla g(z_{k-1})^T (A^T A + I) \nabla g(z_{k-1})}$

This step size ensures the most efficient descent at each iteration.

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Would you like further details on any part of this derivation or have additional questions? Here are some related questions for further exploration:

1. How does the optimal step size $\alpha_k$ affect the convergence rate of gradient descent?
2. What happens if we use a fixed step size instead of the optimal $\alpha_k$?
3. How can the convergence be analyzed using the eigenvalues of $A^T A + I$?
4. What would the gradient descent update look like if we introduced a regularization term in $g(z)$?
5. Can you implement this gradient descent method in a programming language like Python or MATLAB?

Tip: When optimizing functions using gradient descent, the choice of step size is crucial for convergence and efficiency, and can be problem-specific.