Q1) Let A ∈ R
n×n be a constant matrix and b ∈ R
n be a constant vector.
Let z ∈ R
n
. Consider the function g(z) defined as
g(z) = 1
2
∥Az − b∥
2
2 +
1
2
∥z∥
2
2
1. Find the gradient of g(z) with respect to z.
2. Consider the problem of finding the minimum of g(z) using gradient
descent. Derive the optimal step size αk to be used at iteration number
k of the gradient descent scheme assuming estimate zk−1 from the
previous iteration.


### 1. Gradient of \( g(z) \) with respect to \( z \)

Given the function:

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

where \( A \) is an \( n \times n \) matrix, \( b \) is an \( n \)-dimensional vector, and \( z \) is an \( n \)-dimensional vector, we want to find the gradient \( \nabla g(z) \) with respect to \( z \).

#### Step 1: Expand the terms
The first term \( \|Az - b\|_2^2 \) is the squared norm, which can be written as:

\[
\|Az - b\|^2_2 = (Az - b)^\top (Az - b)
\]

Expanding this:

\[
\|Az - b\|^2_2 = z^\top A^\top Az - 2b^\top Az + b^\top b
\]

The second term is:

\[
\|z\|^2_2 = z^\top z
\]

So, the function \( g(z) \) becomes:

\[
g(z) = \frac{1}{2} \left( z^\top A^\top A z - 2b^\top A z + b^\top b \right) + \frac{1}{2} z^\top z
\]

#### Step 2: Differentiate with respect to \( z \)
Now, we differentiate each term with respect to \( z \).

- The derivative of \( \frac{1}{2} z^\top A^\top A z \) with respect to \( z \) is \( A^\top A z \).
- The derivative of \( - b^\top A z \) with respect to \( z \) is \( - A^\top b \).
- The derivative of \( \frac{1}{2} z^\top z \) with respect to \( z \) is \( z \).
- The term \( \frac{1}{2} b^\top b \) is independent of \( z \) and vanishes upon differentiation.

Therefore, the gradient \( \nabla g(z) \) is:

\[
\nabla g(z) = A^\top A z - A^\top b + z
\]

This can be simplified to:

\[
\nabla g(z) = (A^\top A + I)z - A^\top b
\]

where \( I \) is the identity matrix of appropriate dimensions.

### 2. Optimal Step Size \( \alpha_k \) in Gradient Descent

In gradient descent, the update rule is:

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

where \( \alpha_k \) is the step size to be determined.

To find the optimal step size \( \alpha_k \), we want to minimize \( g(z_k) \) with respect to \( \alpha_k \). Substitute the update rule into \( g(z_k) \):

\[
g(z_k) = \frac{1}{2} \|A(z_{k-1} - \alpha_k \nabla g(z_{k-1})) - b\|^2_2 + \frac{1}{2} \|z_{k-1} - \alpha_k \nabla g(z_{k-1})\|^2_2
\]

This expression is generally difficult to minimize directly. However, for quadratic functions like this one, the optimal step size can be obtained by differentiating \( g(z_k) \) with respect to \( \alpha_k \) and setting it to zero.

To derive the optimal step size \( \alpha_k \), consider:

\[
\alpha_k = \frac{\nabla g(z_{k-1})^\top \nabla g(z_{k-1})}{\nabla g(z_{k-1})^\top (A^\top A + I)\nabla g(z_{k-1})}
\]

This choice of \( \alpha_k \) ensures that the step size is optimal at each iteration, given the current gradient.

---

### Do you want details or have any questions?

Here are some related questions:

1. What is the significance of the identity matrix \( I \) in the gradient expression?
2. How does the condition number of \( A \) affect the convergence of gradient descent in this scenario?
3. Can you apply gradient descent to a non-quadratic function? How would the step size change?
4. How would the gradient descent update rule change if \( g(z) \) had an additional regularization term?
5. How does the choice of initial point \( z_0 \) impact the convergence of gradient descent?

**Tip:** When choosing a step size in gradient descent, it's often useful to perform a line search to find the optimal value, particularly when dealing with non-quadratic functions.

Explore the gradient descent method for minimizing a quadratic function involving matrix operations and vector calculus. Learn how to derive gradients and determine optimal step sizes in iterative optimization. Suitable for advanced undergraduate studies.

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