Let f(x) = 1 2 ∥Ax − b∥ 2 + ∥x∥ 2 where the norm is the euclidean norm and A ∈ R 3×4 , b ∈ R 3 is defined as: A =   3 2 0 −1 −1 3 0 2 0 −4 −2 7   , b =   3 1 −4   Find the gradient of f at x = [x1, x2, x3, x4] T . Without actually finding the value or values of x at which the gradient of f is 0, can we say whether it will be unique or not? Justify.

Given the function \( f(x) = \frac{1}{2} \|Ax - b\|^2 + \|x\|^2 \), where \( A \in \mathbb{R}^{3 \times 4} \) and \( b \in \mathbb{R}^3 \), we want to find the gradient of \( f \) with respect to \( x \), and discuss the uniqueness of the solution where the gradient is zero.

### Step 1: Express \( f(x) \) explicitly
The given function is:
\[
f(x) = \frac{1}{2} \|Ax - b\|^2 + \|x\|^2
\]
Expanding the first term:
\[
\|Ax - b\|^2 = (Ax - b)^T(Ax - b)
\]
Thus,
\[
f(x) = \frac{1}{2} (Ax - b)^T(Ax - b) + x^T x
\]
Expanding further:
\[
f(x) = \frac{1}{2} \left[ x^T A^T A x - 2b^T A x + b^T b \right] + x^T x
\]
Since \( b^T b \) is a constant, it doesn't affect the gradient. So, we simplify \( f(x) \) to:
\[
f(x) = \frac{1}{2} x^T A^T A x - b^T A x + x^T x + \text{constant}
\]
\[
f(x) = \frac{1}{2} x^T A^T A x - b^T A x + x^T x
\]

### Step 2: Compute the gradient of \( f(x) \)
To find the gradient \( \nabla f(x) \), take the derivative with respect to \( x \):
\[
\nabla f(x) = \frac{\partial}{\partial x} \left( \frac{1}{2} x^T A^T A x - b^T A x + x^T x \right)
\]

The gradient of each term is:
1. \( \frac{\partial}{\partial x} \left( \frac{1}{2} x^T A^T A x \right) = A^T A x \)
2. \( \frac{\partial}{\partial x} (-b^T A x) = -A^T b \)
3. \( \frac{\partial}{\partial x} (x^T x) = 2x \)

So, the gradient of \( f(x) \) is:
\[
\nabla f(x) = A^T A x - A^T b + 2x
\]
\[
\nabla f(x) = (A^T A + 2I)x - A^T b
\]
where \( I \) is the identity matrix.

### Step 3: Discuss the uniqueness of the solution
The gradient \( \nabla f(x) = 0 \) leads to the equation:
\[
(A^T A + 2I)x = A^T b
\]
To determine if the solution \( x \) is unique, we need to check if the matrix \( A^T A + 2I \) is invertible. A matrix is invertible if it is full rank, which means it has no zero eigenvalues.

- \( A^T A \) is a symmetric \( 4 \times 4 \) matrix.
- \( 2I \) is also symmetric and positive definite with eigenvalues equal to 2.

Adding \( 2I \) to \( A^T A \) shifts the eigenvalues of \( A^T A \) by 2, making \( A^T A + 2I \) positive definite and hence invertible (since all eigenvalues are positive).

Since \( A^T A + 2I \) is invertible, there is a unique solution to the equation \( (A^T A + 2I)x = A^T b \), meaning the solution to \( \nabla f(x) = 0 \) is unique.

### Summary
The gradient of \( f(x) \) is \( \nabla f(x) = (A^T A + 2I)x - A^T b \). The solution where the gradient is zero will be unique because the matrix \( A^T A + 2I \) is invertible.

Would you like more details or have any other questions?

Here are some related questions for further exploration:
1. How do you find the explicit solution for \( x \) where \( \nabla f(x) = 0 \)?
2. What is the geometric interpretation of the gradient in this context?
3. How does the choice of \( A \) affect the curvature of \( f(x) \)?
4. Can you derive the Hessian of \( f(x) \) and discuss its significance?
5. How would the solution change if \( A \) were not full rank?

**Tip:** When dealing with quadratic functions, the positive definiteness of matrices often determines the uniqueness of solutions.

Explore the gradient calculation for the quadratic function f(x) = 1/2 ||Ax - b||^2 + ||x||^2 using matrix A and vector b. Understand the uniqueness of the gradient's zero solution with detailed mathematical analysis. Suitable for advanced undergraduate students and professionals.

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