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 need to find the gradient of \( f \) at \( x = [x_1, x_2, x_3, x_4]^T \). After that, we'll discuss the uniqueness of the point where the gradient is zero.

### Step 1: Expression of the gradient

The function \( f(x) \) can be expanded as follows:

\[
f(x) = \frac{1}{2} (Ax - b)^T (Ax - b) + x^T x
\]

Taking the gradient of \( f(x) \) with respect to \( x \):

\[
\nabla f(x) = \frac{1}{2} \nabla \left( (Ax - b)^T (Ax - b) \right) + \nabla \left( x^T x \right)
\]

For the first term:

\[
\frac{1}{2} \nabla \left( (Ax - b)^T (Ax - b) \right) = \nabla \left( \frac{1}{2} \|Ax - b\|^2 \right)
\]

Using the chain rule:

\[
\nabla \left( \frac{1}{2} \|Ax - b\|^2 \right) = A^T(Ax - b)
\]

For the second term:

\[
\nabla \left( x^T x \right) = 2x
\]

Therefore, the gradient of \( f(x) \) is:

\[
\nabla f(x) = A^T(Ax - b) + 2x
\]

### Step 2: Substituting the given matrices

Given:

\[
A = \begin{bmatrix} 3 & 2 & 0 & -1 \\ -1 & 3 & 0 & 2 \\ 0 & -4 & -2 & 7 \end{bmatrix}, \quad b = \begin{bmatrix} 3 \\ 1 \\ -4 \end{bmatrix}
\]

So,

\[
\nabla f(x) = \begin{bmatrix} 3 & -1 & 0 \\ 2 & 3 & -4 \\ 0 & 0 & -2 \\ -1 & 2 & 7 \end{bmatrix} \begin{bmatrix} 3x_1 + 2x_2 - x_4 - 3 \\ -x_1 + 3x_2 + 2x_4 - 1 \\ -4x_2 - 2x_3 + 7x_4 + 4 \end{bmatrix} + 2 \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}
\]

This can be computed further, but since we're focusing on the form rather than specific values, this expression is enough to understand the structure of the gradient.

### Step 3: Uniqueness of the point where \( \nabla f(x) = 0 \)

The uniqueness of the point where the gradient \( \nabla f(x) = 0 \) can be determined by the properties of \( f(x) \).

- The function \( f(x) \) is a sum of two convex functions: \( \frac{1}{2} \|Ax - b\|^2 \) and \( \|x\|^2 \).
- \( \frac{1}{2} \|Ax - b\|^2 \) is convex since it's a quadratic function with respect to \( x \), and so is \( \|x\|^2 \).
- The function \( f(x) \) is therefore strictly convex because \( \|x\|^2 \) (which is \( x^T x \)) is strictly convex, adding strict convexity to the overall function.

For a strictly convex function, the gradient is zero at most at one point, meaning there will be a unique minimum. Therefore:

**Yes, the point where \( \nabla f(x) = 0 \) will be unique.**

This is because \( f(x) \) is strictly convex due to the presence of the \( \|x\|^2 \) term, ensuring a single global minimum.

---

Would you like any additional details or have any questions?

### Related Questions
1. How do we determine if a matrix is positive definite, and why is this important for convexity?
2. What are the conditions under which a quadratic function has a unique minimum?
3. How does the dimension of matrix \( A \) affect the uniqueness of solutions to \( Ax = b \)?
4. What is the role of the Hessian matrix in determining the convexity of a function?
5. How does the concept of regularization relate to the term \( \|x\|^2 \) in optimization problems?

### Tip
Remember that the strict convexity of a function ensures not just the existence but also the uniqueness of the optimal solution, which is crucial in many optimization problems.

Explore the gradient computation of a quadratic function involving matrix operations and convex optimization principles. Learn about strict convexity and its implications in optimization. Suitable for advanced mathematics enthusiasts and students.

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