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.