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.

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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.