https://file.aiquickdraw.com/mathbot/file/1724970069253tsl7kclc.jpg

Let f(x) =
1
2
∥Ax−b∥2 +∥x∥2 where the norm is the euclidean norm
and A ∈ R3×4, b ∈ R3 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.

The given problem is about finding the gradient of a function \( f(x) \) and determining if the solution to \( \nabla f(x) = 0 \) will be unique.

### Problem Statement Recap:
Let \( f(x) = \frac{1}{2}\|Ax - b\|^2 + \|x\|^2 \), where \( A \in \mathbb{R}^{3 \times 4} \) and \( b \in \mathbb{R}^3 \). The matrices \( A \) and \( b \) are defined as:

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

We need to find the gradient of \( f \) with respect to \( x = [x_1, x_2, x_3, x_4]^T \) and determine whether the solution to \( \nabla f(x) = 0 \) will be unique.

### Step 1: Finding the Gradient \( \nabla f(x) \)

The function \( f(x) \) can be written as:

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

Using the Euclidean norm properties, we have:

\[
\|Ax - b\|^2 = (Ax - b)^T (Ax - b)
\]
\[
\|x\|^2 = x^T x
\]

Thus, \( f(x) \) becomes:

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

The gradient of \( f(x) \) with respect to \( x \) is:

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

### Step 2: Analyzing Uniqueness of Solution

To check whether the solution to \( \nabla f(x) = 0 \) is unique, we rewrite the condition for optimality:

\[
A^T (Ax - b) + 2x = 0
\]

This can be rearranged as:

\[
(A^T A + 2I)x = A^T b
\]

Here, \( I \) is the \( 4 \times 4 \) identity matrix. The uniqueness of the solution depends on whether \( (A^T A + 2I) \) is invertible. For \( (A^T A + 2I) \) to be invertible, it must be a positive definite matrix.

- \( A^T A \) is a \( 4 \times 4 \) matrix, and adding \( 2I \) makes it \( (A^T A + 2I) \) positive definite, since adding a positive definite matrix to a positive semi-definite matrix results in a positive definite matrix.
- A positive definite matrix is always invertible.

Thus, \( (A^T A + 2I) \) being positive definite implies it is invertible. Therefore, the solution \( x \) to the equation \( (A^T A + 2I)x = A^T b \) will be unique.

### Conclusion

Without finding the actual values of \( x \), we can conclude that the solution to \( \nabla f(x) = 0 \) is unique because the matrix \( (A^T A + 2I) \) is positive definite and hence invertible.

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

Here are five related questions that expand on the previous information:

1. What are the eigenvalues of the matrix \( (A^T A + 2I) \), and how do they relate to the matrix's positive definiteness?
2. Can we explicitly compute the solution \( x \) using the normal equations derived?
3. How does the dimensionality of \( A \) (being \( 3 \times 4 \)) affect the solution space of \( Ax = b \) in general?
4. What would change if \( A \) were a square matrix (i.e., \( 4 \times 4 \) instead of \( 3 \times 4 \))?
5. How would the problem formulation and solution change if we used a different norm (e.g., \( L_1 \) norm) instead of the Euclidean norm?

**Tip:** When dealing with optimization problems, checking the positive definiteness of the Hessian matrix or its equivalent ensures that the problem has a unique minimum and that any solution found is stable.

Explore the calculation of gradients using matrix calculus for a function involving matrix operations and the Euclidean norm. Understand the uniqueness of solutions in optimization problems with positive definite matrices.

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