Let f(x) =1/2 llAx - bll^2+ llxll^2 where the norm is the euclidean norm and A € R3x4,b E 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.

To solve this problem, let's first express the function \( f(x) \) and then compute its gradient.

### Step 1: Understanding the Function

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

Here:
- \( A \in \mathbb{R}^{3 \times 4} \)
- \( b \in \mathbb{R}^3 \)
- \( x \in \mathbb{R}^4 \)

The Euclidean norm squared, \( \|v\|^2 \), is given by \( v^T v \).

So, we can rewrite the function as:
\[
f(x) = \frac{1}{2} (Ax - b)^T (Ax - b) + x^T x
\]

### Step 2: Compute the Gradient of \( f(x) \)

Let's compute the gradient of \( f(x) \) with respect to \( x \).

1. **First term**: \( \frac{1}{2} \|Ax - b\|^2 \)

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

   To find the gradient of this term with respect to \( x \), we use the identity:
   \[
   \nabla_x \left( \frac{1}{2} (Ax - b)^T (Ax - b) \right) = A^T(Ax - b)
   \]

2. **Second term**: \( \|x\|^2 = x^T x \)

   The gradient of \( x^T x \) with respect to \( x \) is:
   \[
   \nabla_x (x^T x) = 2x
   \]

Putting these together, the gradient of \( f(x) \) is:
\[
\nabla f(x) = A^T (Ax - b) + 2x
\]

### Step 3: Evaluate the Gradient at Any Point \( x \)

The expression for the gradient \( \nabla f(x) = A^T (Ax - b) + 2x \) is valid for any point \( x \).

### Step 4: Determining the Uniqueness of the Critical Point

The critical points of \( f(x) \) are found by solving:
\[
\nabla f(x) = 0
\]
\[
A^T(Ax - b) + 2x = 0
\]

This can be rewritten as:
\[
(A^T A + 2I)x = A^T b
\]

Here, \( (A^T A + 2I) \) is a \( 4 \times 4 \) matrix. To determine if the solution for \( x \) is unique, we need to check if this matrix is invertible.

- \( A^T A \) is a symmetric \( 4 \times 4 \) matrix.
- Adding \( 2I \) (where \( I \) is the \( 4 \times 4 \) identity matrix) shifts all eigenvalues of \( A^T A \) by 2.

Since \( A^T A \) is positive semi-definite, all its eigenvalues are non-negative. Adding 2 ensures that all eigenvalues of \( (A^T A + 2I) \) are strictly positive, making it **positive definite**.

A positive definite matrix is always invertible. Hence, the matrix \( A^T A + 2I \) is invertible, implying that the equation \( (A^T A + 2I)x = A^T b \) has a **unique solution** for \( x \).

### Conclusion

- The gradient \( \nabla f(x) = A^T(Ax - b) + 2x \).
- The critical point where \( \nabla f(x) = 0 \) will be unique because \( A^T A + 2I \) is positive definite and therefore invertible.

Would you like more details or have any questions?

### Related Questions:
1. What are the eigenvalues of the matrix \( A^T A + 2I \)?
2. How would the solution change if \( A \) were a different size matrix?
3. Can you provide a specific example with numerical values to solve for \( x \)?
4. What does it mean for a matrix to be positive definite?
5. How do we determine if a matrix is invertible in general?

### Tip:
Always check if a matrix is positive definite or positive semi-definite when determining the uniqueness of solutions in optimization problems.

Explore the gradient computation of a quadratic function involving matrix algebra. Understand how to find critical points and determine uniqueness without solving for specific values of x. Detailed explanation suitable for advanced college-level students.

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