Math Problem Statement

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.

Solution

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

Problem Statement Recap:

Let f(x)=12Axb2+x2f(x) = \frac{1}{2}\|Ax - b\|^2 + \|x\|^2, where AR3×4A \in \mathbb{R}^{3 \times 4} and bR3b \in \mathbb{R}^3. The matrices AA and bb are defined as:

A=[320113020427],b=[314]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 ff with respect to x=[x1,x2,x3,x4]Tx = [x_1, x_2, x_3, x_4]^T and determine whether the solution to f(x)=0\nabla f(x) = 0 will be unique.

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

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

f(x)=12Axb2+x2f(x) = \frac{1}{2} \|Ax - b\|^2 + \|x\|^2

Using the Euclidean norm properties, we have:

Axb2=(Axb)T(Axb)\|Ax - b\|^2 = (Ax - b)^T (Ax - b) x2=xTx\|x\|^2 = x^T x

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

f(x)=12(Axb)T(Axb)+xTxf(x) = \frac{1}{2}(Ax - b)^T (Ax - b) + x^T x

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

f(x)=AT(Axb)+2x\nabla f(x) = A^T (Ax - b) + 2x

Step 2: Analyzing Uniqueness of Solution

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

AT(Axb)+2x=0A^T (Ax - b) + 2x = 0

This can be rearranged as:

(ATA+2I)x=ATb(A^T A + 2I)x = A^T b

Here, II is the 4×44 \times 4 identity matrix. The uniqueness of the solution depends on whether (ATA+2I)(A^T A + 2I) is invertible. For (ATA+2I)(A^T A + 2I) to be invertible, it must be a positive definite matrix.

  • ATAA^T A is a 4×44 \times 4 matrix, and adding 2I2I makes it (ATA+2I)(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, (ATA+2I)(A^T A + 2I) being positive definite implies it is invertible. Therefore, the solution xx to the equation (ATA+2I)x=ATb(A^T A + 2I)x = A^T b will be unique.

Conclusion

Without finding the actual values of xx, we can conclude that the solution to f(x)=0\nabla f(x) = 0 is unique because the matrix (ATA+2I)(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 (ATA+2I)(A^T A + 2I), and how do they relate to the matrix's positive definiteness?
  2. Can we explicitly compute the solution xx using the normal equations derived?
  3. How does the dimensionality of AA (being 3×43 \times 4) affect the solution space of Ax=bAx = b in general?
  4. What would change if AA were a square matrix (i.e., 4×44 \times 4 instead of 3×43 \times 4)?
  5. How would the problem formulation and solution change if we used a different norm (e.g., L1L_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.

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Math Problem Analysis

Mathematical Concepts

Matrix Calculus
Optimization
Positive Definite Matrices

Formulas

Gradient formula
Euclidean norm properties

Theorems

Positive Definite Matrix Theorem

Suitable Grade Level

Advanced Mathematics