To find the gradient of the function $f(x) = \text{Tr}(A^2x)$ with respect to the matrix $x$, let's go through the problem step by step.

Given:

• $A$ is a symmetric $2 \times 2$ matrix: $A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$.
• $x$ is a $2 \times 2$ matrix.
• $f(x) = \text{Tr}(A^2x)$, where $\text{Tr}(M)$ denotes the trace of matrix $M$.

Step 1: Compute $A^2$

First, calculate $A^2$:

$A^2 = A \cdot A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$

$A^2 = \begin{pmatrix} 1 \cdot 1 + 2 \cdot 2 & 1 \cdot 2 + 2 \cdot 1 \\ 2 \cdot 1 + 1 \cdot 2 & 2 \cdot 2 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}$

Step 2: Express $f(x)$

The function $f(x)$ can be rewritten as:

$f(x) = \text{Tr}(A^2x) = \text{Tr}\left(\begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix} \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix}\right)$

The product $A^2 x$ is:

$A^2 x = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix} \cdot \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} = \begin{pmatrix} 5x_{11} + 4x_{21} & 5x_{12} + 4x_{22} \\ 4x_{11} + 5x_{21} & 4x_{12} + 5x_{22} \end{pmatrix}$

Thus,

$f(x) = \text{Tr}\left(\begin{pmatrix} 5x_{11} + 4x_{21} & 5x_{12} + 4x_{22} \\ 4x_{11} + 5x_{21} & 4x_{12} + 5x_{22} \end{pmatrix}\right) = (5x_{11} + 4x_{21}) + (4x_{12} + 5x_{22})$

$f(x) = 5x_{11} + 4x_{21} + 4x_{12} + 5x_{22}$

Step 3: Compute the gradient $\nabla_x f(x)$

The gradient of $f(x)$ with respect to $x$ is the matrix of partial derivatives:

$\nabla_x f(x) = \frac{\partial f(x)}{\partial x} = \begin{pmatrix} \frac{\partial f(x)}{\partial x_{11}} & \frac{\partial f(x)}{\partial x_{12}} \\ \frac{\partial f(x)}{\partial x_{21}} & \frac{\partial f(x)}{\partial x_{22}} \end{pmatrix}$

Computing each partial derivative:

$\frac{\partial f(x)}{\partial x_{11}} = 5, \quad \frac{\partial f(x)}{\partial x_{12}} = 4$ $\frac{\partial f(x)}{\partial x_{21}} = 4, \quad \frac{\partial f(x)}{\partial x_{22}} = 5$

Therefore, the gradient is:

$\nabla_x f(x) = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}$

Final Answer:

The gradient of $f(x) = \text{Tr}(A^2 x)$ with respect to $x$ is:

$\nabla_x f(x) = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}$

Do you want details or have any questions?

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Here are five relative questions for further exploration:

1. How would the gradient change if $A$ were not symmetric?
2. What happens to the gradient if the matrix $x$ is constrained to be symmetric?
3. How does the trace function affect the differentiation process when applied to different types of matrices?
4. Can you generalize this result for larger symmetric matrices $A$?
5. How does the gradient $\nabla_x f(x)$ relate to the optimization of $f(x)$?

Tip: When working with matrix calculus, understanding the role of the trace operator and the properties of matrices involved can greatly simplify gradient calculations.