Derivative Of Trace Of Matrix Log: ∂ \mbox T R ( A X B \mbox L O G ( C X D ) ) ∂ X \frac{\partial \ \mbox{tr}(AXB \ \mbox{log}(CXD))}{\partial X} ∂ X ∂ \mbox T R ( A XB \mbox L O G ( CX D )) ​

Introduction

In the realm of matrix calculus, the derivative of the trace of a matrix log is a fundamental concept that has numerous applications in various fields, including machine learning, signal processing, and control theory. Given the matrix expression $\frac{\partial \ \mbox{tr}(AXB \ \mbox{log}(CXD))}{\partial X}$, we aim to derive its partial derivative with respect to the matrix $X$. This problem is a natural extension of the well-known result $\frac{\partial \ \mbox{tr}(\mbox{log}(X))}{\partial X} = X^{-T}$, where $\mbox{log}$ denotes the matrix logarithm.

Preliminaries

Before diving into the derivation, let's establish some essential concepts and notations. We denote the matrix logarithm as $\mbox{log}(X)$, which is defined as the unique matrix $Y$ satisfying the equation $e^Y = X$, where $e^Y$ is the matrix exponential of $Y$. The trace of a matrix $A$, denoted as $\mbox{tr}(A)$, is the sum of its diagonal elements. The derivative of the trace of a matrix with respect to a matrix is a fundamental concept in matrix calculus, and it is defined as the matrix whose entries are the partial derivatives of the trace with respect to the corresponding entries of the matrix.

Derivation

To derive the partial derivative of the given expression, we can utilize the chain rule and the product rule of matrix calculus. Let's start by applying the chain rule to the expression $\mbox{tr}(AXB \ \mbox{log}(CXD))$. We can rewrite this expression as $\mbox{tr}(A \cdot (XB \ \mbox{log}(CXD)))$, where we have factored out the matrix $A$.

\frac{\partial \ \mbox{tr}(AXB \ \mbox{log}(CXD))}{\partial X} = \frac{\partial \ \mbox{tr}(A \cdot (XB \ \mbox{log}(CXD)))}{\partial X}

Now, we can apply the product rule of matrix calculus, which states that the derivative of the product of two matrices is the product of the derivative of the first matrix and the second matrix, plus the product of the first matrix and the derivative of the second matrix. Applying this rule, we get:

\frac{\partial \ \mbox{tr}(A \cdot (XB \ \mbox{log}(CXD)))}{\partial X} = \mbox{tr}(A \cdot \frac{\partial (XB \ \mbox{log}(CXD))}{\partial X})

Next, we can apply the product rule again to the expression $\frac{\partial (XB \ \mbox{log}(CXD))}{\partial X}$. We get:

\frac{\partial (XB \ \mbox{log}(CXD))}{\partial X} = \frac{\partial (X \cdot (B \ \mbox{log}(CXD)))}{\partial X} = B \ \mbox{log}(CXD) + X \cdot \frac{\partial (B \ \mbox{log}(CXD))}{\partial X}

Now, we can apply the chain rule to the expression $\frac{\partial (B \ \mbox{log}(CXD))}{\partial X}$. We get:

\frac{\partial (B \ \mbox{log}(CXD))}{\partial X} = \frac{\partial (B \ \mbox{log}(C \cdot (XD)))}{\partial X} = \frac{\partial (B \ \mbox{log}(C \cdot (XD)))}{\partial (XD)} \cdot \frac{\partial (XD)}{\partial X}

Using the chain rule again, we can rewrite the expression $\frac{\partial (B \ \mbox{log}(C \cdot (XD)))}{\partial (XD)}$ as:

\frac{\partial (B \ \mbox{log}(C \cdot (XD)))}{\partial (XD)} = B \cdot \frac{\partial \ \mbox{log}(C \cdot (XD))}{\partial (XD)}

Now, we can apply the chain rule to the expression $\frac{\partial \ \mbox{log}(C \cdot (XD))}{\partial (XD)}$. We get:

\frac{\partial \ \mbox{log}(C \cdot (XD))}{\partial (XD)} = \frac{\partial \ \mbox{log}(C \cdot (X \cdot D))}{\partial (X \cdot D)} \cdot \frac{\partial (X \cdot D)}{\partial (XD)}

Using the chain rule again, we can rewrite the expression $\frac{\partial \ \mbox{log}(C \cdot (X \cdot D))}{\partial (X \cdot D)}$ as:

\frac{\partial \ \mbox{log}(C \cdot (X \cdot D))}{\partial (X \cdot D)} = D^{-1} \cdot C^{-1}

Now, we can substitute this expression back into the previous equation:

\frac{\partial (B \ \mbox{log}(C \cdot (XD)))}{\partial X} = B \cdot D^{-1} \cdot C^{-1} \cdot \frac{\partial (XD)}{\partial X}

Using the product rule again, we can rewrite the expression $\frac{\partial (XD)}{\partial X}$ as:

\frac{\partial (XD)}{\partial X} = D

Now, we can substitute this expression back into the previous equation:

\frac{\partial (B \ \mbox{log}(C \cdot (XD)))}{\partial X} = B \cdot D^{-1} \cdot C^{-1} \cdot D

Simplifying this expression, we get:

\frac{\partial (B \ \mbox{log}(C \cdot (XD)))}{\partial X} = B \cdot C^{-1}

Now, we can substitute this expression back into the previous equation:

\frac{\partial (XB \ \mbox{log}(CXD))}{\partial X} = B \ \mbox{log}(CXD) + X \cdot B \cdot C^{-1}

Now, we can substitute this expression back into the previous equation:

\frac{\partial \ \mbox{tr}(A \cdot (XB \ \mbox{log}(CXD)))}{\partial X} = \mbox{tr}(A \cdot (B \ \mbox{log}(CXD) + X \cdot B \cdot C^{-1}))

Using the linearity of the trace, we can rewrite this expression as:

\frac{\partial \ \mbox{tr}(A \cdot (XB \ \mbox{log}(CXD)))}{\partial X} = \mbox{tr}(A \cdot B \ \mbox{log}(CXD)) + \mbox{tr}(A \cdot X \cdot B \cdot C^{-1})

Using the cyclic property of the trace, we can rewrite the first term as:

\mbox{tr}(A \cdot B \ \mbox{log}(CXD)) = \mbox{tr}(B \ \mbox{log}(CXD) \cdot A)

Now, we can substitute this expression back into the previous equation:

\frac{\partial \ \mbox{tr}(A \cdot (XB \ \mbox{log}(CXD)))}{\partial X} = \mbox{tr}(B \ \mbox{log}(CXD) \cdot A) + \mbox{tr}(A \cdot X \cdot B \cdot C^{-1})

Using the linearity of the trace, we can rewrite this expression as:

\frac{\partial \ \mbox{tr}(A \cdot (XB \ \mbox{log}(CXD)))}{\partial X} = \mbox{tr}(B \ \mbox{log}(CXD) \cdot A) + \mbox{tr}(A \cdot X \cdot B \cdot C^{-1})

Now, we can use the result $\frac{\partial \ \mbox{tr}(\mbox{log}(X))}{\partial X} = X^{-T}$ to simplify the first term:

\mbox{tr}(B \ \mbox{log}(CXD) \cdot A) = \mbox{tr}(B \cdot A \cdot \mbox{log}(CXD)) = \mbox{tr}(A \cdot \mbox{log}(CXD) \cdot B) = \mbox{tr}(A \cdot (C^{-1} \cdot X \cdot D^{-1}) \cdot B)

Now, we can substitute this expression back into the previous equation:

\frac{\partial \ \mbox{tr}(A \cdot (XB \ \mbox{log}(CXD)))}{\partial X} = \mbox{tr}(A \cdot<br/>
**Derivative of Trace of Matrix Log: A Comprehensive Analysis**
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Q&A
Q: What is the derivative of the trace of a matrix log?
A: The derivative of the trace of a matrix log is a fundamental concept in matrix calculus, and it is defined as the matrix whose entries are the partial derivatives of the trace with respect to the corresponding entries of the matrix.
Q: How do I compute the derivative of the trace of a matrix log?
A: To compute the derivative of the trace of a matrix log, you can use the chain rule and the product rule of matrix calculus. The chain rule states that the derivative of a composite function is the product of the derivatives of the individual functions, while the product rule states that the derivative of a product of two functions is the product of the derivatives of the individual functions.
Q: What is the formula for the derivative of the trace of a matrix log?
A: The formula for the derivative of the trace of a matrix log is:
$$\frac{\partial \ \mbox{tr}(AXB \ \mbox{log}(CXD))}{\partial X} = \mbox{tr}(A \cdot (B \ \mbox{log}(CXD) + X \cdot B \cdot C^{-1}))$$
Q: How do I simplify the formula for the derivative of the trace of a matrix log?
A: To simplify the formula for the derivative of the trace of a matrix log, you can use the result $\frac{\partial \ \mbox{tr}(\mbox{log}(X))}{\partial X} = X^{-T}$ to simplify the first term:
$$\mbox{tr}(B \ \mbox{log}(CXD) \cdot A) = \mbox{tr}(A \cdot \mbox{log}(CXD) \cdot B) = \mbox{tr}(A \cdot (C^{-1} \cdot X \cdot D^{-1}) \cdot B)$$
Q: What is the final formula for the derivative of the trace of a matrix log?
A: The final formula for the derivative of the trace of a matrix log is:
$$\frac{\partial \ \mbox{tr}(AXB \ \mbox{log}(CXD))}{\partial X} = \mbox{tr}(A \cdot (C^{-1} \cdot X \cdot D^{-1}) \cdot B) + \mbox{tr}(A \cdot X \cdot B \cdot C^{-1})$$
Q: How do I use the formula for the derivative of the trace of a matrix log in practice?
A: To use the formula for the derivative of the trace of a matrix log in practice, you can substitute the values of the matrices $A$, $B$, $C$, and $D$ into the formula and simplify the expression to obtain the final result.
Q: What are some common applications of the derivative of the trace of a matrix log?
A: Some common applications of the derivative of the trace of a matrix log include:

Machine learning: The derivative of the trace of a matrix log is used in machine learning algorithms such as gradient descent and stochastic gradient descent.
Signal processing: The derivative of the trace of a matrix log is used in signal processing algorithms such as filtering and convolution.
Control theory: The derivative of the trace of a matrix log is used in control theory algorithms such as optimal control and model predictive control.

Conclusion
In conclusion, the derivative of the trace of a matrix log is a fundamental concept in matrix calculus, and it has numerous applications in various fields. The formula for the derivative of the trace of a matrix log is:
$$\frac{\partial \ \mbox{tr}(AXB \ \mbox{log}(CXD))}{\partial X} = \mbox{tr}(A \cdot (B \ \mbox{log}(CXD) + X \cdot B \cdot C^{-1}))$$
This formula can be simplified using the result $\frac{\partial \ \mbox{tr}(\mbox{log}(X))}{\partial X} = X^{-T}$ to obtain the final formula:
$$\frac{\partial \ \mbox{tr}(AXB \ \mbox{log}(CXD))}{\partial X} = \mbox{tr}(A \cdot (C^{-1} \cdot X \cdot D^{-1}) \cdot B) + \mbox{tr}(A \cdot X \cdot B \cdot C^{-1})$$
This formula can be used in practice to compute the derivative of the trace of a matrix log, and it has numerous applications in various fields.