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Let's solve the problem step by step as outlined:

We are asked to prove that, for matrices \( Y \in \mathbb{R}^{n \times q} \), \( X \in \mathbb{R}^{n \times p} \), and \( B \in \mathbb{R}^{p \times q} \),
\[
\frac{\partial}{\partial B} \operatorname{tr}\left((Y - XB)^\top (Y - XB)\right) = -2 X^\top (Y - XB).
\]
The trace function \( \operatorname{tr}(\cdot) \) simplifies the summation of the diagonal elements of a matrix. To prove the result, we need to compute the gradient of the trace with respect to \( B \).

### Step 1: Expand the trace expression
The expression inside the trace is:
\[
\operatorname{tr}\left((Y - XB)^\top (Y - XB)\right).
\]
First, expand this expression:
\[
\operatorname{tr}\left((Y - XB)^\top (Y - XB)\right) = \operatorname{tr}(Y^\top Y) - \operatorname{tr}(Y^\top XB) - \operatorname{tr}(B^\top X^\top Y) + \operatorname{tr}(B^\top X^\top X B).
\]
Now, we'll differentiate each term with respect to \( B \).

### Step 2: Differentiate each term
1. \( \frac{\partial}{\partial B} \operatorname{tr}(Y^\top Y) \):
   - This term is independent of \( B \), so its derivative is zero:
   \[
   \frac{\partial}{\partial B} \operatorname{tr}(Y^\top Y) = 0.
   \]

2. \( \frac{\partial}{\partial B} \operatorname{tr}(Y^\top XB) \):
   - By the linearity of the trace, we can apply the rule \( \frac{\partial}{\partial B} \operatorname{tr}(A^\top B) = A \):
   \[
   \frac{\partial}{\partial B} \operatorname{tr}(Y^\top XB) = -X^\top Y.
   \]

3. \( \frac{\partial}{\partial B} \operatorname{tr}(B^\top X^\top Y) \):
   - This term is the transpose of the previous term, so:
   \[
   \frac{\partial}{\partial B} \operatorname{tr}(B^\top X^\top Y) = -X^\top Y.
   \]

4. \( \frac{\partial}{\partial B} \operatorname{tr}(B^\top X^\top X B) \):
   - This is a quadratic form, and the derivative is given by:
   \[
   \frac{\partial}{\partial B} \operatorname{tr}(B^\top X^\top X B) = 2 X^\top X B.
   \]

### Step 3: Combine the results
Now, summing the results from all terms, we get:
\[
\frac{\partial}{\partial B} \operatorname{tr}\left((Y - XB)^\top (Y - XB)\right) = -X^\top Y - X^\top Y + 2 X^\top X B.
\]
This simplifies to:
\[
\frac{\partial}{\partial B} \operatorname{tr}\left((Y - XB)^\top (Y - XB)\right) = -2 X^\top (Y - XB).
\]
Thus, we have proved the required result.

---

Would you like a more detailed explanation of any step or related examples? Here are some follow-up questions for better understanding:

1. What is the significance of the trace function in matrix calculus?
2. How do we apply the chain rule in matrix calculus, and why is it important in this problem?
3. Can you explain how matrix derivatives differ from scalar derivatives?
4. How does expanding and simplifying the matrix expression help in finding the derivative?
5. What are some common applications of this derivative in optimization problems?

**Tip**: In matrix calculus, always ensure the dimensions of the matrices align during operations, particularly when transposing or applying the trace function.

For the matrices Y ∈ ℝⁿˣq, X ∈ ℝⁿˣp, B ∈ ℝᵖˣq, without using vector or matrix identities prove that ∂/∂B tr((Y - XB)ᵀ(Y - XB)) = -2Xᵀ(Y - XB).

This problem requires proving the derivative of a trace expression for matrices Y, X, and B without using matrix identities. It focuses on key concepts in matrix calculus, including gradient computation and matrix differentiation techniques. Suitable for advanced undergraduate or graduate students.

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