The question provided discusses a symmetric matrix $A \in \mathbb{R}^{n \times n}$ and its spectral decomposition $A = Q \Lambda Q^T$, where $\Lambda$ is a diagonal matrix consisting of the eigenvalues of $A$, and $Q$ is the orthogonal matrix of eigenvectors. You are asked to compute the gradients of the trace function $\text{tr}(A)$ with respect to $\Lambda$ and $Q$, respectively.

1. Gradient of $\text{tr}(A)$ with respect to $\Lambda$:

Since $\text{tr}(A) = \text{tr}(\Lambda)$ because the trace of a matrix is invariant under similarity transformations (i.e., $\text{tr}(A) = \text{tr}(Q \Lambda Q^T) = \text{tr}(\Lambda)$), and $\Lambda$ is a diagonal matrix containing the eigenvalues of $A$, we can directly differentiate this:

$\nabla_\Lambda \text{tr}(A) = \nabla_\Lambda \text{tr}(\Lambda)$

Since $\text{tr}(\Lambda)$ is simply the sum of the diagonal elements of $\Lambda$, and those diagonal elements are the eigenvalues, the derivative is:

$\nabla_\Lambda \text{tr}(\Lambda) = I$

Thus, the gradient with respect to $\Lambda$ is the identity matrix $I$.

2. Gradient of $\text{tr}(A)$ with respect to $Q$:

The trace function $\text{tr}(A) = \text{tr}(Q \Lambda Q^T)$ does not explicitly depend on $Q$ because the trace is invariant under orthogonal transformations. Therefore, the gradient of $\text{tr}(A)$ with respect to $Q$ is zero:

$\nabla_Q \text{tr}(A) = 0$

Final Results:

1. $\nabla_\Lambda \text{tr}(A) = I$
2. $\nabla_Q \text{tr}(A) = 0$

Would you like further details on how these gradients are computed or related concepts? Here are 5 related questions you could explore:

1. What is the significance of spectral decomposition in matrix analysis?
2. Why is the trace invariant under orthogonal transformations?
3. How does spectral decomposition relate to the eigenvalues and eigenvectors of a matrix?
4. What is the gradient of the determinant with respect to $\Lambda$ and $Q$?
5. Can the results be generalized for non-symmetric matrices?

Tip: When computing matrix gradients, always consider whether the function involves any invariance under transformations like orthogonal or unitary matrices, as this often simplifies the calculation.