The differential of a symmetric matrix in terms of its eigen-decomposition Given a (square) symmetric matrix $A$, I would like to write its first order perturbation in terms of its eigenvalue decomposition
$$A=Q\Lambda Q^T$$
I'm thinking about this problem in terms of perturbing $Q$ and $\Lambda$, and then observing the resulting perturbation in $A$.
For $A+dA$ to remain symmetric, this will require that the perturbation direction lie within the subspace of symmetric matrices.  In particular, this requires that $dQ$ must satisfy $Q^T(dQ)+(dQ^T)Q=0$ and that $d\Lambda$ be diagonal.  Apparently, the resulting differential of $A$ can be written as
$$dA=\prod_{i<j}|\lambda_i-\lambda_j|(d\Lambda)Q^T(dQ)$$
I went ahead expanded it using the product rule
$$dA=d(Q\Lambda Q^T)=(dQ)\Lambda Q^T + Q(d\Lambda)Q^T + Q\Lambda(dQ^T)$$
But I'm not really sure what to do from here.

If $Q\in O(n)$ and the diagonal matrix $\Lambda$ (of the eigenvalues) vary with $t$, then it is not difficult. $A'=Q'\Lambda Q^T+Q\Lambda'Q^T+Q\Lambda Q'^T$; here $QQ^T=I$ implies $Q'Q^T+QQ'^T=0$, that is $Q'Q^T$ is skew-symmetric or $Q'=KQ$ where $K(t)$ is skew-symm. and $\Lambda'$ is diagonal.
Finally $A'=KQ\Lambda Q^T+Q\Lambda'Q^T-Q\Lambda Q^TK=KA-AK+Q\Lambda'Q^T$. 
If $A$ varies with $t$, then it is more difficult; cf. the wikipedia reference cited by user1952009; yet, this ref. is very disputed.
