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Consider $\frac{dX}{dt}(t)=AX(t) + X(t)A^{T} + BX(t)B^{T}$ with $X(0)=X_0$, where $X$ is a square matrices, and $A$, $B$ and $X_0$ are diagonal matrices.

When is $X$ diagonal?

For the equation $\frac{dX}{dt}=AX + XA^{T} + BB^{T}$, if the initial condition $X_0$ is diagonal, then what we have is just multiple scalar valued ODEs, so it is clear. But in the presence of $BXB^{T}$ I am not really sure. I thought of Euler methods, and argue each step $\hat{X}$ is diagonal, and then argue a convergence etc., but I want to think about an infinite dimensional setting as well, and wondering if there is a neat way of proving/disproving this.

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    Surely $AX_0$ has off-diagonal elements hence $X(t)$ will develop them. – 2017-02-02
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    Oops sorry I don't know why I did that. $A$ is assumed to be diagonal now. – 2017-02-02
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    Then $X$ will stay diagonal. A more interesting question would be if $X_0$ or $A$ or $B$ has small off-diagonal elements does the off-diagonal elements of $X(t)$ grow to be a problem. – 2017-02-02

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A newbye answer: the subspace of the diagonal matrices is a linear subspace. if $X_0$ is diagonal then the right hand side of the ODE is diagonal (as long as B and A are diagonal). Hence the subspace of the diagobal matrices is invariant under the flow of that ODE. Thus yes, if A B are diagonal then as long as the initial condition is diagonal you should have a diagonal solution.

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    Sorry, this sounds like an interesting way of seeing this, but how can you claim "if $X_0$ is diagonal then the right hand side of the ODE is diagonal (as long as $B$ and $A$ are diagonal). "? – 2017-02-02
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    Well it is a sum of products of diagonal matrices. The product of diagonal matrices is diagonal, the sum of diagonal matrices is diagonal – 2017-02-02
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    Sure. But my question is, there is $X(t)$, which is what we want to know if it is diagonal. I guess now I understand it as, if we consider the diagonal terms separately, they are just linear ODEs, and if we put them back to the matrix form we see that they solve the original problem, and shouldn't be difficult to check the uniqueness and done? – 2017-02-02
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    I don't really see the causation from the second sentence to the third sentence. Sorry, currently, although being interesting, those looks to me equivalent statements to something I wish to understand. – 2017-02-02
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    Yes you can also see it that way. In any case, if the derivative stays in a subspace for almost all t, then the solution stays in that space – 2017-02-02
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    @user412313 What LJSilver meant in the causation from the second sentence to the third is the following. To check that some domain or subspace is invariant under the flow you only have to check how the vector field is aligned with boundary of this domain or this subspace. This is the content of [Bony-Brezis theorem](https://en.wikipedia.org/wiki/Bony%E2%80%93Brezis_theorem). The subspace of diagonal matrix is a manifold, and LJSilver checked that vector field is tangent to this subspace, hence it is invariant. – 2017-02-04
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    @Evgeny Ah thank you very much for this pointer. This is very helpful even though I cannot upvote it cuz I haven't got enough reputation. – 2017-02-04
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    @Evgeny What should I read to know more about infinite dimension cases? E.g., suppose $dX/dt=AX+B(X)$, where $A$ generates a $C_0$-semigroup on a Banach space, and $B$ is nice. What are the condition such that the solution remains in a certain subspace? I think if $A$ and $B$ restricted to a closed subspace maps to that subspace itself, and the initial condition is in the subspace, the solution stays in the subspace because it is just another initial value problem on a different Banach space (our subspace is assumed to be closed). But not sure about more general cases. – 2017-02-04
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    @user412313 Sorry, I'm not very competent here, I'm doing finite-dimensional dynamics. But I think that condition that $\forall v \in V \; \; Av + B(v) \subseteq V$ should also serve as condition for invariance of $V$. – 2017-02-04