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Consider a real matrix $A$ of dimension $n \times n$. Assume $k \leq n$ is given. I am looking for ways to describe the following set of matrices in terms of properties of $A$. $\mathcal{S}(A) = \{B \in \mathbb{R}^{n \times k}: \exists F \in \mathbb{R}^{k \times k}. AB = BF\}$.

One example of a matrix $B \in \mathcal{S}(A)$ is any matrix whose columns are eigenvectors of $A$. In this case, $F$ is diagonal and contains the eigenvalues. But maybe there are other choices of $B$, where $F$ is not diagonal.

I am looking for an answer which would tell me something like: all members of $\mathcal{S}(A)$ are related to, say, the eigenvectors of $A$ in some way (or to some other property of $A$, not just eigenvectors).

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1 Answer 1

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One possible description of this set is the following. Let $\Im(C)$ denote the image of the matrix $C$. Thus, $S(A)=\{B,\ \Im(AB)\subset\Im(B)\}$.

Before proving it, notice that we can provide many examples of matrices in $S(A)$ with this description. For example, any matrix $B$ such that $\Im(A)\subset\Im(B)$ belongs to $S(A)$, because $\Im(AB)\subset\Im(A)\subset\Im(B)$.

Now let's prove that $S(A)=\{B,\ \Im(AB)\subset\Im(B)\}$.

One direction is obvious.

If $AB=BF$ then $\Im(AB)=\Im(BF)\subset\Im(B)$. Thus, $S(A)\subset \{B,\ \Im(AB)\subset\Im(B)\}$.

Now, suppose $\Im(AB)\subset\Im(B)$.

Let $B^{+}$ be the pseudo inverse of $B$.

Since $BB^{+}$ is a orthogonal projection onto the $\Im(B)$ and $\Im(AB)\subset\Im(B)$ then $BB^{+}AB=AB$. Define $F=B^{+}AB$ and we obtain $BF=AB$. Thus, $\{B,\ \Im(AB)\subset\Im(B)\}\subset S(A)$.

Edit: Notice that $\Im(AB)=A(\Im(B))$. Thus, $S(A)=\{B,\ \Im(B)\text{ is an invariant subspace of A}\}$. In order to construct the set $S(A)$, we must find the invariant subspaces of $A$. The columns of $B\in S(A)$ must be a generating set for the invariant subspace.

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Thank you for the answer, but unfortunately it seems a bit tautological. I was looking for a way to constructively describe this set. I probably didn't make this explicit in the question. –  ziutek Aug 30 at 14:10
    
@ziutek Notice that $\Im(B)$ must be an invariant subspace of $A$. The columns of $B$ must be a generating set for an invariant subspace. –  Daniel Aug 31 at 16:17

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