In this post, my goal is to show that for any \(n\) x \(n\) matrix \(A \in \operatorname{Mat}_{n \times n}(\mathbb{F})\) and \(n\) x \(m\) matrix \(X \in \operatorname{Mat}_{n \times m}(\mathbb{F})\) with entries in a field \(\mathbb{F}\), the column space of \(X\) is \(A\)-invariant if and only if \(AX = XY\) for some \(m\) x \(m\) matrix \(Y \in \operatorname{Mat}_{m \times m}(\mathbb{F})\).

Firstly, suppose that we are given matrices \(A \in \operatorname{Mat}_{n \times n}(\mathbb{F})\) and \(X \in \operatorname{Mat}_{n \times m}(\mathbb{F})\) where the column space of \(X\) is \(A\)-invariant. Then, for all \(1 \leq j \leq n\), we can construct a vector \(y_j \in \mathbb{F}^{m}\) such that:

\[A (X e_j)\] \[= \sum\limits_{k = 1}^{m} a_{k,j} (X e_k)\]

[ for some \(a_{k, j} \in \mathbb{F}\) for all \(1 \leq k \leq n\), since the column space of \(X\) is \(A\)-invariant ]

\[= X \left ( \sum\limits_{k = 1}^{m} a_{k,j} e_k \right )\] \[= X y_j\]

[ by letting \(y_j = \sum\limits_{k = 1}^{m} a_{k,j} e_k\) ]

. From there, we can construct a matrix \(Y \in \operatorname{Mat}_{m \times m}(\mathbb{F})\) such that:

\[AX\] \[= \left [ A (X e_1) \mid \cdots \mid A (X e_m) \right ]\] \[= \left [ X y_1 \mid \cdots \mid X y_m \right ]\] \[= XY\]

[ by letting \(Y = \left [ y_1 \mid \cdots \mid y_m \right ]\) ]. Therefore, since our particular choices of \(A\) and \(X\) in this context are arbitrary, we have that: for any matrices \(A \in \operatorname{Mat}_{n \times n}(\mathbb{F})\) and \(X \in \operatorname{Mat}_{n \times m}(\mathbb{F})\), if the column space of \(X\) is \(A\)-invariant, then there exists a matrix \(Y \in \operatorname{Mat}_{m \times m}(\mathbb{F})\) such that \(AX = XY\) as desired. Secondly, suppose that we are given matrices \(A \in \operatorname{Mat}_{n \times n}(\mathbb{F})\) and \(X \in \operatorname{Mat}_{n \times m}(\mathbb{F})\) where \(AX = XY\) for some \(Y \in \operatorname{Mat}_{m \times m}(\mathbb{F})\). Then, for all \(1 \leq j \leq m\), we have that:

\[A(X e_j)\] \[= X (Y e_j)\] \[= X \left( \sum\limits_{k=1}^{m} Y_{k, j} e_k \right )\] \[= \sum\limits_{k=1}^{m} Y_{k, j} (X e_k)\]

. This means, in particular, that \(A (X e_j) \in \operatorname{Col}(X)\) for all \(1 \leq j \leq m\).

Now, select any vector \(u \in \operatorname{Col}(X)\); then we know that \(u = X v = \sum\limits_{j=1}^{m} v_j (X e_j)\) for some vector \(v \in \mathbb{F}^{m}\). From there, we can deduce that:

\[A u\] \[= A \left ( \sum\limits_{j=1}^{m} v_j (X e_j) \right )\] \[= \sum\limits_{j=1}^{m} v_j (A (X e_j) )\] \[= \sum\limits_{j=1}^{m} v_j \left ( \sum\limits_{k=1}^{m} Y_{k, j} (X e_k) \right )\] \[= \sum\limits_{j=1}^{m} \sum\limits_{k=1}^{m} v_j Y_{k, j} (X e_k)\] \[= \sum\limits_{k=1}^{m} \left ( \sum\limits_{j=1}^{m} v_j Y_{k, j} \right ) (X e_k)\]

. This means that \(A u \in \operatorname{Col}(X)\), and hence that \(\operatorname{Col}(X)\) is \(A\)-invariant, since our choice of \(u\) in this inner context is arbitrary. Therefore, since our contextual choices of \(A\) and \(X\) are likewise arbitrary, we have that: for any matrices \(A \in \operatorname{Mat}_{n \times n}(\mathbb{F})\) and \(X \in \operatorname{Mat}_{n \times m}(\mathbb{F})\), if \(AX = XY\) for some \(Y \in \operatorname{Mat}_{m \times m}(\mathbb{F})\), then the column of space of \(X\) is \(A\)-invariant as desired. Hence we can conclude that for any \(n x n\) matrix \(A \in \operatorname{Mat}_{n \times n}(\mathbb{F})\) and \(n\) x \(m\) matrix \(X \in \operatorname{Mat}_{n \times m}(\mathbb{F})\) with entries in a field \(\mathbb{F}\), the column space of \(X\) is \(A\)-invariant if and only if \(AX = XY\) for some \(m\) x \(m\) matrix \(Y \in \operatorname{Mat}_{m \times m}(\mathbb{F})\), as desired.