I have a conjecture:
If two finite-dimensional matrix spaces $\mathcal{A},\mathcal{B}\subseteq \mathrm{M}(n,\mathbb{F})$ have the same dimension, then there would exist $P,Q\in\operatorname{GL}(n,\mathbb{F})$ such that $\mathcal{B}=Q^{-1}\mathcal{A}P$. (Reminder: Matrix spaces refer to linear spaces spanned by matrices.)
Is it true or false? It is well-known that two finite-dimensional linear spaces are isomorphic if and only if they have the same dimension. Also, this means there is a bijection from a basis of one space to a basis of the other. Does it complete the proof of my conjecture?
It seems that if the bases are vectors instead of matrices, then the statement would be true. So I'm very confused if the two cases can be the same. Thank you!
Best Answer
No this is not true.
Let $T\in GL(n,\mathbb F)$ be invertible and let $S \in M(n,\mathbb F)$ be non-invertible. Let $\mathcal A = \text{span}(T)$ and $\mathcal B = \text{span}(S)$ so that $\dim \mathcal A = \dim \mathcal B = 1$. Note that every element of $\mathcal B$ is not invertible.
Then for any $P,Q \in GL(n,\mathbb F)$ we have that $Q^{-1} TP$ is invertible (as $T$ is invertible) and hence cannot be in $\mathcal B$ (as every element of $\mathcal B$ is not invertible), so $\mathcal B \ne Q^{-1} \mathcal A P$.