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If you want to determine truth in this language with real or complex entries, then Yes. All this is expressible in the language of real-closed fields, simply by using components, and is therefore expressible in the complete theory of $\langle\mathbb{R},+,\cdot,0,1,\lt\rangle$, which is decidable by Tarski's theorem on real-closed fields. For example, quantifying over $n\times 1$ vectors is just $n$ quantifiers over reals (or $2n$ if you want complex numbers).

You mentioned that for each particular $n$, it is decidable by interpreting in the real-closed field, but my point is that this algorithm is uniform in $n$, and so you get a full decision procedure for the multi-sorted logic. That is, given a sentence in the multi-sorted language, we can tell which sorts are quantified over, and so we know how to translate it into a question about real-closed fields, which we can then answer. (I assume that you use a set-up as usual in the multi-sorted logic where each sort gets its own variables and quantifiers.)

If you intend to interpret it over the rationals, then No, since even the $1$-dimensional ring theory of $\langle\mathbb{Q},+,\cdot,0,1,\lt\rangle$ is not decidable, as the integers are definable there, and so you can express the halting problem.

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If you want to determine truth in this language with real or complex entries, then Yes. All this is expressible in the language of real-closed fields, simply by using components, and is therefore expressible in the complete theory of $\langle\mathbb{R},+,\cdot,0,1,\lt\rangle$, which is decidable by Tarski's theorem on real-closed fields. For example, quantifying over $n\times 1$ vectors is just $n$ quantifiers over reals (or $2n$ if you want complex numbers).

If you intend to interpret it over the rationals, then No, since even the $1$-dimensional ring theory of $\langle\mathbb{Q},+,\cdot,0,1,\lt\rangle$ is not decidable, as the integers are definable there, and so you can express the halting problem.