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You can have $m=n+1$. Take the vertices of a regular simplex with centre at the origin. You can't have $m=n+2$. There is at least a two-dimensional space of vectors $(a_1,\ldots,a_{n+2})$ such that $$ …

If every matrix has a Smith normal form, then every finitely generated $R$-submodule $M$ of $R^n$ satisfies $R^n/M$ is a finite direct sum of modules isomorphic to $R/aR$. If $R$ is Noetherian this im …

I shouldn't expect there to be exact results; compare the similar problem with matrices with entries $\pm1$. For an $n$-by-$n$ matrix with entries $\pm1$ one gets an upper bound for the determinant of …

The general element is $\pm\exp(A)$ where $A$ is skew-symmetric. (This gives each element infinitely often). This trick essentially works for all compact Lie groups. There is also the Cayley paramete …

If a vector space had bases of two different finite sizes $m < n$ say, then expressing one in terms of the other gives $m$ by $n$ and $n$ by $m$ matrices $A$ and $B$ such that $BA=I_n$. Now use Gaussi …

For a start the accepted usage for "rational canonical form" in the literature is for a diagonal sum $C(f_1)\oplus C(f_2)\oplus\cdots\oplus C(f_k)$ where $C(f_i)$ is the companion matrix for a monic p …

The answer is always yes. Indeed the set is path-connected. Let $C(f)$ denote the companion matrix associated to the monic polynomial $f$. Every matrix $A$ is similar to a matrix in rational canonica …

For real/complex vector spaces, this is Theorem 1.21 in Rudin's Functional Analysis (2nd ed.). I believe the proof works for any complete field, but haven't checked in detail.

For your first question, I presume you also wish to insist that $k$ be the least integer such that $A^k=I$. The matrix $A$ is then similar over your field to a direct sum $B_1,\ldots,B_m$ of companion …

Yes. The case where $v=0$ is trivial so suppose $v\ne0$. Consider the projection map from $V$ to the hyperplane orthogonal to $v$ and let $a'$ and $b'$ be the images of $a$ and $b$ under this projecti …

What carries over? As Peter pointed out, a submodule of a free $\mathbb{Z}$-module though free need not have a complement. Indeed each submodule of a free $\mathbb{Z}$-module is free, but a quotient …

This occurs if and only if the matrices $Q^r$ and $Q^s$ are conjugate. This is the case if and only if these matrices are conjugate over the algebraic closure of $\mathbb{F}_p$. If $Q$ iis diagonaliza …

This is the question of finding maximal isotropic subspaces of an inner-product space. The results for finite fields of odd characteristic are well-known and can be found in Serre's Course in Arithmet …

Let $A$ and $B$ be infinite sets. Let $M$ be a rank $|B|$ module with basis $e_b$ for $b\in B$. If we take $|A|$ elements $m_a$ of $M$, then each can be expressed in terms of finitely many of the $e_b …

Any transformation $$(v_1,\ldots,v_n)\mapsto (v_1,\ldots,v_{j-1},v_j+av_k,v_{j+1},\ldots,v_n)$$ for $j\ne k$ is achievable by means of some such matrix. It suffices to reduce an admissible vector to $ …