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I totally lied about this not being a natural thing to ask! As loup blanc alludes to, in fact $n \times n$ matrices such that $M^{-1} = \overline{M}$ can be interpreted as Galois descent data for desc …

von Neumann thought about this; the keyword is continuous geometry.

Let me expand a little on Ben's matrix remark, since all this talk about Grassmannians might give you the impression that the answer is complicated! Any $d \times d$ matrix has a rowspace of dimensio …

$k$ must be at least as large as $n$. Otherwise you have a system of less than $n$ linear equations in the variables $\log |x_i|$ (assuming they are real or complex), which is underdetermined or whic …

I can get an upper bound of $C \sqrt{n} \sqrt[4]{\log n}$ and it should be possible to push this technique to get $C_l \sqrt{n} \sqrt[2l]{\log n}$ for any $l$, although Fedor's approach might be a lot …

Let $V$ be an inner product space. The tensor product $V^{\otimes k}$ is equipped with an inner product induced from that of $V$ (obtained by contracting via the inner product $k$ times). Concretely, …

The space being defined is the span of the functions fn, and in the definition of the span we only allow finite sums of the basis vectors. Edit: I should also mention that the notion of infinite sum …

Linear algebra is relevant to pretty much every branch of mathematics. I think it'll make more sense once you see examples of it being relevant to every branch of mathematics. Here's an example that …

Suppose $v_j$ is an eigenbasis of $A$ with eigenvalues $\lambda_j$, so that $A v_j = \lambda_j v_j$. Then for all dual vectors $f$ we have $$\langle f, A v_j \rangle = \lambda_j \langle f, v_j \rangl …

Let me elaborate a little on what Steve Huntsman is talking about. A matrix is just a list of numbers, and you're allowed to add and multiply matrices by combining those numbers in a certain way. Wh …

I guess my comment is worth expanding into an answer. Over an arbitrary field $k$, an invariant polynomial on $\mathcal{M}_n(k)$ extends to an invariant polynomial on $\mathcal{M}_n(\bar{k})$. Since t …

Perhaps it would be most appropriate to answer your question with another question: how do you distinguish a finite-dimensional vector space from an infinite-dimensional one without talking about base …

We have $M^T = B^T A^T = -BA$. Since $AB$ and $BA$ have the same characteristic polynomial, and so do $M$ and $M^T$, it follows that $M$ and $-M$ have the same characteristic polynomial. Equivalently, …

The answer to Question 1 is yes, although I don't think I can extract a reasonable bound from the argument I have in mind. First observe that the question reduces to a question about largest (in absol …

Here are the details on Faisal's suggestion in the comments. Lemma (Dixmier): Let $k$ be an algebraically closed field, let $A$ be a $k$-algebra with $\dim A < |k|$, and let $V$ be a simple left $A$- …