Intereting Posts

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Let $f:A \to B$ and $g:B\to A$ be arbitrary functions.
$\mathbb{Q}/\mathbb{Z}$ has a unique subgroup of order $n$ for any positive integer $n$?
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If this problem is not unbounded, what's wrong in this dual derivation?
What is an irreducible character of a finite group?
Irreducibility issue
Why is this more-detailed proof more acceptable than its trivial counterpart?
Finding a function from the given functional equation .

Let’s take $n$ vectors in $\mathbb{R}^n$ at random. What is the probability that these vectors are linearly independent? (i.e. they form a basis of $\mathbb{R}^n$)

(of course the problem is equivalent of “taken a matrix at random from $M_{\mathbb{R}}(n,n)$, what is the probability that its determinant $\neq 0$)

Don’t know if this question is difficult to answer or not. Please share any information about it! ðŸ™‚

- an exercise problem in the book “linear algebra” by Kenneth Hoffman and Ray Kunze, pg #5, Q.2
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- Trace minimization with constraints

(the $n$ vectors are meant with real values, I’m interested in solutions in $\mathbb{N}$ or $\mathbb{Q}$ or whatever fields you like)

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- Prove $\mathbb R$ vector space over $\mathbb Q$
- The number of esquares of idempotents in the rank 2 $\mathcal{D}$-class of $M_n(\mathbb{Z}_2)$.
- Showing equality in Cauchy-Schwarz inequality
- Is it a Ring? Is it PID?
- Calculating norm of a $2 \times 3$ matrix

As others have pointed out the main problem is what “taking a vector at random” means. Probability theory requires that one specifies a certain probability measure on ${\mathbb R}^n$ before one can make any predictions about outcomes of experiments concerning chosen vectors. E.g., if it is totally unlikely, meaning: the probability is zero, that a vector with $x_n\ne 0$ is chosen, then the probability that $n$ vectors chosen independently are linearly independent is $\>=0$, since with probability $1$ they all lie in the plane $x_n=0$.

A reasonable starting point could be installing a *rotational invariant* probability measure. As the length of the $n$ chosen vectors does not affect their linear dependence or independence this means that we are chosing $n$ independent vectors uniformly distributed on the sphere $S^{n-1}$. (This informal description has a precise mathematical meaning.)

Under this hypothesis the probability that the $n$ chosen vectors $X_k$ are linearly independent is $=1$.

*Proof.* The first vector $X_1$ is linearly independent with probability $1$, as $|X_1|=1$. Assume that $1< r\leq n$ and that the first $r-1$ vectors are linearly independent with probability $1$. Then with probability $1$ these $r-1$ vectors span a subspace $V$ of dimension $r-1$, which intersects $S^{n-1}$ in an $(r-2)$-dimensional “subsphere” $S_V^{r-2}$. This subsphere has $(n-1)$-dimensional measure $0$ on $S^{n-1}$. Therefore the probability that $X_r$ lies in this subsphere is zero. It follows that with probability $1$ the vectors $X_1$, $\ldots$, $X_{r-1}$, $X_r$ are linearly independent.

This depend of how you takes the vectors. For the case $n=2$, for example, if you have a vector $v_1$, the probability that you get a vector LD with $v_1$ is $0$. So, the probability that you get two vector LI is 1.

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