How to prove randomly-generated high-dimensional $0$-$1$ vectors are probably independent? Today our teacher told us that if you randomly generated ten 100-dimensional $0$-$1$ vectors, it's very unlikely that they are dependent in $\mathbb{R} ^ {100}$. To be specific, every entry has equal chance to be $0$ or $1$.  
I have thought about this, but I can't get a nice upper bound for the probability that they are dependent. Can someone help me?
 A: I assume stochastic independence.
Using determinants, we easily see that if the vectors are dependent over $\mathbb{R}$, then they (in fact, their projections) also are dependent over $\mathbb{F_2}=\mathbb{Z}/2\mathbb{Z}$. But the probability that 10 independent randomly choosen (with uniform probability) vectors of $(\mathbb{F}_2)^{100}$  are dependent over $\mathbb{F_2}$ is
$$
1 - \frac{(2^{100}-1)(2^{100} - 2)\times(2^{100}-2^9)}{(2^{100})^{10}}= 1 - \prod_{i=91}^{100} (1-2^{-i}) \leq 1 - (1-2^{-91})^{10} \leq 10\times  2^{-91}
$$
(where we used Bernoulli's inequality $(1-p)^n \geq 1 - np$).
Note that this bound is very crude.

The passage from $\mathbb{R}$ to $\mathbb{F}_2$ is a special case of the following lemma:

Let $A$ be a $n\times m$ matrix with coefficients in $\mathbb{Z}$, and $p$ be a prime number.
  Note $\overline{A}$ the $n\times m$ matrix with coefficients in $\mathbb{F}_p$ given by $\overline{A}_{i,j} = A_{i,j} \bmod p$. We have, $\mathrm{rank}_\mathbb{R}\,A \geq \mathrm{rank}_{\mathbb{F}_p}\,\overline{A}$.

Proof. Remember that the rank of a matrix is the size of the largest non-vanishing minor. Furthermore, notice that for every square matrix $B$ with coefficients in $\mathbb{Z}$, we have $\det_\mathbb{R} B \in \mathbb{Z}$ and $$ \det_{\mathbb{F_p}} (\overline{B}) = \det_{\mathbb{R}} (B) \bmod p$$
since the determinant is a polynomial function of the coefficients.
A: This was a thought which is possibly wrong. Please read the Comments to it. 
This is a thought. Let us stack all this vectors in the matrix $X$ whose size is $10 \times 100$. Your question is equivalent to asking if $X$ is a full-rank matrix. When is it full rank?  One way is to look at its singular values, if the lowest singular value is non-zero, then it is full rank. What is the probability for this? Thus, you are looking at the distribution of the lowest singular value of a given random binary rectangular matrix. It is a continuous random variable. To see this, note that the lowest singular value of $X$ is the square root of the smallest eigenvalue of $X^TX$. Eigenvalues are a continuous function of the entries of the matrix. 
So the probability that it will take a discrete point $0$ is zero. So the matrix should be full rank with probability one. 
If you are interested in learning more about the distribution of lowest singular value of a matrix, this blog post by terence tao would be a good starting point. 
