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Let $A$ be an $n\times n$ gaussian matrix whose entries are i.i.d. copies of a gaussian variable, and $\left\{ a_{j}\right\} _{j=1}^{n}$ be the column vectors of $A$. How to show that the probability $\mathbb{P}\left(d\geq t\right)\leq Ce^{-ct}$ for some $c,C>0$ and every $t>0$, where $d$ is the distance between $a_{1}$ and the $n-1$-dimensional subspace spanned by $a_{2},\cdots,a_{n}$.

Thanks a lot!

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Is this Gaussian variable assumed to have mean 0? –  Harald Hanche-Olsen Mar 13 '10 at 13:23
Yes, we assume it is in stardard normal distribution. –  user4606 Mar 14 '10 at 3:46

1 Answer 1

symmetry shows that you can suppose that $$\text{span}(a_2, \ldots, a_n) = ( x \in \mathbb{R}^n: x_1=0 ) = H.$$ Hence you just want to show that $d(a_1, H) = |A_{1,1}|$ is exponentially small - there is a closed-form expression for that: $$ P(d>t) = P(|\mathcal{N}(0,1)|>t) = \sqrt{\frac{2}{\pi}} \int_t^{\infty} e^{-\frac{x^2}{2}} dx.$$

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I don't see how to use symmetry to get $$\text{span}(a_2, \ldots, a_n) = ( x \in \mathbb{R}^n: x_1=0 ) = H$$. Could you please explain a little more on that? –  user4606 Mar 14 '10 at 3:25
You can see it using the concepts of conditional probability. For each possible $H$ (using Alekk's notation) there is a probability distribution of $d$ conditional on that $H$, and the total (unconditional) distribution of $d$ is obtained from that by integrating over $H$. Now the conditional distribution of $d$ given $H$ is independent of $H$, because the joint probability distribution of all the $a$'s is invariant under rotation. So integrating over $H$ is unnecessary – just use one fixed $H$. –  Harald Hanche-Olsen Mar 14 '10 at 4:40
If you want $\ell_1$ distance, I think the formula (if one can be found) will be rather complex, and its derivation horrendously so. You might experiment with a computer algebra system and see how it works out for small values of $n$. –  Harald Hanche-Olsen Mar 14 '10 at 21:53

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