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Consider the following random matrix \begin{equation} \mathbf{G}=\left( \begin{array}{cc} g_{1,1} & g_{1,2} \\ g_{2,1} & g_{2,2} \\ \end{array} \right) \end{equation} where the entries of $\mathbf{G}$ are i.i.d standard Gaussian random variables. Then what is the rank of this matrix?

I kind feel that $\text{Probablity}(\text{Rank}(\mathbf{G})=2)=1$. But not very confident. Can anyone give a judgement or a proof?

Thank!

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  • $\begingroup$ The set of singular matrices has measure zero (more generally, see this MO question). So, if entries of $\mathrm{G}$ follow a continuous joint distribution (Gaussian or not), the probability that $\mathrm{G}$ is singular is zero. $\endgroup$
    – user1551
    Oct 18, 2013 at 15:46

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Intuition says that $P(A\in\mathbb{R}^{n\times n}\text{ is singular})=0$ as the set of nonsingular matrices is dense in $\mathbb{R}^{n\times n}$. Nice asymptotic results can be found, e.g., in this talk.

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  • $\begingroup$ "Intuition" should have nothing to do with the result. Density does not imply full measure, in general. $\endgroup$
    – Did
    Jan 17, 2015 at 22:13

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