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Is the cardinality of GL$_{n}$ ($\mathbb{R}$) and $\mathbb{R} \setminus\{0\}$ the same? If so what is the bijection and if not then why not?

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    Sorry, didn't know how to use MathJax. I have edited the question, please don't downvote. – 2017-02-12

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The cardinalities are the same (they're both uncountable), but I'm not sure how to construct an explicit bijection. But the case for $n = 1$ is easy. Pick a basis vector for $\mathbb R$. Then every element of $GL_1(\mathbb R)$ is represented by a $1 \times 1$ matrix whose determinant is nonzero, i.e. a nonzero real number.

In general, an element of $GL_n(\mathbb R)$ is represented by a $n \times n$ matrix with nonzero determinant. This gives an injection from $GL_n(\mathbb R)$ into the set $\mathbb R^{n^2}$. So the cardinalities of $GL_n(\mathbb R)$ and $\mathbb R^{n^2}$ are the same. It's then a standard result that the cardinality of $\mathbb R^{n^2}$ is the same as the cardinality of $\mathbb R$.

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    When you write : "an element of $GL_n(\mathbb{R}$ **is represented** by a $n\times n$ matrix with nonzero determinant", you mean **is equal to**, right ? – 2017-02-12
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    @Adren Yes, if that's your definition of $GL_n(\mathbb R)$. I always define $GL_n(\mathbb R)$ as the set of invertible linear maps $\mathbb R^n \to \mathbb R^n$, so the particular $n \times n$ matrix depends on a choice of basis. – 2017-02-12
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    Ok, I see. I thought that the standard notation was $GL(\mathbb{R}^n)$ for the set of linear automorphisms of $\mathbb{R}^n$ and $GL_n(\mathbb{R})$ for the set of invertible $n\times n$ real square matrix. – 2017-02-12