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The theorem that projective spaces are not affine varieties is a theorem over the complex numbers. As you note, your construction fails over the complex numbers, so there is no contradiction.

To give an even simpler example over the reals, $\mathbb {RP}^1=S^1$ is the vanishing set of $x^2+y^2-1=0$.

I think you're slightly confused about what the functions of your conditions are. Every matrix satisfying condition $1$ are is diagonalizable. Indeed, all matrices satisfying a polynomial equation without repeated roots are diagonalizable, and $A^2-A=0$ certainly has no repeated roots. The second condition is ensuring ensures that the kernel is the orthogonal complement of the image, making sure that the projection which is determined by the only way to ensure that there is a unique projection with a given image. Remove it, and you have an affine bundle on $\mathbb P^n$, which is of course perfectly alright.

If you take $AA^T=0$ to not be the conjugate-transpose in the complex case, then you get an affine subvariety of $\mathbb P^n$ - the complement of the hypersurface of points corresponding to lines where a certain bilinear form on $\mathbb C^{n+1}$ is nonzero, that is, the vanisihing set of a degree two polynomial equation.

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The theorem that projective spaces are not affine varieties is a theorem over the complex numbers. As you note, this your construction fails over the complex numbers, so there is no contradiction.

To give an even simpler example over the reals, $\mathbb {RP}^1=S^1$ is the vanishing set of $x^2+y^2-1=0$.

I think you're slightly confused about what the functions of your conditions are. Every matrix satisfying condition $1$ are diagonalizable. Indeed, all matrices satisfying a polynomial equation without repeated roots are diagonalizable, and $A^2-A=0$ certainly has no repeated roots. The second condition is ensuring that the kernel is the orthogonal complement of the image, making sure that the projection is determined by the image. Remove it, and you have an affine bundle on $\mathbb P^n$, which is of course perfectly alright.

If you take $AA^T=0$ to not be the conjugate-transpose in the complex case, then you get an affine subvariety of $\mathbb P^n$ - the complement of the hypersurface of points corresponding to lines where a certain bilinear form on $\mathbb C^{n+1}$ is nonzero, that is, the vanisihing set of a degree two polynomial equation.

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The theorem that projective spaces are not affine varieties is a theorem over the complex numbers. As you note, this fails over the real complex numbers.

To give an even simpler example over the reals, $\mathbb {RP}^1=S^1$ is the vanishing set of $x^2+y^2-1=0$.

I think you're slightly confused about what the functions of your conditions are. Every matrix satisfying condition $1$ are diagonalizable. Indeed, all matrices satisfying a polynomial equation without repeated roots are diagonalizable, and $A^2-A=0$ certainly has no repeated roots. The second condition is ensuring that the kernel is the orthogonal complement of the image, making sure that the projection is determined by the image. Remove it, and you have an affine bundle on $\mathbb P^n$, which is of course perfectly alright.

If you take $AA^T=0$ to not be the conjugate-transpose in the complex case, then you get an affine subvariety of $\mathbb P^n$ - the complement of the hypersurface of points corresponding to lines where a certain bilinear form on $\mathbb C^{n+1}$ is nonzero, that is, the vanisihing set of a degree two polynomial equation.

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