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I hope you could give a hint for proving that the following set is compact $(k<n)$:

$X=\left\{A\in \mathbb{R}^{n\times n}:A=A^{t},A^{2}=A,rank(A)=k\right\}$

I can proof that $X$ is bounded(not so difficult, to be honest). But to prove that $X$ is closed, the rank $k$ property stucks me.

Thanks in advance.

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Hint: Consider the characteristic polynomial.

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A projection has rank $k$ iff its characteristic polynomial is $x^{n-k}(x-1)^k$ (since it can be diagonalized so its diagonal entries are $n-k$ $0$s and $k$ $1$s). The coefficients of the characteristic polynomial are polynomials and hence continuous in the entries of the matrix, so the set of matrices with any given characteristic polynomial is closed.

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one approach to show openness of the complement:

0.) Select a norm easy to work with-- I suggest Frobenius norm
1.) The result is immediate for non symmetric matrices
2.) for any matrix $B$ in the complementary set, it is symmetric and orthogonally diagonalizable, and the Frobenius Norm is orthogonally invariant in reals, so assume WLOG that your matrix $B$ is diagonal with eigenvalues in the usual ordering $\lambda_1 \geq \lambda_2 \geq .... \geq \lambda_n$. Now show for any $B$, $$\big \Vert B-A\big \Vert_F \geq 0$$

and the inequality is strict unless we select $A:=B$ which is allowable iff diagonal matrix $B$ has $k$ ones and $n-k$ zeros on the diagonal, i.e. iff $B$ is a rank k projector. When the inequality is strict, then there is some $\delta \gt 0$ neighborhood around $B$ such that all matrices in this neighborhood are not in the set containing $A$, hence the set containing $B$'s is open and your set containing $A$'s is closed. Put differently the complementary set of symmetric matrices, in effect (up to orthogonal similarity), is the set of all diagonal matrices that aren't purely $k$ oness and $n-k$ zeros.

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