I was solving a GRE-like test and came across the following question (it's argued that the only answer is correct)
Suppose $A$ is a projector with the symmetric $n \times n \ (n > 2)$ matrix. If $A$ is neither $I$ nor zero, then
(i) Is $\{x \in \mathbb{R}^n: x^TAx = 1\}$ bounded?
(ii) Is $\{x \in \mathbb{R}^n: x^TAx = 0\}$ a linear subspace?
As I know a projector with symmetric matrix is defined by a positive semidefinite matrix hence $$x^TAx = 0 \iff Ax = 0$$ but $Ax = 0$ defines a linear subspace therefore (ii) is true. But why (i) isn't true?
If $A$ is symmetric then it's diagonalizable hence there exists an orthogonal $Q$ such that $A = Q^TDQ$ ($D$ is diagonal) so
$$\{x^TAx = 1\} = \{x^TQ^TDxQ = 1\} = \{(Qx)^TDQx = 1\}.$$ But since $A$ is a projector then its eigenvalues are $0$ or $1$ hence $\{(Qx)^TDQx = 1\}$ is something like $\xi_1^2 + \dots+ \xi_k^2 = 1$ but this equation defines a bounded set, as I understand.
Could you please help me to figure out what option is true? Thanks a lot in advance!
Your reasoning for statement (i) is almost correct except for the last step. Given the equation$$ ξ_1^2 + \cdots + ξ_k^2 = 1, $$ if $k < n$, then this equation defines an unbounded set in $\mathbb{R}^n$ since $ξ_{k + 1}, \cdots, ξ_n$ can be any real numbers.