Discussion
Let \(P : \mathbf{R}^{3} \to \mathbf{R}^{3}\) represent the projection of R3 onto xy-plane along z-axis. Then :
Answer: B
The projection satisfies P3 = P.
P sends (x, y, z) to (x, y, 0). Applying it again changes nothing, so \(P^{2} = P\), and then \(P^{3} = P^{2}P = P \cdot P = P\).
\(P^{2} = I\) is false, because \(P^{2} = P \neq I\). P is not invertible, since it sends (0, 0, 1) to 0. And P is diagonalizable: in the standard basis its matrix is already diagonal, diag(1, 1, 0), with eigenvalues 1, 1, 0.
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