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)Let \(\lambda\) be an eigenvalue of an idempotent matrix \(A\). Therefore there is an eigenvector \(x\) such that \(Ax = \lambda x\). So \[ \begin{aligned} Ax &= A^2x = \lambda x\\ &= A(Ax)\\ &= A(\lambda x)\rightarrow x\:eigenvector\:of\:A\\ &= \lambda(Ax)\rightarrow scalar\:matrix\:multiplication\\ &= \lambda(\lambda x)\rightarrow x\:eigenvector\:of\:A\\ &= \lambda^2x = \lambda x\\ \end{aligned} \]
We have just shown \(\lambda x = \lambda^2x\) Hence; \[ \begin{aligned} \lambda^2x - \lambda x &= 0\\ (\lambda^2 - \lambda)x &= 0\rightarrow distribution\:property\\ \lambda(\lambda - 1)x &= 0 \end{aligned} \]
\(x\) is an eigenvector and thus a nonzero vector. Therefore the solution of the above equation and subsequently the eigenvalues of \(A\), are \(1\) and \(0\).
For example let \(A = \begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}\) \[ \begin{aligned} A^2 &= \begin{bmatrix} 1 & 0\\0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0\\0 & 0 \end{bmatrix}\\ &=\begin{bmatrix} 1 & 0\\0 & 0 \end{bmatrix} = A \end{aligned} \] Thus \(A\) is idempotent For the eigenvalues of \(A\): \[ \begin{aligned} \det(A - \lambda I) &= 0\\ \det( \begin{bmatrix} 1 & 0\\0 & 0 \end{bmatrix} - \lambda\begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix} ) &= 0\\ \det( \begin{bmatrix} 1 & 0\\0 & 0 \end{bmatrix} - \begin{bmatrix} \lambda & 0\\0 & \lambda \end{bmatrix} ) &= 0\\ \begin{bmatrix} 1 - \lambda & 0\\0 & - \lambda \end{bmatrix} ) &= 0\\ -\lambda(1-\lambda) & = 0 \end{aligned} \] The solution of the above equation and thus the eigenvalues of \(A\) are \(1\) and \(0\). Therefore for idempotent matrix \(A\) the eigenvalues are; \(\lambda = 1\) and = 0$.