91Ó°ÊÓ

We want our matrix to have eigenvalues 0 and 1 (with 1 having an algebraic multiplicity of 2). For a diagonal matrix, the eigenvalues are the diagonal elements. Let's set up a diagonal matrix \(A\) with eigenvalues \(\lambda_{1} = 0\), \(\lambda_{2} = 1\), and \(\lambda_{3} = \lambda_{4} = 1\). The matrix \(A\) will look like: \[A = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\]

To satisfy the minimal polynomial, the matrix \(A\) must display the property \(z(z-1)^{2}\), which means the matrix should satisfy the equation \(A(A-I)^{2}=0\), where \(I\) is the identity matrix. In this case, let's modify the matrix \(A\) by changing one of its elements to create a desired minimal polynomial: \[B = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\] Now, let's compute the required terms and verify if this operator satisfies the desired minimal polynomial.

Compute \((B-I)\): \[(B-I) = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\] Now, compute \((B-I)^2\): \[(B-I)^2 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\] Finally, compute \(B(B-I)^2\): \[B(B-I)^2 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\] Since \(B(B-I)^2 = 0\), the matrix \(B\) satisfies the minimal polynomial \(z(z-1)^{2}\), and it acts on the complex vector space \(\mathrm{C}^{4}\). Therefore, we have found an operator (matrix) on \(\mathrm{C}^{4}\) whose minimal polynomial equals \(z(z-1)^{2}\): \[B = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\]