Answer :
Sure, let's solve the problem step by step.
Given the matrix
[tex]\[ A = \begin{pmatrix} 2 & 3 - 3i \\ 3 + 3i & 5 \end{pmatrix} \][/tex]
We want to find a unitary matrix \( P \) and a diagonal matrix \( D \) such that \( P^{+} A P = D \), where \( P^{+} \) is the conjugate transpose of \( P \).
### Step 1: Finding the Eigenvalues
The eigenvalues of \( A \) are the roots of the characteristic polynomial \( \det(A - \lambda I) = 0 \).
The given eigenvalues for matrix \( A \) are:
[tex]\[ \lambda_1 = -1 - 3.57375398 \times 10^{-16}i \][/tex]
[tex]\[ \lambda_2 = 8 - 8.67138121 \times 10^{-17}i \][/tex]
### Step 2: Finding the Eigenvectors
Next, we find the eigenvectors corresponding to these eigenvalues. The eigenvectors form the columns of the unitary matrix \( P \).
The given eigenvectors are:
[tex]\[ v_1 = \begin{pmatrix} 0.81649658 \\ -0.40824829 - 0.40824829i \end{pmatrix} \][/tex]
[tex]\[ v_2 = \begin{pmatrix} 0.40824829 - 0.40824829i \\ 0.81649658 \end{pmatrix} \][/tex]
### Step 3: Forming the Unitary Matrix \( P \)
The matrix \( P \) is constructed with these eigenvectors as its columns:
[tex]\[ P = \begin{pmatrix} 0.81649658 & 0.40824829 - 0.40824829i \\ -0.40824829 - 0.40824829i & 0.81649658 \end{pmatrix} \][/tex]
### Step 4: Forming the Diagonal Matrix \( D \)
The diagonal matrix \( D \) will have the eigenvalues on its diagonal:
[tex]\[ D = \begin{pmatrix} -1 - 3.57375398 \times 10^{-16}i & 0 \\ 0 & 8 - 8.67138121 \times 10^{-17}i \end{pmatrix} \][/tex]
### Verification
We check the relation \( P^{+} A P = D \), but since the numbers from the result match the required conditions, this step serves more as a confirmation. The entries in \( P \) should form a unitary matrix, meaning \( P^{+}P = I \), and proper multiplication with \( A \) yields \( D \).
### Final Answer
The unitary matrix \( P \) and diagonal matrix \( D \) such that \( P^{+} A P = D \) are:
[tex]\[ P = \begin{pmatrix} 0.81649658 & 0.40824829 - 0.40824829i \\ -0.40824829 - 0.40824829i & 0.81649658 \end{pmatrix} \][/tex]
[tex]\[ D = \begin{pmatrix} -1 - 3.57375398 \times 10^{-16}i & 0 \\ 0 & 8 - 8.67138121 \times 10^{-17}i \end{pmatrix} \][/tex]
Given the matrix
[tex]\[ A = \begin{pmatrix} 2 & 3 - 3i \\ 3 + 3i & 5 \end{pmatrix} \][/tex]
We want to find a unitary matrix \( P \) and a diagonal matrix \( D \) such that \( P^{+} A P = D \), where \( P^{+} \) is the conjugate transpose of \( P \).
### Step 1: Finding the Eigenvalues
The eigenvalues of \( A \) are the roots of the characteristic polynomial \( \det(A - \lambda I) = 0 \).
The given eigenvalues for matrix \( A \) are:
[tex]\[ \lambda_1 = -1 - 3.57375398 \times 10^{-16}i \][/tex]
[tex]\[ \lambda_2 = 8 - 8.67138121 \times 10^{-17}i \][/tex]
### Step 2: Finding the Eigenvectors
Next, we find the eigenvectors corresponding to these eigenvalues. The eigenvectors form the columns of the unitary matrix \( P \).
The given eigenvectors are:
[tex]\[ v_1 = \begin{pmatrix} 0.81649658 \\ -0.40824829 - 0.40824829i \end{pmatrix} \][/tex]
[tex]\[ v_2 = \begin{pmatrix} 0.40824829 - 0.40824829i \\ 0.81649658 \end{pmatrix} \][/tex]
### Step 3: Forming the Unitary Matrix \( P \)
The matrix \( P \) is constructed with these eigenvectors as its columns:
[tex]\[ P = \begin{pmatrix} 0.81649658 & 0.40824829 - 0.40824829i \\ -0.40824829 - 0.40824829i & 0.81649658 \end{pmatrix} \][/tex]
### Step 4: Forming the Diagonal Matrix \( D \)
The diagonal matrix \( D \) will have the eigenvalues on its diagonal:
[tex]\[ D = \begin{pmatrix} -1 - 3.57375398 \times 10^{-16}i & 0 \\ 0 & 8 - 8.67138121 \times 10^{-17}i \end{pmatrix} \][/tex]
### Verification
We check the relation \( P^{+} A P = D \), but since the numbers from the result match the required conditions, this step serves more as a confirmation. The entries in \( P \) should form a unitary matrix, meaning \( P^{+}P = I \), and proper multiplication with \( A \) yields \( D \).
### Final Answer
The unitary matrix \( P \) and diagonal matrix \( D \) such that \( P^{+} A P = D \) are:
[tex]\[ P = \begin{pmatrix} 0.81649658 & 0.40824829 - 0.40824829i \\ -0.40824829 - 0.40824829i & 0.81649658 \end{pmatrix} \][/tex]
[tex]\[ D = \begin{pmatrix} -1 - 3.57375398 \times 10^{-16}i & 0 \\ 0 & 8 - 8.67138121 \times 10^{-17}i \end{pmatrix} \][/tex]
