Question

Verify the spectral theorem for each of the following symmetric matrices, by finding an orthonormal basis of the appropriate vector space, the change of basis matrix to this basis, and the spectral decomposition.

A = [2 3 3 2]
λ = E_ -1,5. E.₁ = Span({(-√2/2, √2/2))) and Es Span({(√2/2, √2/2))}). alpha = {(-√2/2, √2/2), (√2/2, √2/2)) is an orthonormal basis for R².
Q [-V/2 -√2/2 √2/2 √2/2 √2/2 A = (-1)PE-₁ + 5P₁ = (-1) = (-1) ¹ [-1/2-1/2] + [1/2 1/2] 5
Solution is given. Not need to solve the problem complelety.Explain how to find PE_₁ and PE . Please, dont find the eigen vector and orthonormal basis

Answer 1

The projection matrices \
`( P_{E_(-1)} \) and \( P_(E_1) \)` for the given eigenvalues lambda
`_(-1) = -1 \) and \( \lambda_1 = 5 \)`are found as follows:

`\[ P_{E_(-1)} = (1)/(2) \`begin bmatrix
`1 & -1 \\ -1 & 1 \end{bmatrix} \]`

`\[ P_(E_1) = (1)/(2) \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \]`

Step-by-step explanation:

To find the projection matrices
`\( P_{E_(-1)} \) and \( P_(E_1) \)` we utilize the formula
`P_{E_(\lambda)} = (1)/(m_(\lambda))(v_iv_i^T) \), where \( m_(\lambda)` is the multiplicity of the eigenvalue lambda and v_i is a corresponding eigenvector. Given the eigenvalues lambda
`_(-1) = -1 \) and \( \lambda_1 = 5 \)`with eigenvectors
`\( v_(-1) =`\begin{bmatrix} -
`(√(2))/(2) \\ (√(2))/(2) \end{bmatrix} \) and \( v_1 = \begin{bmatrix} (√(2))/(2) \\ (√(2))/(2) \end{bmatrix} \)`the projection matrices are computed as follows:

`\[ P_{E_(-1)} = (1)/(2) \begin{bmatrix} -(√(2))/(2) \\ (√(2))/(2) \end{bmatrix} \begin{bmatrix} -(√(2))/(2) & (√(2))/(2) \end{bmatrix} = (1)/(2) \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \]\[ P_(E_1) = (1)/(2) \begin{bmatrix} (√(2))/(2) \\ (√(2))/(2) \end{bmatrix} \begin{bmatrix} (√(2))/(2) & (√(2))/(2) \end{bmatrix} = (1)/(2) \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \]`

These projection matrices are essential components in the spectral decomposition of the symmetric matrix A .