Answer:
the P matrix you are looking for is P=(1/[tex]\sqrt{2}[/tex]) · [[1 1 0 0],[1 -1 0 0],[0 0 1 1],[0 0 1 -1]]
Step-by-step explanation:
Answer:
For an orthogonal diagonalization of any matrix you have to:
1º) Find the matrix eigenvalues in a set order.
2º) Find the eigenvectors of each respective eigenvalues.
Tip: You can write the matrix A like A = [tex]P^{t}[/tex] D P
3º) D is the diagonal matrix with each eigenvalue (in order) in the diagonal.
4º) Write P as the normalized eigenvectors in order (in columns).
Tip 2: Remember, [tex]P^{t}[/tex]·P = I, so if A = [tex]P^{t}[/tex] D P, then:
P A [tex]P^{t}[/tex] = P [tex] P^{t}[/tex] D P [tex]P^{t}[/tex] = I D I = D
So the P we are looking for is the [tex]P^{t}[/tex] of the diagonalization.
Tip 3: In this case, A is a block matrix with null nondiagonal submatrixes, therefore its eigenvalues can be calculated by using the diagonal submatrixes. The problem is reduced to calculate the eigenvalues of A₁₁ = A ₂₂ = [[5 3],[3 5]]
Solving:
1º)the eigenvalues of A₁₁ are {8,2}, therefore the D matrix is [tex]\left[\begin{array}{cccc}8&0&0&0\\0&2&0&0\\0&0&8&0\\0&0&0&2\end{array}\right][/tex]
2º) the eigenvectors of A₁₁ are P₈= {[1 1]T} P₂= {[1 -1]T}, therefore normalizing the eigenvectors you obtain P = (1/[tex]\sqrt{2}[/tex]) · [[1 1 0 0],[1 -1 0 0],[0 0 1 1],[0 0 1 -1]] (you can see that P = [tex]P^{t}[/tex] in this case).
As said in "tip 2": P A [tex]P^{t}[/tex] = P [tex] P^{t}[/tex] D P [tex]P^{t}[/tex] = I D I = D
So the P obtained is the one you are looking for.
