(a) Maybe this one ?
For every symmetric real matrix A there exists a real orthogonal matrix Q such that D = (QT)AQ is a diagonal matrix,
where (QT) denotes the transpose of matrix Q.
(b) If A is symmetric, then <AX,Y> = <X,AY> for all X,Y in R^n.
Suppose now that X is an eigenvector associated to eigenvalue λ, and Y is an eigenvector associated to eigenvalue μ (μ≠λ).
Then λ.<X,Y> = <λX,Y> = <AX,Y> = <X,AY> = <X,μY> = μ.<X,Y>
If <X,Y> ≠ 0, then λ = μ, which is false.
Hence : <X,Y> = 0 (X, Y are orthogonal).
(c) Not sure if this is what you're calling "spectral factorization" :
| 1-x 2 |
| 2 4-x | = (1-x)(4-x) - 4 = x² - 5x = x(x-5)
hence λ = 0 and μ = 5 are the two eigenvalues of A.
(d)
Solve AX = 0 : X = (-2t , t ), then choose the eigenvector ( -2 , 1 ).
Solve AY = 5Y : Y = (t,2t), then choose the eigenvector ( 1 , 2 ).
Let Q = ( X Y ) = matrix ( -2 1 // 1 2 )
Q =
( -2 1 )
( 1 2 )
Compute Q⁻¹ = (1/5) * Q
then check that Q⁻¹ A Q = Diag(0,5).