Eigenvalue Decomposition of a 2 times 2 Symmetric Matrix

Let
 A := left(begin{array}{cc} 3/2 & -1/2  -1/2 & 3/2 end{array}right) .
We solve for the characteristic equation:
 0 = det (lambda I - A) = (lambda - (3/2))^2- (1/4) = (lambda-1)(lambda - 2) .
Hence the eigenvalues are lambda_1 = 1, lambda_2 =2. For each eigenvalue lambda, we look for a unit-norm vector u such that A u = lambda u. For lambda = lambda_1, we obtain the equation in u=u_1
 0 = (A-lambda_1)u_1 = left(begin{array}{cc} 1/2 & -1/2  -1/2 & 1/2 end{array}right) u_1 ,
which leads to (after normalization) to an eigenvector u_1 := (1/sqrt{2})[1,1]. Similarly for lambda_2 we obtain the eigenvector u_2 :=(1/sqrt{2})[1,-1]. Hence, A admits the SED
 A = left(frac{1}{sqrt{2}} left(begin{array}{cc} 1 & 1  1 & -1 end{array}right) right)^T left(begin{array}{cc} 1 & 0  0 & 2 end{array}right) left(frac{1}{sqrt{2}} left(begin{array}{cc} 1 & 1  1 & -1 end{array}right) right).

See also: Sums-of-squares for a quadratic form.