Let

[latex]\begin{align*} A:=\left(\begin{array}{cc} 3/2 & -1/2 \\ -1/2 & 3/2 \end{array}\right). \end{align*}[/latex]

We solve for the characteristic equation:

[latex]\begin{align*} 0 &= \det(\lambda I-A) = (\lambda -(3/2))^2 -(1/4) = (\lambda-1)(\lambda-2). \end{align*}[/latex]

Hence the eigenvalues are [latex]\lambda_1 =1[/latex], [latex]\lambda_2 =2[/latex]. For each eigenvalue [latex]\lambda[/latex], we look for a unit-norm vector [latex]u[/latex] such that [latex]Au = \lambda u[/latex]. For [latex]\lambda =\lambda_1[/latex], we obtain the equation in [latex]u= u_1[/latex]

[latex]\begin{align*} 0 &= (A-\lambda_1) u_1 = \left(\begin{array}{cc} 1/2 & -1/2 \\ -1/2 & 1/2 \end{array}\right) u_1 \end{align*}[/latex]

which leads to (after normalization) an eigenvector [latex]u_1: := (1/\sqrt{2})[1,1][/latex]. Similarly for [latex]\lambda_2[/latex] we obtain the eigenvector [latex]u_2: := (1/\sqrt{2})[1,-1][/latex]. Hence, [latex]A[/latex] admits the SED (Symmetric Eigenvalue Decomposition)

[latex]\begin{align*} A=&\left(\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right)\right)^T\left(\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right)\left(\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right)\right). \end{align*}[/latex]

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