Question 5.2.4: Show that every 2 × 2 real symmetric matrix is diagonalizabl...

Recall that the matrix A is symmetric if and only if A = A^{t} . Every 2 × 2 symmetric matrix has the form

A= \begin{bmatrix} a& b \\b &d  \end{bmatrix}

See Example 5 of Sec. 1.3. The eigenvalues are found by solving the characteristic equation

det(A − λI) = \begin{vmatrix} a − λ& b \\ b& d− λ \end{vmatrix} = λ² − (a + d)λ + ad − b² = 0
By the quadratic formula, the eigenvalues are

Since the discriminant (a − d)² + 4b² ≥ 0, the characteristic equation has either one or two real roots. If (a − d)² + 4b² = 0, then (a − d)² = 0 and b² = 0, which holds if and only if a = d and b = 0. Hence, the matrix A is diagonal. If (a − d)² + 4b² > 0, then A has two distinct eigenvalues; so by Corollary 1, the matrix A is diagonalizable.