Suppose that \(A\) is a \(2 \times 2\) matrix with real entries which is symmetric (\(A^t = A\)). Prove that \(A\) is similar over \(\RR\) to a diagonal matrix.
Solution
Write \(A=\begin{bmatrix}a&b\\b&d\end{bmatrix}\). Its characteristic polynomial is \[ x^2-(a+d)x+(ad-b^2), \] with discriminant \[ (a-d)^2+4b^2\ge0. \] Thus it splits over \(\mathbb R\). If the roots are distinct, eigenvectors for them are linearly independent and diagonalize \(A\). If the root is repeated, the discriminant is zero, so \(a=d\) and \(b=0\); hence \(A=aI\) is already diagonal. Therefore \(A\) is diagonalizable over \(\mathbb R\).