Prove that if a matrix A is symmetric, then it is diagonalisable I need to prove that if a matrix $A_{2 \times 2}$ is symmetric, i.e., $A^t = A$, then it is diagonalisable.
I know that a matrix $M_{n \times n}$ is diagonalisable, if and only if there is a basis of $K^n$ consisting of its eigenvectors. That is, I must have $n$ linear independent eigenvectors of $M$.
So, in order to prove that the matrix $A$ is diagonalisable, I must prove that it has $n$ linear independent eigenvector.
Matrix A has the following form:
$$
A =
\left[ {\begin{array}{cc}
 a & b \\
 b & c
\end{array} } \right]
$$
Therefore, its characteristic equation is:
$$
(\lambda -a)(\lambda -c) -b^2 =0
$$
Now I need to solve this equation in order to find its eigenvalues in terms of $a$, $b$ and $c$.
I'm not sure how to follow from here.
 A: As has been noted, the conclusion holds even for $n\times n$ matrices, $n\geq 2$. In fact, a symmetric $n\times n$ matrix with real coefficients is not just diagonalizable, it is orthogonally diagonalizable: you can find an orthonormal basis of eigenvectors.
But for your purposes it suffices to show that a $2\times 2$ symmetric matrix is either diagonal (when $b=0$), or else has two distinct eigenvalues.
If $b=0$, there is nothing to do, so assume $b\neq 0$. The characteristic polynomial is, as you note,
$$(t-a)(t-c)-b^2 = t^2 - (a+c)t + (ac-b^2).$$
The discriminant of this quadratic is
$$(a+c)^2 - 4(ac-b^2) = a^2+2ac+c^2 - 4ac + 4b^2 = a^2-2ac+c^2 +4b^2 = (a-c)^2+(2b)^2.$$
Since this is a sum of squares, it is nonnegative. Since $b\neq 0$, it is actually positive. Since the discriminant of the quadratic is positive, the polynomial has two distinct real roots, so the matrix has two distinct eigenvalues, hence is diagonalizable.
In conclusion, a symmetric $2\times 2$ matrix is either diagonal (and hence diagonalizable), or has two distinct eigenvalues (and hence is diagonalizable).
