# Trick to diagonalize symmetric matrices?

I will write an exam on Quantum Mechanics soon. I was wondering whether there is any smart and fast way to determine the eigenvalues/eigenvectors of a symmetric 3x3 matrix other than by calculating the characteristic polynomial?

So I am only intersted in fast techniques that one can use by hand to get those things.

• the characteristic polynomial is strategic to get its zeros a.k.a the eigenvalues – janmarqz Jan 24 '14 at 21:34
• yes sure, but maybe there is anything better available for this particular type of matrices. – user66906 Jan 24 '14 at 21:35
• Create a characteristic polynomial and work backwards. And for $3 \times 3$ matrices, that's quick. You could always use Maple if you don't want to do the calculations by hand. – vertical.void Jan 24 '14 at 21:36
• yeah, but after that, i will get kicked out of the exam ;-) – user66906 Jan 24 '14 at 21:37
• For $2\times2$ real symmetric matrices there is Mohr's circle, and every exercise to get fluent with that is well spent IMHO. Furthermore, the Wikipedia article contains an adaptation for $3$D, but first make sure that you can do the $2$D case confidently. – ccorn Jan 24 '14 at 22:31

There is no special trick to find the eigenvalues/vectors of your matrix $A$ (instead of you just "see them" or more special cases of $A$) but if you want to diagonalize your matrix you can make benefit of the symmetry of the matrix when finding an transformation matrix $S$ with $\Delta = S A S^{-1}$
If $A$ is symmetric you know, that eigenvectors to different eigenvalues are ortthogonal, so it's most time very easy to find even an orthonormal basis of eigenvalues. The benefit for your calculation then is, that you don't have to find the inverse of your transformation matrix $S$ because it is given by its transposed matrix $S^T$
Now you are able to calculate the diagonal matrix $\Delta$ simply by $\Delta = S^T A S$