14.3 Eigenvectors of symmetric matrices

Real symmetric matrices (or more generally, complex Hermitian matrices) always have real eigenvalues, and they are never defective. Their eigenvectors can, and in this class must, be taken orthonormal. (Mutually orthogonal and of length 1.)

For real symmetric matrices, initially find the eigenvectors like for a nonsymmetric matrix. However, after that, you must make the eigenvectors orthonormal in this class.

• For eigenvectors belonging to single eigenvalues, orthogonality to all the other eigenvectors is automatic. Therefore it is enough to just divide the eigenvector by its length:
  

  $$\fbox{$\displaystyle \vec e_i = \vec e_i^{ \rm BAD} / \vert\vec e_i^{ \rm BAD}\vert $}$$ (14.1)

  
  
  Clean up, and that is it.

• However, eigenvectors belonging to a multiple eigenvalue are usually not mutually orthogonal if correctly found using the class procedure (as basis of the $A-\lambda_iI$ null space). You must make them orthogonal.

  The most general method to do that is using the Gram-Schmidt procedure. This works as follows. Start with the nonorthogonal eigenvectors corresponding to the multiple eigenvalue,
  

  $$\vec e_1^{ \rm BAD},\vec e_2^{ \rm BAD},\vec e_3^{ \rm BAD}, \ldots$$
  
  
  obtained using the same procedure as for nonsymmetric matrices. For an eigenvalue of multiplicity $m$, there will be exactly $m$ of these eigenvectors (since symmetric matrices can never be defective). I am assuming here that you have numbered these eigenvectors as 1, 2, .... If not, just replace 1, 2, ...by $i_1$, $i_2$, ....

  You create your first good eigenvector $\vec e_1$ as before, by simply dividing $\vec e_1^{ \rm BAD}$ by its length:
  

  $$\vec e_1 = \vec e_1^{ \rm BAD} / \vert\vec e_1^{ \rm BAD}\vert$$
  
  

  Now you need to make the remaining vectors in the set of $m$ orthogonal to $\vec e_1$. You do that by removing from each vector its vector component in the direction of $\vec e_1$, as follows:
  

  $$\vec e_2^{ \rm BAD,2} = \vec e_2^{ \rm BAD} - (\vec e_... ...(\vec e_3^{ \rm BAD}\cdot\vec e_1)\vec e_1 \quad \ldots$$
  
  
  Vectors $\vec e_2^{ \rm BAD,2},\vec e_3^{ \rm BAD,2},\ldots$ are now orthogonal to $\vec e_1$.

  Now you create your second good eigenvector $\vec e_2$ by simply dividing $\vec e_2^{ \rm BAD,2}$ by its length
  

  $$\vec e_2 = \vec e_2^{ \rm BAD,2} / \vert\vec e_2^{ \rm BAD,2}\vert$$
  
  

  If there are still vectors left, you need to make each orthogonal to $\vec e_2$. In the same way as before
  

  $$\vec e_3^{ \rm BAD,3} = \vec e_3^{ \rm BAD,2} - (\vec e_3^{ \rm BAD,2}\cdot\vec e_2)\vec e_2 \quad \ldots$$
  
  

  Now you create your third good eigenvector $\vec e_3$ by simply dividing $\vec e_3^{ \rm BAD,3}$ by its length
  

  $$\vec e_3 = \vec e_3^{ \rm BAD,3} / \vert\vec e_3^{ \rm BAD,3}\vert$$
  
  

  If there are still vectors $\vec e_4^{ \rm BAD,3},\ldots$ left, you now have to make them orthogonal to $\vec e_3$. Then you can find $\vec e_4$ by dividing $\vec e_4^{ \rm BAD,4}$ by its length.

  Continue like this for multiplicities greater than 4. The general formula is
  

  $$\fbox{$\displaystyle \vec e_j^{ \rm BAD,new} = \vec e... ... e_i \quad \mbox{for} \quad i=j{+}1, j{+}2, \ldots, m $}$$ (14.2)

  
  
  where $e_i$ is the last good eigenvector obtained so far and $j$ $\raisebox{.3pt}{$>$}$ $i$.