10.3.  Diagonanalisation of real symmetric matrices

In the previous section we observed that a  n× n real symmetric matrix has n- eigenvalues . The
eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix have a special property,
as given in the next theorem.

 

10.3.1 Theorem:

Let A be a real symmetric matrix and be distinct eigenvalues of A .Let
be nonzero such that
.                             , 1ik.
Then forms an orthonormal set.

                                                                                                                                                   Proof

 
10.3.2 Corollary:
Let A be n × n real symmetric matrix such that all its eigenvales are distinct. Then, there
exists an orthogonal matrix P such that
                                  
,
where D is a diagonal matrix with diagonal entries being the eigenvalues of A.

                                                                                                                                                   Proof

 

We prove in the next theorem that  the claim of the above corollary always leads, even if not all the
eigenvalues of A are distinct.

 
10.3.3 Examples:

Let  
                
                                       = ( 3 - λ )[( 3 - λ)( 5 - λ ) - 1] -1 [ 5 - λ - 1 ] -1 [ -1 + 3 - λ ]
                                       = [( 3 - λ )( 3 - λ)( 5 - λ ) - ( 3 - λ )] - ( 4 - λ ) - ( 2 - λ )
                                       = [( 3 - λ )( 3 - λ)( 5 - λ ) - ( 3 - λ )] - 2 ( 3 - λ )
                                       = ( 3 - λ )[( 3 - λ )( 5 - λ ) - 1 - 2 ]
                                       = ( 3 - λ )[ 15 - 5λ -3λ + -3 ]
                                       = ( 3 - λ )( - 8λ + 12 )
                                       = ( 3 - λ )( λ - 6 )( λ - 2 )
Thus, λ = 2, 3, 6 are the eigenvalues of A .Let us find an eigenvector corresponding to each
eigenvalue . For the eigenvalue λ = 2, since
                            
A vector will be an eigenvector for eigenvalue λ = 2 if
                                       
i.e.,      = 0
           +== 0.
If we choose =1, then = -1. Hence
                              
is an eigenvector for λ = 2  For the eigenvalue λ = 3,
                         
Thus
                           
will be an eigenvector for the eigenvalue λ = 3, if
                            
i.e.,   = and  -= 0. Hence
                              is an eigenvector for λ = 3.

Finally, for λ = 6
                       
                       
Thus ,    
                      will be an eigenvector for eigenvalue  λ = 6 if 

 -4-2 = 0,   --+= 0.
Thus, if we take = 2 , then = -1 and =-+=-1.
Hence
                      is an eigenvector for the eigenvalue λ = 6.


Note that for
                        the columns of P are orthogonal.

To make P orthogonal, we normalize each  , and  , and define
                        
It is easy to verify that , where
                         
In fact, . Thus, we checks .                                 

 
10.3.4 Theorem:
Let A be any real symmetric matrix . Then there exists an orthogonal matrix P such that
the  following holds :
(i)   AP = D , a diagonal matrix.
(ii)  The diagonal entries of D are the eigenvalues of A .
(iii) The column vectors of P are the eigenvectors of the eigenvalues of A.

                                                                                                                                                   Proof

 
10.3.5 Note:

The above theorem not only tells us that a real symmetric matrix is diagonalizable, it also gives
us a procedure of finding the matrix P such that  
                         
In fact, P is orthogonal  and its column vectors are the eigenvectors for the n-eigenvalues of A.
To find P, we proceed as follows:
(i)   Compute the eigenvalues of A by solving its characteristics equation. If are
       the distinct eigenvalues of A ,for each eigenvalues , find an orthonormal basis of the
       eigensubspace .
(ii)  Then will be an orthonormal basis of and the required matrix
      
       P
is given by
              
       Further  , a diagonal matrix and diagonal entries of D are ,
       each repeated as many times as the dimensions of .
 

10.3.6 Example:

 Let
        
Then A is a real symmetric matrix with eigenvalues given by
         
                                     = - λ ( - 4 ) - 2 ( -2λ - 4 ) + 2 ( 4 + 2λ )
                                     = - λ ( λ - 2 )( λ + 2 ) +  4 ( λ + 2 ) + 4 ( λ + 2 )
                                     = ( λ + 2 )( - + 2λ + 8 )
                                     = - ( λ + 2 )( λ + 2 )( λ - 4 ).
Hence, λ = -2 , λ = 4 are two distinct eigenvalues of A. To find eigenvectors for λ = -2, since
          

If we choose = 0 and = 1 then = -1. And for  = 1 and = 0 we get = -1.

Thus
                
are two mutually orthogonal eigenvector for λ = -2  .

For the eigenvalue λ = 4, since
           ,
           
Thus, if we choose
              = == 1, then
                  
is an eigenvector for the eigenvalue λ = 4. Thus, the eigenvector for A are
             
To make orthonormal, we use the Gram-Schmidt process.
Define
              
We normalize also to get
               
are orthonormal basis of of eigenvectors of A.  Thus, for
                
one checks that
.               

 

10.3.7 Theorem ( Spectral Decomposition ):
Let A be a n × n real symmetric matrix . Then there exist real numbers and
real matrices   such that
the  following holds :
(i)    Each
is an eigenvalue of A.
(ii)   Each is an orthogonal projection matrix, of rank
1, i.e.,
        is real symmetric, = P and rank
( ) = 1.

                                                                                                                                                   Proof

 
10.3.8 Example:

Let
         
It is easy to check that A has eigenvalues λ = -2 , 4 , -1 with orthonormal eigenvectors:
         
Thus, for
          
Let
          
           
and
           
We can verify that
           
This, is the spectral decomposition of A.
 

 
 
 
 
10.3.9 Exercises:
Click here to take a Quiz: Quiz 10.3