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Course by: Steven J. Cox. E-mail the author

# The Spectral Representation of a Symmetric Matrix

Module by: Steven J. Cox. E-mail the author

Summary: Hermitian transposes and the spectral representation ofsymettric matrices are explained.

## Introduction

Our goal is to show that if B B is symmetric then

• each λ j λ j is real,
• each P j P j is symmetric and
• each D j D j vanishes.
Let us begin with an example.

### Example 1

The transfer function of B=( 111 111 111 ) B 1 1 1 1 1 1 1 1 1 is Rs=1s(s3)( s211 1s21 11s2 ) R s 1 s s 3 s 2 1 1 1 s 2 1 1 1 s 2 Rs=1s( 2/3-1/3-1/3 -1/32/3-1/3 -1/3-1/3-1/3 )+1s3( 1/31/31/3 1/31/31/3 1/31/31/3 ) R s 1 s 23 -13 -13 -13 23 -13 -13 -13 -13 1 s 3 13 13 13 13 13 13 13 13 13 Rs=1s λ 1 P 1 +1s λ 2 P 2 R s 1 s λ 1 P 1 1 s λ 2 P 2

and so indeed each of the bullets holds true. With each of the D j D j falling by the wayside you may also expect that the respective geometric and algebraic multiplicities coincide.

## The Spectral Representation

We have amassed anecdotal evidence in support of the claim that each D j D j in the spectral representation

B= j =1h λ j P j + j =1h D j B j 1 h λ j P j j 1 h D j
(1)
is the zero matrix when B B is symmetric, i.e., when B=BT B B , or, more generally, when B=BH B B where BHB¯T B B Matrices for which B=BH B B are called Hermitian. Of course real symmetric matrices are Hermitian.

Taking the conjugate transpose throughout Equation 1 we find

BH= j =1h λ j ¯ P j H+ j =1h D j H B j 1 h λ j P j j 1 h D j
(2)
That is, the λ j ¯ λ j are the eigenvalues of BH B with corresponding projections P j H P j and nilpotents D j H D j Hence, if B=BH B B , we find on equating terms that λ j = λ j ¯ λ j λ j P j = P j H P j P j and D j = D j H D j D j The former states that the eigenvalues of an Hermitian matrix are real. Our main concern however is with the consequences of the latter. To wit, notice that for arbitrary x x, D j m j 1x2=xH D j m j 1H D j m j 1x D j m j 1 x 2 x D j m j 1 D j m j 1 x D j m j 1x2=xH D j m j 1 D j m j 1x D j m j 1 x 2 x D j m j 1 D j m j 1 x D j m j 1x2=xH D j m j 2 D j m j x D j m j 1 x 2 x D j m j 2 D j m j x D j m j 1x2=0 D j m j 1 x 2 0 As D j m j 1x=0 D j m j 1 x 0 for every x x it follows (recall this previous exercise) that D j m j 1=0 D j m j 1 0 . Continuing in this fashion we find D j m j 2=0 D j m j 2 0 and so, eventually, D j =0 D j 0 . If, in addition, B B is real then as the eigenvalues are real and all the D j D j vanish, the P j P j must also be real. We have now established

### Proposition 1

If B B is real and symmetric then

B= j =1h λ j P j B j 1 h λ j P j
(3)
where the λ j λ j are real and the P j P j are real orthogonal projections that sum to the identity and whose pairwise products vanish.

#### Proof

One indication that things are simpler when using the spectral representation is

B100= j =1h λ j 100 P j B 100 j 1 h λ j 100 P j
(4)
As this holds for all powers it even holds for power series. As a result, eB= j =1he λ j P j B j 1 h λ j P j It is also extremely useful in attempting to solve Bx=b B x b for x x. Replacing B B by its spectral representation and b b by Ib I b or, more to the point by j j P j b j j P j b we find j =1h λ j P j x= j =1h P j b j 1 h λ j P j x j 1 h P j b Multiplying through by P 1 P 1 gives λ 1 P 1 x= P 1 b λ 1 P 1 x P 1 b or P 1 x= P 1 b λ 1 P 1 x P 1 b λ 1 . Multiplying through by the subsequent P j P j 's gives P j x= P j b λ j P j x P j b λ j . Hence,
x= j =1h P j x= j =1h1 λ j P j b x j 1 h P j x j 1 h 1 λ j P j b
(5)
We clearly run in to trouble when one of the eigenvalues vanishes. This, of course, is to be expected. For a zero eigenvalue indicates a nontrivial null space which signifies dependencies in the columns of B B and hence the lack of a unique solution to Bx=b B x b .

Another way in which Equation 5 may be viewed is to note that, when B B is symmetric, this previous equation takes the form zIB-1= j =1h1z λ j P j z I B j 1 h 1 z λ j P j Now if 0 0 is not an eigenvalue we may set z=0 z 0 in the above and arrive at

B-1= j =1h1 λ j P j B j 1 h 1 λ j P j
(6)
Hence, the solution to Bx=b B x b is x=B-1b= j =1h1 λ j P j b x B b j 1 h 1 λ j P j b as in Equation 5. With Equation 6 we have finally reached a point where we can begin to define an inverse even for matrices with dependent columns, i.e., with a zero eigenvalue. We simply exclude the offending term in Equation 6. Supposing that λ h =0 λ h 0 we define the pseudo-inverse of B B to be B + j =1h11 λ j P j B + j 1 h 1 1 λ j P j Let us now see whether it is deserving of its name. More precisely, when bB b B we would expect that x= B + b x B + b indeed satisfies Bx=b B x b . Well B B + b=B j =1h11 λ j P j b= j =1h11 λ j B P j b= j =1h11 λ j λ j P j b= j =1h1 P j b B B + b B j 1 h 1 1 λ j P j b j 1 h 1 1 λ j B P j b j 1 h 1 1 λ j λ j P j b j 1 h 1 P j b It remains to argue that the latter sum really is b b. We know that b= j =1h P j b  ,   bB    b b B b j 1 h P j b The latter informs us that bNBT b N B . As B=BT B B , we have, in fact, that bNB b N B . As P h P h is nothing but orthogonal projection onto NB N B it follows that P h b=0 P h b 0 and so B( B + b)=b B B + b b , that is, x= B + b x B + b is a solution to Bx=b B x b . The representation Equation 4 is unarguably terse and in fact is often written out in terms of individual eigenvectors. Let us see how this is done. Note that if x P 1 x P 1 then x= P 1 x x P 1 x and so, Bx=B P 1 x= j =1h λ j P j P 1 x= λ 1 P 1 x= λ 1 x B x B P 1 x j 1 h λ j P j P 1 x λ 1 P 1 x λ 1 x i.e., x x is an eigenvector of B B associated with λ 1 λ 1 . Similarly, every (nonzero) vector in P j P j is an eigenvector of B B associated with λ j λ j .

Next let us demonstrate that each element of P j P j is orthogonal to each element of P k P k when jk j k . If x P j x P j and y P k y P k then xTy=( P j x)T P k y=xT P j P k y=0 x y P j x P k y x P j P k y 0 With this we note that if x j , 1 x j , 2 x j , n j x j , 1 x j , 2 x j , n j constitutes a basis for P j P j then in fact the union of such bases, x j , p (1jh) and (1p n j ) x j , p 1 j h 1 p n j forms a linearly independent set. Notice now that this set has j =1h n j j 1 h n j elements. That these dimensions indeed sum to the ambient dimension, n n, follows directly from the fact that the underlying P j P j sum to the n n-by- n n identity matrix. We have just proven

### Proposition 2

If B B is real and symmetric and n n-by- n n, then B B has a set of n n linearly independent eigenvectors.

#### Proof

Getting back to a more concrete version of Equation 4 we now assemble matrices from the individual bases E j x j , 1 x j , 2 x j , n j E j x j , 1 x j , 2 x j , n j and note, once again, that P j = E j E j T E j -1 E j T P j E j E j E j E j , and so B= j =1h λ j E j E j T E j -1 E j T B j 1 h λ j E j E j E j E j I understand that you may feel a little overwhelmed with this formula. If we work a bit harder we can remove the presence of the annoying inverse. What I mean is that it is possible to choose a basis for each P j P j for which each of the corresponding E j E j satisfy E j T E j =I E j E j I As this construction is fairly general let us devote a separate section to it (see Gram-Schmidt Orthogonalization).

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