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# 9.1: The Spectral Representation of a Symmetric Matrix


## Introduction

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

• each $$\lambda_{j}$$ is real,
• each $$P_{j}$$ is symmetric and
• each $$D_{j}$$ vanishes.

Let us begin with an example.

Example

The transfer function of

$B = \begin{pmatrix} {1}&{1}&{1}\\ {1}&{1}&{1}\\ {1}&{1}&{1} \end{pmatrix} \nonumber$

is

$R(s) = \frac{1}{s(s-3)} \begin{pmatrix} {s-2}&{1}&{1}\\ {1}&{s-2}&{1}\\ {1}&{1}&{s-2} \end{pmatrix} \nonumber$

$R(s) = \frac{1}{s} \begin{pmatrix} {2/3}&{-1/3}&{-1/3}\\ {-1/3}&{2/3}&{-1/3}\\ {-1/3}&{-1/3}&{-1/3} \end{pmatrix}+\frac{1}{s-3} \begin{pmatrix} {1/3}&{1/3}&{1/3}\\ {1/3}&{1/3}&{1/3}\\ {1/3}&{1/3}&{1/3} \end{pmatrix}\nonumber$

$R(s) = \frac{1}{s-\lambda_{1}}P_{1}+\frac{1}{s-\lambda_{2}}P_{2} \nonumber$

and so indeed each of the bullets holds true. With each of the $$D_{j}$$ 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 $$D_{j}$$ in the spectral representation

$B = \sum_{j=1}^{h} \lambda_{j}P_{j}+\sum_{j=1}^{h}D_{j} \nonumber$

is the zero matrix when $$B$$ is symmetric, i.e., when $$B = B^T$$, or, more generally, when $$B = B^H$$ where $$B^H \equiv \overline{B}^T$$ Matrices for which $$B = B^H$$ are called Hermitian. Of course real symmetric matrices are Hermitian.

Taking the conjugate transpose throughout we find,

$B^{H} = \sum_{j=1}^{h} \overline{\lambda_{j}}P_{j}^{H}+\sum_{j=1}^{h}D_{j}^{H} \nonumber$

That is, the $$\overline{\lambda_{j}}$$ are the eigenvalues of $$B^H$$ with corresponding projections $$P_{j}^H$$ and nilpotents $$D_{j}^{H}$$ Hence, if $$B = B^H$$, we find on equating terms that

$\lambda_{j} = \overline{\lambda_{j}} \nonumber$

$P_{j} = P_{j}^H \nonumber$

and

$D_{j} = D_{j}^H \nonumber$

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 $$x$$

$(||D_{j}^{m_j-1}x||)^2 = x^{H}(D_{j}^{m_j-1})^{H}D_{j}^{m_j-1}x \nonumber$

$(||D_{j}^{m_j-1}x||)^2 = x^{H}D_{j}^{m_j-1}D_{j}^{m_j-1}x \nonumber$

$(||D_{j}^{m_j-1}x||)^2 = x^{H}D_{j}^{m_j-2}D_{j}^{m_j}x \nonumber$

$(||D_{j}^{m_j-1}x||)^2 = 0 \nonumber$

As $$D_{j}^{m_j-1}x = 0$$ for every $$x$$ it follows (recall the previous exercise) that $$D_{j}^{m_j-1} = 0$$. Continuing in this fashion we find $$D_{j}^{m_j-2} = 0$$, and so, eventually, $$D_{j} = 0$$. If, in addition, $$B$$ is real then as the eigenvalues are real and all the $$D_{j}$$ vanish, the $$P_{j}$$ must also be real. We have now established

If $$B$$ is real and symmetric then

$B = \sum_{j=1}^{h} \lambda_{j} P_{j} \nonumber$

where the $$\lambda_{j}$$ are real and the $$P_{j}$$ are real orthogonal projections that sum to the identity and whose pairwise products vanish.

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

$B^{100} = \sum_{j=1}^{h} \lambda_{j}^{100} P_{j} \nonumber$

As this holds for all powers it even holds for power series. As a result,

$e^B = \sum_{j=1}^{h} e^{\lambda_{j}} P_{j} \nonumber$

It is also extremely useful in attempting to solve

$Bx = b \nonumber$

for $$x$$. Replacing $$B$$ by its spectral representation and $$b$$ by $$I⁢b$$ or, more to the point by $$\sum_{j} P_{j}b$$ we find

$\sum_{j=1}^{h} \lambda_{j} P_{j} x = \sum_{j=1}^{h} P_{j} b \nonumber$

Multiplying through by $$P_{1}$$ gives $$\lambda_{1}P_{1}x = P_{1}b$$ or $$P_{1}x = \frac{P_{1}b}{\lambda_{1}}$$. Multiplying through by the subsequent $$P_{j}$$'s gives $$P_{j}x = \frac{P_{j}b}{\lambda_{j}}$$

Hence,

$x = \sum_{j=1}^{h} P_{j}x \nonumber$

$\sum_{j=1}^{h} \frac{1}{\lambda_{j}}P_{j}b \nonumber$

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 $$B$$ and hence the lack of a unique solution to $$Bx = b$$.

Another way in which may be viewed is to note that, when $$B$$ is symmetric, this previous equation takes the form

$(zI-B)^{-1} = \sum_{j=1}^{h} \frac{1}{z-\lambda_{j}} P_{j} \nonumber$

Now if 0 is not an eigenvalue we may set $$z=0$$ in the above and arrive at

$B^{-1} = \sum_{j=1}^{h} \frac{1}{\lambda_{j}} P_{j} \nonumber$

Hence, the solution to $$Bx = b$$B is

$x = B^{-1}b = \sum_{j=1}^{h} \frac{1}{\lambda_{j}} P_{j} b \nonumber$

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 link. Supposing that $$\lambda_{h} = 0$$ we define the pseudo-inverse of $$B$$ to be

$B^{+} = \sum_{j=1}^{h-1} \frac{1}{\lambda_{j}} P_{j} \nonumber$

Let us now see whether it is deserving of its name. More precisely, when $$b \in \mathscr{R}(B)$$ we would expect that $$x = B^{+} b$$ indeed satisfies $$Bx = b$$. Well

$BB^{+} b = B \sum_{j=1}^{h-1} \frac{1}{\lambda_{j}} P_{j} b = \sum_{j=1}^{h-1} \frac{1}{\lambda_{j}} BP_{j}b = \sum_{j=1}^{h-1} \frac{1}{\lambda_{j}} \lambda_{j}P_{j}b = \sum_{j=1}^{h-1}P_{j}b \nonumber$

It remains to argue that the latter sum really is $$b$$. We know that

$\forall b,b \in \mathscr{R}(B) : (b = \sum_{j=1}^{h} P_{j}b) \nonumber$

The latter informs us that $$b \perp N(B^T)$$. As $$B = B^T$$, we have, in fact, that $$b \perp N(B)$$. AS $$P_{h}$$ is nothing but orthogonal projection onto $$N(B)$$ it follows that $$P_{h}b = 0$$ and so $$B(B^{+}b)=b$$, that is, $$x = B^{+}b$$ is a solution to $$Bx = b$$ The representation 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 $$x \in \mathscr{R}(P_{1})$$ then $$x=P_{1}x$$ and so,

$Bx = BP_{1}x = \sum_{j=1}^{h} \lambda_{j} P_{j}P_{1}x = \lambda_{1}P_{1}x = \lambda_{1}x \nonumber$

i.e., $$x$$ is an eigenvector of $$B$$ associated with $$\lambda_{1}$$. Similarly, every (nonzero) vector $$\mathscr{R}(P_{j})$$ is an eigenvector of $$B$$ associated with $$\lambda_{j}$$.

Next let us demonstrate that each element of $$\mathscr{R}(P_{j})$$ is orthogonal to each element of $$\mathscr{R}(P_{k})$$ when $$j \ne k$$. If $$x \in \mathscr{R}(P_{j})$$ and $$x \in \mathscr{R}(P_{k})$$ then

$x^{T}y = (P_{j}x)^{T}P_{k}y = x^{T}P_{j}P_{k}y = 0 \nonumber$

With this we note that if $$\{x_{j,1}, x_{j,2}, \cdots, x_{j,n_{j}}\}$$ constitutes a basis for $$\mathscr{R}(P_{j})$$ then in fact the union of such bases,

$\{x_{j,p} | (1 \le j \le h) \wedge (1 \le p \le n_{j})\} \nonumber$

forms a linearly independent set. Notice now that this set has

$\sum_{j=1}^{h} n_{j} \nonumber$

elements. That these dimensions indeed sum to the ambient dimension, $$n$$, follows directly from the fact that the underlying $$P_{j}$$

sum to the $$n-by-n$$ identity matrix. We have just proven.

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

Getting back to a more concrete version of link we now assemble matrices from the individual bases

$E_{j} \equiv \{x_{j,1}, x_{j,2}, \cdots, x_{j,n_{j}}\} \nonumber$

and note, once again, that $$P_{j} = E_{j}(E_{j}^{T}E_{j})^{-1}E_{j}^{T} \nonumber$$, and so,

$B = \sum_{j=1}^{h} \lambda_{j}E_{j}(E_{j}^{T}E_{j})^{-1}E_{j}^{T} \nonumber$

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 $$\mathscr{R}⁢(P_{j})$$ for which each of the corresponding $$E_{j}$$ satisfy $$E_{j}^{T}E_{j} = I$$. As this construction is fairly general let us devote a separate section to it (see Gram-Schmidt Orthogonalization).