
# 9.3: The Diagonalization of a Symmetric Matrix


By choosing an orthogonal basis $$\{q_{j,k} | 1 \le k \le n_{j}\}$$ for each $$\mathbb{R}(P_{j})$$ and collecting the basis vectors in

$Q_{j} = \begin{pmatrix} {q_{j,1}}&{q_{j,2}}&{\cdots}&{q_{j,n_{j}}} \end{pmatrix} \nonumber$

We find that

$P_{j} = Q_{j}Q_{j}^{T} = \sum_{k=1}^{n_{j}} q_{j,k}q_{j,k}^{T} \nonumber$

As a result, the spectral representation takes the form

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

$\sum_{j=1}^{h} \lambda_{j} \sum_{k=1}^{n_{j}} q_{j,k}q_{j,k}^{T} \nonumber$

This is the spectral representation in perhaps its most detailed dress. There exists, however, still another form! It is a form that you are likely to see in future engineering courses and is achieved by assembling the $$Q_{j}$$ into a single $$n-by-n$$ orthonormal matrix

$Q = \begin{pmatrix} {Q_{1}}&{\cdots}&{Q_{h}} \end{pmatrix} \nonumber$

Having orthonormal columns it follows that $$Q^{T}Q = I$$. $$Q$$ being square, it follows in addition that $$Q^{T} = Q^{-1}$$. Now,

$Bq_{j,k} = \lambda_{j}q_{j,k} \nonumber$

may be encoded in matrix terms via

$BQ = Q \Lambda \nonumber$

where $$\Lambda$$ is the $$n-by-n$$ diagonal matrix whose first $$n_{1}$$ diagonal terms are $$\lambda_{1}$$, whose next $$n_{2}$$ diagonal terms are $$\lambda_{2}$$, and so on. That is, each $$\lambda_{j}$$ is repeated according to its multiplicity. Multiplying each side of Equation, from the right, by $$Q^{T}$$ we arrive at

$B = Q \Lambda Q^{T} \nonumber$

Because one may just as easily write

$Q^{T} BQ = \Lambda \nonumber$

one says that $$Q$$ diagonalizes $$B$$.

Let us return the our example

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

of the last chapter. Recall that the eigenspace associated with $$\lambda_{1} = 0$$ had

$e_{1, 1} = \begin{pmatrix} {-1}\\ {1}\\ {0} \end{pmatrix} \nonumber$

and

$e_{1, 2} = \begin{pmatrix} {-1}\\ {0}\\ {1} \end{pmatrix} \nonumber$

for a basis. Via Gram-Schmidt we may replace this with

$q_{1, 1} = \frac{1}{\sqrt{2}} \begin{pmatrix} {-1}\\ {1}\\ {0} \end{pmatrix} \nonumber$

$q_{1, 2} = \frac{1}{\sqrt{6}} \begin{pmatrix} {-1}\\ {-1}\\ {2} \end{pmatrix} \nonumber$

Normalizing the vector associated with $$\lambda_{2} = 3$$ we arrive at

$q_{2, 1} = \frac{1}{\sqrt{3}} \begin{pmatrix} {1}\\ {1}\\ {1} \end{pmatrix} \nonumber$

and hence

$Q = \begin{pmatrix} {q_{1}^{1}}&{q_{2}^{1}}&{q_{2}} \end{pmatrix} = \frac{1}{\sqrt{6}} \begin{pmatrix} {-\sqrt{3}}&{-1}&{\sqrt{2}}\\ {\sqrt{3}}&{-1}&{\sqrt{2}}\\ {0}&{2}&{\sqrt{2}} \end{pmatrix} \nonumber$

and

$\begin{pmatrix} {0}&{0}&{0}\\ {0}&{0}&{0}\\ {0}&{0}&{3} \end{pmatrix} \nonumber$