9.3: The Diagonalization of a Symmetric Matrix
- Page ID
- 21856
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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
\[\begin{pmatrix} {0}&{0}&{0}\\ {0}&{0}&{0}\\ {0}&{0}&{3} \end{pmatrix} \nonumber\]