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Mathematics LibreTexts

9.3: The Diagonalization of a Symmetric Matrix

  • Page ID
    21856
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    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

    \[\begin{pmatrix} {0}&{0}&{0}\\ {0}&{0}&{0}\\ {0}&{0}&{3} \end{pmatrix} \nonumber\]