Matrix of a Quadratic Form
For \(Q=ax_1^2+2hx_1x_2+bx_2^2\), write
\[Q=X^TAX,\qquad A=\left[\begin{array}{cc}a&h\\h&b\end{array}\right].\]
The coefficient of \(x_1x_2\) is divided equally between the two symmetric positions.
Short Quadratic Form Reduction
Orthogonal Reduction and Canonical Form
Short quadratic form reduction is developed through one IOE past-paper question. Use the symmetric matrix, the S1 and S2 characteristic equation, orthonormal eigenvectors and the standard orthogonal transformation.
UnitMatrices
TopicQuadratic Form
Question Type2 Mark Problems
Theory Required for This Problem
The \(S_1,S_2,S_3\) Method
For a \(3\times3\) matrix, use
\[S_1=\mathrm{trace}(A).\]
\[S_2=\mathrm{sum\ of\ principal\ minors}.\]
\[S_3=|A|.\]
\[\lambda^3-S_1\lambda^2+S_2\lambda-S_3=0.\]
\[\lambda^3-S_1\lambda^2\]\[{}+S_2\lambda-S_3=0.\]
Rule for a \(2\times2\) Matrix
For the matrix on this page,
\[S_1=\mathrm{trace}(A),\qquad S_2=|A|.\]
Hence,
\[\lambda^2-S_1\lambda+S_2=0.\]
Orthogonal Reduction
Choose unit eigenvectors as the columns of \(P\). Then
\[P^TP=I,\qquad P^TAP=D.\]
With \(X=PY\), the canonical form is \(Q=Y^TDY\).
Calculator check: Use the matrix mode to verify the trace, determinant and eigenvalues. Keep the \(S_1,S_2,S_3\) working in the written solution.
Method used: Form the symmetric matrix, find the characteristic equation using \(S_1,S_2,S_3\), obtain orthonormal eigenvectors, and apply the orthogonal transformation.
Continue through the IOE Engineering Mathematics II Digital Workbook. This short quadratic form reduction page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
58
2 Marks
Reduction to Canonical Form
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 5(b) • 2 Marks
Reduce the quadratic form
\[Q(X)=17x_1^2-30x_1x_2+17x_2^2\]
to canonical form.
Solution
Orthogonal reduction using \(S_1\) and \(S_2\)The symmetric matrix is
\[A=\left[\begin{array}{cc}17&-15\\-15&17\end{array}\right].\]
Using the \(S_1,S_2\) method for this \(2\times2\) matrix,
\[S_1=\mathrm{trace}(A)=17+17=34.\]Also,
\[S_2=|A|=\left|\begin{array}{cc}17&-15\\-15&17\end{array}\right|=289-225=64.\]
Therefore, the characteristic equation is
\[\lambda^2-S_1\lambda+S_2=0.\]Substituting \(S_1=34\) and \(S_2=64\),
\[\lambda^2-34\lambda+64=0.\]Hence,
\[(\lambda-2)(\lambda-32)=0,\qquad \lambda_1=2,\quad\lambda_2=32.\]
\[(\lambda-2)(\lambda-32)=0.\]\[\lambda_1=2,\qquad\lambda_2=32.\]
For \(\lambda_1=2\), a unit eigenvector is
\[u_1=\frac{1}{\sqrt2}\left[\begin{array}{c}1\\1\end{array}\right].\]For \(\lambda_2=32\), a unit eigenvector is
\[u_2=\frac{1}{\sqrt2}\left[\begin{array}{c}1\\-1\end{array}\right].\]Take
\[P=\frac{1}{\sqrt2}\left[\begin{array}{cc}1&1\\1&-1\end{array}\right],\qquad X=PY.\]
Then
\[P^TAP=\left[\begin{array}{cc}2&0\\0&32\end{array}\right].\]
Therefore,
\[Q=Y^T(P^TAP)Y=2y_1^2+32y_2^2.\]Calculator check:
\[S_1=34,\qquad S_2=64,\qquad \lambda=2,32.\]\[Q=2y_1^2+32y_2^2\]