Two-Variable Quadratic Form
For
\[Q=ax_1^2+2hx_1x_2+bx_2^2,\]
the symmetric matrix is
\[A=\left[\begin{array}{cc}a&h\\h&b\end{array}\right].\]
Note: Divide the coefficient of \(x_1x_2\) by \(2\).
Quadratic Forms and Canonical Form
Symmetric Matrices and Orthogonal Reduction
Quadratic forms are converted into symmetric matrices and reduced to canonical form. The cross-term rule, repeated eigenvalues and orthogonal transformations are shown clearly.
UnitMatrices
TopicQuadratic Forms
Question Type4 Mark Problems
Theory Required for These Problems
Three-Variable Quadratic Form
\[Q=ax_1^2+bx_2^2+cx_3^2+2hx_1x_2+2gx_1x_3+2fx_2x_3.\]
\[Q=ax_1^2+bx_2^2+cx_3^2\]\[{}+2hx_1x_2+2gx_1x_3+2fx_2x_3.\]
\[A=\left[\begin{array}{ccc}a&h&g\\h&b&f\\g&f&c\end{array}\right].\]
Coefficient Placement
Square termsPlace their coefficients on the main diagonal.
Cross termsDivide each coefficient by \(2\) and place it in both matching positions.
The S1, S2, S3 Method
\[S_1=\mathrm{trace}(A).\]
\[S_2=\mathrm{sum\ of\ principal\ minors}.\]
\[S_3=|A|.\]
\[\lambda^3-S_1\lambda^2+S_2\lambda-S_3=0.\]
\[\lambda^3-S_1\lambda^2\]\[{}+S_2\lambda-S_3=0.\]
Calculator check: Use matrix mode to verify \(S_3=|A|\) and the eigenvalues.
Canonical Form
For a real symmetric matrix, choose orthonormal eigenvectors as the columns of \(P\). Then
\[P^TAP=D.\]
If \(X=PY\), then \(Q=Y^TDY\).
Method used: Convert the expression into a symmetric matrix, find its eigenvalues and orthonormal eigenvectors, form \(P\), and use \(P^TAP=D\).
Continue through the IOE Engineering Mathematics II Digital Workbook. This quadratic forms and canonical form page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
57
4 MarksReduction of a Quadratic Form
PAST-PAPER REFERENCE: TU IOE • 2083 Baisakh • Back (New Course) • ENSH 151 • Q. 14 • 4 Marks
Reduce the quadratic form
\[Q(X)=5x_1^2+2x_2^2+2x_3^2+4x_1x_2+2x_2x_3+4x_1x_3\]
\[Q(X)=5x_1^2+2x_2^2+2x_3^2\]\[{}+4x_1x_2+2x_2x_3+4x_1x_3\]
to canonical form.
Solution
Symmetric matrix and orthogonal reductionThe square-term coefficients form the diagonal. Half of every cross-term coefficient is placed symmetrically.
Therefore,
\[A=\left[\begin{array}{ccc}5&2&2\\2&2&1\\2&1&2\end{array}\right].\]
Now,
\[S_1=\mathrm{trace}(A)=5+2+2=9.\]The principal minors are
\[\left|\begin{array}{cc}2&1\\1&2\end{array}\right|=3,\quad\left|\begin{array}{cc}5&2\\2&2\end{array}\right|=6,\quad\left|\begin{array}{cc}5&2\\2&2\end{array}\right|=6.\]
Therefore,
\[S_2=3+6+6=15.\]Also,
\[S_3=|A|=7.\]The characteristic equation is
\[\lambda^3-9\lambda^2+15\lambda-7=0.\]Hence,
\[(\lambda-7)(\lambda-1)^2=0.\]Therefore,
\[\lambda_1=7,\quad\lambda_2=1,\quad\lambda_3=1.\]For \(\lambda_1=7\), choose
\[u_1=\frac1{\sqrt6}(2,1,1)^T.\]For the repeated eigenvalue \(\lambda=1\), choose
\[u_2=\frac1{\sqrt2}(0,1,-1)^T,\quad u_3=\frac1{\sqrt3}(-1,1,1)^T.\]
\[u_2=\frac1{\sqrt2}(0,1,-1)^T.\]\[u_3=\frac1{\sqrt3}(-1,1,1)^T.\]
Form
\[P=[u_1\ u_2\ u_3]=\left[\begin{array}{ccc}\frac2{\sqrt6}&0&-\frac1{\sqrt3}\\\frac1{\sqrt6}&\frac1{\sqrt2}&\frac1{\sqrt3}\\\frac1{\sqrt6}&-\frac1{\sqrt2}&\frac1{\sqrt3}\end{array}\right].\]
Then
\[P^TAP=\left[\begin{array}{ccc}7&0&0\\0&1&0\\0&0&1\end{array}\right].\]
Let \(X=PY\). Therefore,
\[Q=Y^T(P^TAP)Y=7y_1^2+y_2^2+y_3^2.\]Hence,
\[\mathrm{rank}=3,\quad\mathrm{index}=3,\quad\mathrm{signature}=3.\]\[Q=7y_1^2+y_2^2+y_3^2\]
59
4 MarksReduction of a Quadratic Form
PAST-PAPER REFERENCE: TU IOE • 2081 Ashwin • Regular (New Course, 2080 Batch) • SH 151 • Q. 14 • 4 Marks
Reduce the quadratic form
\[Q(X)=6x_1^2+3x_2^2+3x_3^2-4x_1x_2-2x_2x_3+4x_1x_3\]
\[Q(X)=6x_1^2+3x_2^2+3x_3^2\]\[{}-4x_1x_2-2x_2x_3+4x_1x_3\]
to canonical form.
Solution
Symmetric matrix and orthogonal reductionHalving the cross-term coefficients gives
\[A=\left[\begin{array}{ccc}6&-2&2\\-2&3&-1\\2&-1&3\end{array}\right].\]
Now,
\[S_1=\mathrm{trace}(A)=6+3+3=12.\]The principal minors are
\[\left|\begin{array}{cc}3&-1\\-1&3\end{array}\right|=8,\quad\left|\begin{array}{cc}6&2\\2&3\end{array}\right|=14,\quad\left|\begin{array}{cc}6&-2\\-2&3\end{array}\right|=14.\]
Therefore,
\[S_2=8+14+14=36.\]Also,
\[S_3=|A|=32.\]The characteristic equation is
\[\lambda^3-12\lambda^2+36\lambda-32=0.\]Hence,
\[(\lambda-8)(\lambda-2)^2=0.\]Therefore,
\[\lambda_1=8,\quad\lambda_2=2,\quad\lambda_3=2.\]For \(\lambda_1=8\), choose
\[u_1=\frac1{\sqrt6}(2,-1,1)^T.\]For the repeated eigenvalue \(\lambda=2\), choose
\[u_2=\frac1{\sqrt2}(0,1,1)^T,\quad u_3=\frac1{\sqrt3}(1,1,-1)^T.\]
\[u_2=\frac1{\sqrt2}(0,1,1)^T.\]\[u_3=\frac1{\sqrt3}(1,1,-1)^T.\]
Form
\[P=[u_1\ u_2\ u_3]=\left[\begin{array}{ccc}\frac2{\sqrt6}&0&\frac1{\sqrt3}\\-\frac1{\sqrt6}&\frac1{\sqrt2}&\frac1{\sqrt3}\\\frac1{\sqrt6}&\frac1{\sqrt2}&-\frac1{\sqrt3}\end{array}\right].\]
Then
\[P^TAP=\left[\begin{array}{ccc}8&0&0\\0&2&0\\0&0&2\end{array}\right].\]
Let \(X=PY\). Therefore,
\[Q=Y^T(P^TAP)Y=8y_1^2+2y_2^2+2y_3^2.\]Hence,
\[\mathrm{rank}=3,\quad\mathrm{index}=3,\quad\mathrm{signature}=3.\]\[Q=8y_1^2+2y_2^2+2y_3^2\]
A Few Important Questions
Conversion into a Two by Two Matrix
Write the symmetric matrix of
\[Q=4x_1^2-6x_1x_2+7x_2^2.\]
Show Solution
The cross-term coefficient is divided by \(2\): \(-6\div2=-3\).
\[A=\left[\begin{array}{cc}4&-3\\-3&7\end{array}\right].\]
Conversion into a Three by Three Matrix
Write the symmetric matrix of
\[Q=3x_1^2+5x_2^2+2x_3^2-4x_1x_2+6x_1x_3-8x_2x_3.\]
\[Q=3x_1^2+5x_2^2+2x_3^2\]\[{}-4x_1x_2+6x_1x_3-8x_2x_3.\]
Show Solution
Place \(3,5,2\) on the diagonal. Use \(-2,3,-4\) in the matching symmetric positions.
\[A=\left[\begin{array}{ccc}3&-2&3\\-2&5&-4\\3&-4&2\end{array}\right].\]