Eigenvalue and Eigenvector
A non-zero column matrix \(X\) is an eigenvector of \(A\) when
\[AX=\lambda X.\]
Therefore,
\[(A-\lambda I)X=0.\]
Eigenvalues and Eigenvectors
Characteristic Equation and the Rule of Cross Multiplication
Eigenvalues and eigenvectors are developed through a complete IOE past-paper problem. Use the S1, S2, S3 method for the characteristic equation and the rule of cross multiplication for each eigenvector.
UnitMatrices
TopicEigenvalues and Eigenvectors
Question Type4 Mark Problems
Theory Required for This Problem
The S1, S2, S3 Method
\[S_1=\mathrm{trace}(A).\]
\[S_2=\mathrm{sum\ of\ principal\ minors}.\]
\[S_3=|A|.\]
\[\lambda^3-S_1\lambda^2+S_2\lambda-S_3=0.\]
\[\lambda^3-S_1\lambda^2\]\[{}+S_2\lambda-S_3=0.\]
Rule of Cross Multiplication
Use two independent equations from \((A-\lambda I)X=0\). Their coefficient rows give a vector perpendicular to both rows.
Note: Any non-zero scalar multiple of an eigenvector is also an eigenvector.
Calculator Check
Use matrix mode to check \(S_1\), \(S_3\), and the eigenvalues. In the solution, show \(S_1\), \(S_2\), \(S_3\), the characteristic equation, and every eigenvector.
Method used: Find \(S_1,S_2,S_3\), form the characteristic equation, calculate each eigenvalue, and use \((A-\lambda I)X=0\) separately for every eigenvector.
Continue through the IOE Engineering Mathematics II Digital Workbook. This eigenvalues and eigenvectors page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
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4 MarksEigenvalues and Corresponding Eigenvectors
PAST-PAPER REFERENCE: TU IOE • 2081 Ashwin • Regular (New Course, 2080 Batch) • SH 151 • Q. 13 • 4 Marks
Find the eigenvalues and eigenvectors of the matrix
\[A=\left[\begin{array}{ccc}3&2&2\\1&4&1\\-2&-4&-1\end{array}\right].\]
Solution
The S1, S2, S3 method and cross multiplicationGiven
\[A=\left[\begin{array}{ccc}3&2&2\\1&4&1\\-2&-4&-1\end{array}\right].\]
First,
\[S_1=\mathrm{trace}(A)=3+4-1=6.\]The principal minors are
\[\left|\begin{array}{cc}4&1\\-4&-1\end{array}\right|=0,\quad\left|\begin{array}{cc}3&2\\-2&-1\end{array}\right|=1,\quad\left|\begin{array}{cc}3&2\\1&4\end{array}\right|=10.\]
Therefore,
\[S_2=0+1+10=11.\]Also,
\[S_3=|A|=6.\]The characteristic equation is
\[\lambda^3-S_1\lambda^2+S_2\lambda-S_3=0.\]Substituting the values,
\[\lambda^3-6\lambda^2+11\lambda-6=0.\]Therefore,
\[(\lambda-1)(\lambda-2)(\lambda-3)=0.\]Hence,
\[\lambda_1=1,\quad\lambda_2=2,\quad\lambda_3=3.\]For \(\lambda_1=1\),
\[(A-I)X=\left[\begin{array}{ccc}2&2&2\\1&3&1\\-2&-4&-2\end{array}\right]\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=0.\]
Using two independent rows and cross multiplication,
\[X_1=(1,0,-1)^T.\]For \(\lambda_2=2\),
\[(A-2I)X=\left[\begin{array}{ccc}1&2&2\\1&2&1\\-2&-4&-3\end{array}\right]\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=0.\]
By cross multiplication,
\[X_2=(2,-1,0)^T.\]For \(\lambda_3=3\),
\[(A-3I)X=\left[\begin{array}{ccc}0&2&2\\1&1&1\\-2&-4&-4\end{array}\right]\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=0.\]
By cross multiplication,
\[X_3=(0,1,-1)^T.\]Verification:
\[AX_1=X_1,\quad AX_2=2X_2,\quad AX_3=3X_3.\]
\[AX_1=X_1.\]\[AX_2=2X_2.\]\[AX_3=3X_3.\]
Calculator check:
\[S_1=6,\quad S_2=11,\quad S_3=6,\quad\lambda=1,2,3.\]
\[S_1=6,\quad S_2=11,\quad S_3=6.\]\[\lambda=1,2,3.\]
\[(\lambda,X)=(1,(1,0,-1)^T),(2,(2,-1,0)^T),(3,(0,1,-1)^T).\]
\[\lambda=1,\quad X_1=(1,0,-1)^T.\]\[\lambda=2,\quad X_2=(2,-1,0)^T.\]\[\lambda=3,\quad X_3=(0,1,-1)^T.\]
A Few Important Questions
Eigenvalues and Eigenvectors with Three Distinct Roots
Find the eigenvalues and eigenvectors of
\[\left[\begin{array}{ccc}2&2&0\\2&1&1\\-7&2&-3\end{array}\right].\]
Show Solution
Using the S1, S2, S3 method,
\[S_1=0,\quad S_2=-13,\quad S_3=-12.\]The characteristic equation is
\[\lambda^3-13\lambda+12=0.\]Therefore,
\[(\lambda+4)(\lambda-3)(\lambda-1)=0.\]Using cross multiplication for each eigenvalue,
\[X_1=(1,-3,13)^T,\quad X_2=(2,1,-2)^T,\quad X_3=(-2,1,4)^T.\]
\[\lambda=-4,\quad X_1=(1,-3,13)^T.\]\[\lambda=3,\quad X_2=(2,1,-2)^T.\]\[\lambda=1,\quad X_3=(-2,1,4)^T.\]
Eigenvalues Including Zero
Find the eigenvalues and eigenvectors of
\[\left[\begin{array}{ccc}4&-20&-10\\-2&10&4\\6&-30&-13\end{array}\right].\]
Show Solution
Using the S1, S2, S3 method,
\[S_1=1,\quad S_2=-2,\quad S_3=0.\]The characteristic equation is
\[\lambda^3-\lambda^2-2\lambda=0.\]Therefore,
\[\lambda(\lambda-2)(\lambda+1)=0.\]Using cross multiplication for each eigenvalue,
\[X_1=(5,1,0)^T,\quad X_2=(0,1,-2)^T,\quad X_3=(2,0,1)^T.\]
\[\lambda=0,\quad X_1=(5,1,0)^T.\]\[\lambda=2,\quad X_2=(0,1,-2)^T.\]\[\lambda=-1,\quad X_3=(2,0,1)^T.\]