Diagonalisation
A matrix with enough independent eigenvectors can be written as
\[P^{-1}AP=D.\]
Equivalently,
\[A=PDP^{-1}.\]
Matrix Diagonalisation
Eigenvectors, the Modal Matrix and the Diagonal Matrix
Matrix diagonalisation is developed through a complete IOE past-paper problem. Form the modal matrix from the eigenvectors and verify the diagonal matrix.
UnitMatrices
TopicDiagonalisation
Question Type4 Mark Problems
Theory Required for This Problem
Modal Matrix
Place the independent eigenvectors as the columns of \(P\). Keep the same order of eigenvalues in \(D\).
Diagonal Matrix
If the columns of \(P\) correspond to \(\lambda_1,\lambda_2,\ldots\), then
\[D=\left[\begin{array}{cc}\lambda_1&0\\0&\lambda_2\end{array}\right].\]
Verification
Calculate \(P^{-1}AP\). The result must equal \(D\).
Method used: Find the eigenvalues and eigenvectors, form \(P\) and \(D\) in matching order, calculate \(P^{-1}\), and verify \(P^{-1}AP=D\).
Continue through the IOE Engineering Mathematics II Digital Workbook. This matrix diagonalisation page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
56
4 MarksDiagonalisation of a Two by Two Matrix
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 14 • 4 Marks
Diagonalise the matrix
\[A=\left[\begin{array}{cc}1&4\\3&2\end{array}\right].\]
Solution
Eigenvectors and the modal matrixGiven
\[A=\left[\begin{array}{cc}1&4\\3&2\end{array}\right].\]
First,
\[S_1=\mathrm{trace}(A)=1+2=3.\]Also,
\[S_2=|A|=(1)(2)-(4)(3)=-10.\]The characteristic equation is
\[\lambda^2-S_1\lambda+S_2=0.\]Substituting the values,
\[\lambda^2-3\lambda-10=0.\]Factorising,
\[(\lambda-5)(\lambda+2)=0.\]Therefore,
\[\lambda_1=5,\quad\lambda_2=-2.\]For \(\lambda_1=5\),
\[(A-5I)X=\left[\begin{array}{cc}-4&4\\3&-3\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]=0.\]
Thus,
\[x=y\quad\Rightarrow\quad X_1=(1,1)^T.\]For \(\lambda_2=-2\),
\[(A+2I)X=\left[\begin{array}{cc}3&4\\3&4\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]=0.\]
Thus,
\[3x+4y=0\quad\Rightarrow\quad X_2=(4,-3)^T.\]Form the modal and diagonal matrices in matching order:
\[P=\left[\begin{array}{cc}1&4\\1&-3\end{array}\right],\quad D=\left[\begin{array}{cc}5&0\\0&-2\end{array}\right].\]
Now,
\[P^{-1}=\frac17\left[\begin{array}{cc}3&4\\1&-1\end{array}\right].\]
Verification:
\[P^{-1}AP=\left[\begin{array}{cc}5&0\\0&-2\end{array}\right]=D.\]
Equivalently,
\[A=PDP^{-1}.\]Calculator check:
\[S_1=3,\quad S_2=-10,\quad\lambda=5,-2.\]\[D=\left[\begin{array}{cc}5&0\\0&-2\end{array}\right],\quad P=\left[\begin{array}{cc}1&4\\1&-3\end{array}\right].\]