Linear Independence
The vectors \(\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_n\) are linearly independent when
\[c_1\mathbf v_1+c_2\mathbf v_2+\cdots+c_n\mathbf v_n=\mathbf0\]
has only the trivial solution \(c_1=c_2=\cdots=c_n=0\).
Linear Independence and Eigenvalues
Determinant Test and the Characteristic Equation
Linear independence and eigenvalues are developed through two IOE past-paper questions. Use the determinant test and the characteristic equation in the standard examination method.
UnitMatrices
TopicIndependence and Eigenvalues
Question Type2 Mark Problems
Theory Required for These Problems
Determinant Test
For \(n\) vectors in \(\mathbf R^n\), place the vectors as columns of a square matrix \(A\).
If \(|A|\ne0\), the vectors are linearly independent. If \(|A|=0\), they are linearly dependent.
Eigenvalues
A non-zero vector \(\mathbf x\) is an eigenvector of \(A\) when
\[A\mathbf x=\lambda\mathbf x.\]
The corresponding scalar \(\lambda\) is an eigenvalue.
Characteristic Equation
Find the eigenvalues from
\[|A-\lambda I|=0.\]
A repeated factor gives a repeated eigenvalue. Keep every repetition in the final answer.
Method used: Use a determinant to test linear independence. For eigenvalues, form \(A-\lambda I\), expand the determinant, factor the characteristic equation, and list every root with its multiplicity.
Continue through the IOE Engineering Mathematics II Digital Workbook. This linear independence and eigenvalues page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
51
2 Marks
Test of Linear Independence
PAST-PAPER REFERENCE: TU IOE • 2081 Ashwin • Regular (New Course, 2080 Batch) • SH 151 • Q. 5(b) • 2 Marks
Test whether the vectors
\[(1,1,1),\qquad(1,-1,1),\qquad(2,0,3)\]
are linearly independent or dependent.
Solution
Determinant testPlace the vectors as columns of the matrix
\[A=\left[\begin{array}{ccc}1&1&2\\1&-1&0\\1&1&3\end{array}\right].\]
Now,
\[|A|=1\left|\begin{array}{cc}-1&0\\1&3\end{array}\right|-1\left|\begin{array}{cc}1&0\\1&3\end{array}\right|+2\left|\begin{array}{cc}1&-1\\1&1\end{array}\right|.\]
Therefore,
\[|A|=-3-3+4=-2\ne0.\]Since \(|A|\ne0\), the equation
\[c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3=\mathbf0\]has only the solution
\[c_1=c_2=c_3=0.\]The vectors are linearly independent.
52
2 Marks
Eigenvalues from the Characteristic Equation
PAST-PAPER REFERENCE: TU IOE • 2083 Baisakh • Back (New Course) • ENSH 151 • Q. 5(b) • 2 Marks
Find the eigenvalues of
\[A=\left[\begin{array}{ccc}3&1&0\\1&3&0\\0&0&2\end{array}\right].\]
Solution
Characteristic equationThe characteristic equation is
\[|A-\lambda I|=\left|\begin{array}{ccc}3-\lambda&1&0\\1&3-\lambda&0\\0&0&2-\lambda\end{array}\right|=0.\]
Expanding the determinant,
\[[(3-\lambda)^2-1](2-\lambda)=0.\]Factor the difference of two squares:
\[[(3-\lambda)-1][(3-\lambda)+1](2-\lambda)=0.\]Therefore,
\[(2-\lambda)^2(4-\lambda)=0.\]Hence,
\[\lambda=4,\ 2,\ 2.\]\[\lambda=4,\ 2,\ 2\]