StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q5.2-7E

Expert-verifiedFound in: Page 267

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.**

** **

**7. \(\left[ {\begin{array}{*{20}{c}}5&3\\- 4&4\end{array}} \right]\) **

Characteristic polynomial: \({\lambda ^2} - 9\lambda + 32\).

\(A\) has no real Eigen values.** **

If is an \(n \times n\) matrix, then \(det\left( {A - \lambda I} \right)\), which is a polynomial of degree \(n\), is called the characteristic polynomial of \(A\).

It is given that \(A = \left[ {\begin{array}{*{20}{c}}5&3\\- 4&4\end{array}} \right]\) and \(I = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\) is identity matrix. Find the matrix\(\left( {A - \lambda I} \right)\) as shown below:

\[\begin{array}A - \lambda I = \left[ {\begin{array}{*{20}{c}}3&- 4\\4&8\end{array}} \right] - \lambda \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 4}\\4&{8 - \lambda }\end{array}} \right]\end{array}\]

Now, calculate the determinant of the matrix \(\left( {A - \lambda I} \right)\) as shown below:

\[\begin{array}det\left( {A - \lambda I} \right) = \det \left[ {\begin{array}{*{20}{c}}5&3\\{ - 4}&4\end{array}} \right]\\ = \left( {5 - \lambda } \right)\left( {4 - {\rm{\lambda }}} \right) + 12\\ = {\lambda ^2} - 9\lambda + 20 + 12\\ = {\lambda ^2} - 9\lambda + 32\end{array}\]

So, the characteristic polynomial of is \({\lambda ^2} - 9\lambda + 32\).

To find the eigenvalues of the matrix, we must calculate all the scalars such that \(\left( {A - \lambda I} \right)x = 0\) has a non-trivial solution which is equivalent to finding all such that the matrix \(\left( {A - \lambda I} \right)\)is not invertible, that is, when determinant of \(\left( {A - \lambda I} \right)\)is zero.

Thus, the eigenvalues of \(A\) are the solutions of the characteristic equation\(\det \left( {A - \lambda I} \right) = 0\). So, find the characteristic equation \(\det \left( {A - \lambda I} \right) = 0\).

\[\begin{array}det\left[ {\begin{array}{*{20}{c}}5&3\\- 4&4\end{array}} \right] = 0\\\left( {5 - \lambda } \right)\left( {4 - {\rm{\lambda }}} \right) + 12 = 0\\{\lambda ^2} - 9\lambda + 20 + 12 = 0\\{\lambda ^2} - 9\lambda + 32 = 0\end{array}\]

For the quadratic equation, \(a{x^2} + bx + c = 0\) , the general solution is given as\(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} \;\;}}{{2a}}\).

Thus, the solution of the characteristic equation \({\lambda ^2} - 9\lambda + 32 = 0\) is obtained as follows:

\[\begin{array}{\lambda ^2} - 9\lambda + 32 = 0\\\lambda = \frac{{ - \left( { - 9} \right) \pm \sqrt {{{\left( { - 9} \right)}^2} - 4\left( {32} \right)} }}{2}\\ = \frac{{ - 9 \pm \sqrt { - 47} }}{2}\end{array}\]

The eigenvalues of \(A\) are complex. So, \(A\) has no real Eigen values. This implies that no real vector \(x\)in \({\mathbb{R}^2}\)satisfies the characteristic equation \(\left( {A - \lambda I} \right)x = 0\).

94% of StudySmarter users get better grades.

Sign up for free