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Expert-verified Found in: Page 267 ### Linear Algebra and its Applications

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.

See the step by step solution

## Step 1: Find the characteristic polynomial

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 $${\lambda ^2} - 9\lambda + 32$$.

## Step 2: Describe the characteristic equation

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 $$\det \left( {A - \lambda I} \right) = 0$$.

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

## Step 3: Find roots of characteristic equation

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 $$\left( {A - \lambda I} \right)x = 0$$. ### Want to see more solutions like these? 