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Q5.2-3E

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 1-8.3. $$\left[ {\begin{array}{*{20}{c}}3&-2\\1&-1\end{array}} \right]$$

Characteristic polynomial: $${\lambda ^2} - 2\lambda - 1$$.

Eigenvalues: $$\lambda = 1 + \sqrt 2$$ and $$\lambda = 1 - \sqrt 2$$.

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 $$A = \left[ {\begin{array}{*{20}{c}}3&- 2\\1&- 1\end{array}} \right]$$ 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:

$\begin{array} - \lambda I = \left[ {\begin{array}{*{20}{c}}3&- 2\\1&- 1\end{array}} \right] - \lambda \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 2}\\1&{ - 1 - \lambda }\end{array}} \right]\end{array}$

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

$\begin{array}det\left( {A - \lambda I} \right) = det\left[ {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 2}\\1&{ - 1 - \lambda }\end{array}} \right]\\ = \left( {3 - \lambda } \right)\left( { - 1 - \lambda } \right) + 2\\ = {\lambda ^2} - 2\lambda - 1\end{array}$

So, the characteristic polynomial of is $${\lambda ^2} - 2\lambda - 1$$.

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

$\begin{array}det\left[ {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 2}\\1&{ - 1 - \lambda }\end{array}} \right] = 0\\\left( {3 - \lambda } \right)\left( { - 1 - \lambda } \right) + 2 = 0\\{\lambda ^2} - 2\lambda - 3 + 2 = 0\\{\lambda ^2} - 2\lambda - 1 = 0\end{array}$

## 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 ${\lambda ^2} - 2\lambda - 1 = 0$ is obtained as follows:

$\begin{array}{\lambda ^2} - 2\lambda - 1 = 0\\\lambda = \frac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\left( { - 1} \right)} }}{2}\\ = \frac{{2 \pm 2\sqrt 2 }}{2}\\ = 1 \pm \sqrt 2 \end{array}$

The eigenvalues of $$A$$ are $$\lambda = 1 + \sqrt 2$$ and $$\lambda = 1 - \sqrt 2$$.