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Expert-verified Found in: Page 336 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # For each of the matrices in Exercises 1 through 13, find all real eigenvalues, with their algebraic multiplicities. Show your work. Do not use technology.$\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$

Eigenvalues are:${\lambda }_{1}=\frac{5+\sqrt{33}}{2}$ with multiplicity of 1

with multiplicity of 1

See the step by step solution

## Step 1: Eigenvalues

• In linear algebra, an eigenvector or characteristic vector of a linear transformation is a nonzero vector that changes at most by a scalar factor when that linear transformation is applied to it. The corresponding eigenvalue, often denoted by ${\mathbit{\lambda }}$, is the factor by which the eigenvector is scaled.
• Eigenvalues of a triangular matrix are its diagonal matrix.

## Step 2: Finding all real eigenvalues, with their algebraic multiplicities

We can clearly see that,

$\mathrm{det}\left({\mathrm{\lambda I}}_{\mathrm{n}}\right)=\mathrm{det}\left[\begin{array}{cc}1-\mathrm{\lambda }& 2\\ 3& 4-\mathrm{\lambda }\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left(1-\mathrm{\lambda }\right)\left(4-\mathrm{\lambda }\right)-2.3\phantom{\rule{0ex}{0ex}}={\mathrm{\lambda }}^{2}-5\mathrm{\lambda }-2$

Find zeroes of characteristic equation:

role="math" localid="1659585122780" ${\mathrm{\lambda }}^{2}-5\mathrm{\lambda }-2=0\phantom{\rule{0ex}{0ex}}\lambda =\frac{5+\sqrt{25+8}}{2}\phantom{\rule{0ex}{0ex}}\lambda =\frac{5+\sqrt{33}}{2}$

Eigenvalues are:

${\lambda }_{1}=\frac{5+\sqrt{33}}{2}$ with multiplicity of 1

${\lambda }_{2}=\frac{5+\sqrt{33}}{2}$with multiplicity of 1

Hence, the answer is $\lambda =\frac{5+\sqrt{33}}{2}$. ### Want to see more solutions like these? 