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Q6E

Expert-verifiedFound in: Page 336

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

- 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 ${\mathit{\lambda}}$, is the factor by which the eigenvector is scaled.
- Eigenvalues of a triangular matrix are its diagonal matrix.

** **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 $\lambda =\frac{5+\sqrt{33}}{2}$.

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