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Q12E

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.**

**${\left[\begin{array}{cccc}2& -2& 0& 0\\ 1& -1& 0& 0\\ 0& 0& 3& -4\\ 0& 0& 2& -3\end{array}\right]}$**

Eigenvalues are:

$\begin{array}{l}{\mathrm{\lambda}}_{1}=0,\mathrm{almuu}\left(0\right)=1\\ {\mathrm{\lambda}}_{2}=-1,\mathrm{almu}(-1)=2\\ {\mathrm{\lambda}}_{4}=1,\mathrm{almu}\left(1\right)=1\end{array}$

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

Since, given matrix is triangular its eigenvalues are the entries on the main diagonal.

Find eigenvalues using characteristic equation as:

$det(A-\lambda l)=0\phantom{\rule{0ex}{0ex}}\left|\begin{array}{cccc}2-\lambda & -2& 0& 0\\ 1& -1-\lambda & 0& 0\\ 0& 0& 3-\lambda & -4\\ 0& 0& 2& -3\end{array}\right|=0\phantom{\rule{0ex}{0ex}}\left(\left(2-\lambda \right)\left(-1-\lambda \right)+2\right)\left(\left(3-\lambda \right)\left(-3-\lambda \right)+8\right)=0$

$\lambda =0,{\lambda}_{2,3}=-1,{\lambda}_{4}=1$

Hence, the answer is:

$\begin{array}{l}{\mathrm{\lambda}}_{1}=0,\mathrm{almuu}\left(0\right)=1\\ {\mathrm{\lambda}}_{2}=-1,\mathrm{almu}(-1)=2\\ {\mathrm{\lambda}}_{4}=1,\mathrm{almu}\left(1\right)=1\end{array}$

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