Suggested languages for you:

Americas

Europe

Q15E

Expert-verifiedFound in: Page 345

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**For a given eigenvalue, find a basis of the associated eigenspace. Use the geometric multiplicities of the eigenvalues to determine whether a matrix is diagonalizable. For each of the matrices A in Exercises 1 through 20, find all (real) eigenvalues. Then find a basis of each eigenspace, and diagonalize A, if you can. Do not use technology**

**${\mathbf{15}}{\mathbf{.}}\mathbf{\left(}\begin{array}{ccc}\mathbf{-}\mathbf{1}& \mathbf{0}& \mathbf{1}\\ \mathbf{-}\mathbf{3}& \mathbf{0}& \mathbf{1}\\ \mathbf{4}& \mathbf{0}& \mathbf{3}\end{array}\mathbf{\right)}$ **

The given matrix A is not diagonalizable.

**Algebraic versus geometric multiplicity If λ is an eigenvalues of a square matrix A, **

**then **

${\mathbf{gemu}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}{\mathbf{<}}{\mathbf{almu}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

$\mathrm{det}\left(A-\lambda I\right)=0\phantom{\rule{0ex}{0ex}}\left(\begin{array}{ccc}-1-\lambda & 0& 1\\ -3& -\lambda & 1\\ -4& 0& 3-\lambda \end{array}\right)=0\phantom{\rule{0ex}{0ex}}-\lambda (1-\lambda )\left(3-\lambda \right)-4\lambda =0\phantom{\rule{0ex}{0ex}}\lambda =\left({\lambda}^{2}-2\lambda -3+4\right)=0$

$-\lambda ({\lambda}^{2}-2\lambda -3+4)=0\phantom{\rule{0ex}{0ex}}-\lambda ({\lambda}^{2}-2\lambda +1)=0\phantom{\rule{0ex}{0ex}}-\lambda (\lambda -1{)}^{2}=0\phantom{\rule{0ex}{0ex}}{\lambda}_{1}=0,{\lambda}_{2,3}=1$

For $\lambda =0$

$Ax=0\phantom{\rule{0ex}{0ex}}\left(\begin{array}{ccc}-1& 0& 1\\ -3& 0& 1\\ -4& 0& 3\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\phantom{\rule{0ex}{0ex}}{x}_{1}+{x}_{3}=0,-3{x}_{1}+{x}_{3}=0,-4{x}_{1}+3{x}_{3}=0\phantom{\rule{0ex}{0ex}}{x}_{1}={x}_{2}=0$

The basic of the eigenspace is $\left\{\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]\right\}=\left\{{v}_{1}\right\}$

For $\lambda =1$ , we get

$\left(A-I\right)x=0\phantom{\rule{0ex}{0ex}}\left(\begin{array}{ccc}-2& 0& 1\\ -3& -1& 1\\ -4& 0& 2\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\phantom{\rule{0ex}{0ex}}-2{x}_{1}+{x}_{3}=0,-3{x}_{1}-{x}_{2}+{x}_{3}=0$

The basic of this eigenspace is $\left\{\left[\begin{array}{c}1\\ -1\\ 2\end{array}\right]\right\}=\left\{{v}_{2}\right\}$

Therefore, the $\mathrm{gemu}\left(1\right)<\mathrm{almu}\left(1\right)$. So A is not diagonalizable.

94% of StudySmarter users get better grades.

Sign up for free