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Q49E

Expert-verifiedFound in: Page 325

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**Give an example of a matrix ****A**** of rank 1 that fails to be diagonalizable.**

The required example is $A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$

The matrix A is diagonalizable if there exists an eigenbasis for A . The ${\overrightarrow{v}}_{{1}}{,}{.}{.}{.}{,}{\overrightarrow{v}}_{{n}}$ is an eigenbasis for A , with ${A}{\overrightarrow{v}}_{{1}}{=}{{\lambda}}_{{1}}{\overrightarrow{v}}_{{1}}{,}{.}{.}{.}{,}{A}{\overrightarrow{v}}_{{n}}{\overrightarrow{V}}_{{n}}$ , then the matrices

${S}{=}\left[\begin{array}{ccc}|& |& ||\\ {\overrightarrow{v}}_{1}& {\overrightarrow{v}}_{2}& {\overrightarrow{v}}_{n}\\ |& |& ||\end{array}\right]$ and ${B}{=}\left[\begin{array}{cccc}{\lambda}_{1}& o& \dots & 0\\ 0& {\lambda}_{2}& \dots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0& 0& \dots & {\lambda}_{n}\end{array}\right]$

Will be diagonalize A , meaning that ${{S}}^{-1}{A}{S}{=}{B}$ .

For example,

$A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$

It’s only eigenvalue is $\lambda =0$ ,with its corresponding eigenvectors being $v=[10]$ .

However, this does not make for an eigenbasis, so A is not diagonalizable.

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