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Found in: Page 325

### Linear Algebra With Applications

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

See the step by step solution

## Step 1: Definition of diagonalizable

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

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

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

## Step 2: Finding the suitable example

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 $v=\left[10\right]$ .

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