031 Review Part 1, Problem 3

True or false: If  $A$  is a  $4\times 4$  matrix with characteristic equation  $\lambda (\lambda -1)(\lambda +1)(\lambda +e)=0,$  then  $A$  is diagonalizable.

Solution:
First, we switch to the limit to $x$ so that we can use L'Hopital's rule.
So, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \infty }{\frac {3-2x^{2}}{5x^{2}+x+1}}}&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {-4x}{10x+1}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\frac {-4}{10}}\\&&\\&=&\displaystyle {-{\frac {2}{5}}}.\end{array}}$

Final Answer:
False

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