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## Q. 3.11

Test if

$A= \left[\begin{matrix} 0 & -1 & 0 \\ 3 & 3 & 1 \\ 1 & 1 & 1 \end{matrix} \right]$ and $B= \left[\begin{matrix} -1 & 4 & 2 \\ -1 & 3 & 1 \\ -1 & 2 & 2 \end{matrix} \right]$

are similar.

## Verified Solution

The characteristic polynomials are

$det\left(A-tI_{3} \right) = \left|\begin{matrix} -t & -1 & 0 \\ 3 & 3-t & 1 \\ 1 & 1 & 1-t \end{matrix} \right| = -t\left(3-t\right)\left(1-t\right)-1+3\left(1-t\right)+t$

$=-\left(t-1\right) ^{2}\left(t-2\right),$

$det\left(B-tI_{3} \right) =-\left(t-1\right) ^{2}\left(t-2\right).$

So both are the same and A and B have common eigenvalues 1, 1 and 2.

By computation,

$\left(A-I_{3} \right)\left(A-2I_{3} \right)= \left[\begin{matrix} -1 & -1 & 0 \\ 3 & 2 & 1 \\ 1 & 1 & 0 \end{matrix} \right]\left[\begin{matrix} -2 & -1 & 0 \\ 3 & 1 & 1 \\ 1 & 1 & -1 \end{matrix} \right]$

$=\left[\begin{matrix} -1 & \ldots & \ldots \\ \ldots & \ldots & \ldots \\ \ldots & \ldots & \ldots \end{matrix} \right] \neq O,$

$\left(B-I_{3} \right)\left(B-2I_{3} \right)= \left[\begin{matrix} -2 & 4 & 2 \\ -1 & 2 & 1 \\ -1 & 2 & 1 \end{matrix} \right]\left[\begin{matrix} -3 & 4 & 2 \\ -1 & 1 & 1 \\ -1 & 2 & 1 \end{matrix} \right]=O.$

So A has the minimal polynomial $\left(t-1\right) ^{2}\left(t-2\right),$ while $B has \left(t-1\right)\left(t-2\right).$

Therefore A and B are not similar and B is diagonalizable while A is not.

Equivalently , $r\left(A-I_{3} \right)=2$ implies that dim $E_{1} =3-r\left(A-I_{3} \right)=1$ which is not equal to the algebraic multiplicity 2 of 1, meanwhile $r\left(B-I_{3} \right)=1$ implies that dim $E_{1} =3-r\left(B-I_{3} \right)=2.$ Thus A is not diagonalizable while B is and they are not similar.