# Let A=begin{bmatrix}1 & 0 0 & 1 end{bmatrix} text{ and } B=begin{bmatrix}1 & 2 0 & 1 end{bmatrix} , show that A and B are not similar.

Let , show that A and B are not similar.
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Step 1
To check the similarity of the matrices, calculate the determinant of the matrices
$|\begin{array}{c}A\end{array}|=|\begin{array}{cc}1& 0\\ 0& 1\end{array}|$
$=1-0$
$=1$
$|\begin{array}{c}B\end{array}|=|\begin{array}{cc}1& 2\\ 0& 1\end{array}|$
$=1-0$
$=1$
Here, the determinants of the matrices A and B are equal.
Step 2
Now, find the eigen values of the matrices A and B.
$|\begin{array}{c}A-\lambda I\end{array}|=0$
$|\begin{array}{c}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}|=0$
$|\begin{array}{cc}1-\lambda & 0\\ 0& 1-\lambda \end{array}|=0$
$\left(1-\lambda {\right)}^{2}=0$
$1-\lambda =0$
${\lambda }_{1}=1$
${\lambda }_{2}=1$
$|\begin{array}{c}B-\lambda I\end{array}|=0$
$|\begin{array}{c}\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}|=0$
$|\begin{array}{cc}1-\lambda & 2\\ 0& 1-\lambda \end{array}|=0$
$\left(1-\lambda {\right)}^{2}=0$
$1-\lambda =0$
${\lambda }_{3}=1$
${\lambda }_{4}=1$
Thus, the matrices A and B have equal eigen values.
Step 3
Use the definition of similar matrices $B={S}^{-1}AS$, where S is a nonsingular matrix, to check the similarity of matrices A and B.
$B={S}^{-1}AS$
$={S}^{-1}{I}_{2}S$
$={S}^{-1}S$
$={I}_{2}$

Hence, the matrix B is not diagonalizable, $B\ne {S}^{-1}AS$
Therefore, the matrices A and B are not similar.
Jeffrey Jordon