91影视

Let us consider two 2 x 2 square matrices A and B such that

$A=\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right],B=\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]$

Clearly, rank of A = 2 = rank of B.

A square matrix A is said to be similar to a square matrix B of the same order if there exists an invertible matrix P such that-

$A=P^{-1}BP$

Let us assume $P=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$be an invertible matrix such that matrix A is similar to matrix B. Then we have,

$AP=PA$

$鈬�\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]$

$鈬�\left[\begin{array}{cc}2a+c& 2b+d\\ 3c& 3d\end{array}\right]=\left[\begin{array}{cc}a& 2b\\ c& 2d\end{array}\right]$

$$\begin{gathered} 鈬�2a+c=a \\ 2b+d=2b \\ 3c=c \\ 3d=2d \end{gathered}$$

$$\begin{gathered} 鈬�d=0,c=0 \\ \text{and} \\ a+c=0 \\ 鈬�a=0 \end{gathered}$$

Thus, the matrix$P=\left[\begin{array}{cc}0& b\\ 0& 0\end{array}\right]$ is not invertible for any value of 鈥榖鈥�.

So, there doesn鈥檛 exist any invertible matrix P such that$A=P^{-1}BP.$

If we take two 2 x 2 square matrices A and B such that rank(A) is equal to rank(B) then there doesn鈥檛 exist any invertible matrix P such that$A=P^{-1}BP.$

Since it is true for 2 x 2 matrices, then it is true for n x n matrices.

Hence, if two n x n matrices A and B have the same rank, then they may not be similar.