Please illustrate and show me that how to find equivalence matrix of a given matrix A.
Two Matrices A and B are called equivalence if for any two non-singular matrices P and Q such that, A=PBQ.
Pl help..
Well, the easiest and dumbest answer is to let $\displaystyle P=Q=I \text{ and } B = A$. One way to find a similar matrix is by making the matrix B be a matrix with the eigenvalues of $\displaystyle A$ as the diagonal of $\displaystyle B$. Then the matrix $\displaystyle P$ would have the corresponding eigenvectors as columns and $\displaystyle Q = P^{-1}$. That is, if $\displaystyle (\lambda_i, v_i)$ is an eigenpair then
$\displaystyle B = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & & \ddots & \vdots \\ 0 & \cdots & 0 & \lambda_n\end{pmatrix}$
and
$\displaystyle P = \begin{pmatrix} v_1 \cdots v_n\end{pmatrix}$.
Somehow I doubt this is what you are looking for.
But i think it is dioganalised matrix..
Can we take similar marices as equivalent matrices?
As I know for similar matrices we have a nonsingular matrix P s.t. $\displaystyle A=P^{-1}BP$ and for equivalent matrices we have two nonsingular matrices s.t. $\displaystyle A=PBQ$..
You said before that you wanted "any" matrix equivalent to A. As Ivleph said, take any (invertible) 3 by 3 matrix, P, find [tex]P^{-1}[/math ]and calculate $\displaystyle PAP^{-1}$.
If you specifically want a diagonal matrix equivalent to A, as you seem to be saying now, you must first prove that such at thing exists- that A is "diagonalizable" which is true if and only if A has 3 independent eigenvectors. Once you have found those eigenvectors, form the matrix P having the eigenvectors as columns. The $\displaystyle D= P^{-1}AP$ is a diagonal matrix equivalent to A.