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..

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- Dec 4th 2009, 05:36 PMkjchauhanEquivalence Matrices
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.. - Dec 4th 2009, 05:51 PMBruno J.
What do you want exactly? All matrices equivalent to $\displaystyle A$?

- Dec 4th 2009, 06:09 PMkjchauhan
Any one..

suppose the matrix $\displaystyle A=\left[\begin{matrix} 1 & 2 & 3\\ 1 & 0 & 2\\ 3 & -1 & 0 \\\end{matrix}\right]$.

Then how we can find an equivalent matrix of A? - Dec 4th 2009, 06:11 PMlvleph
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. - Dec 4th 2009, 06:17 PMkjchauhan
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. $\displaystyle A=PBQ$.. - Dec 4th 2009, 06:26 PMlvleph
Well, similar matrices are equivalent by definition. Since $\displaystyle P,P^{-1}$ are non-singular and taking $\displaystyle P=P,Q=P^{-1}$ we fulfill the definition.

- Dec 5th 2009, 04:51 AMHallsofIvy
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.