# Thread: Problem of diagonal matrix..

1. ## Problem of diagonal matrix..

i found the diagonal matrix =

$\displaystyle D = \left(\begin{array}{ccc}0&0&0\\0&1&0\\0&0&1\end{ar ray}\right)$
but i dont know how to find X^2= A......

2. Originally Posted by kin

i found the diagonal matrix =

$\displaystyle D = \left(\begin{array}{ccc}0&0&0\\0&1&0\\0&0&1\end{ar ray}\right)$
but i dont know how to find X^2= A......
Dear kin,

Can you tell me what you obtained for matrix P? Then I shall try to help you.

3. Originally Posted by Sudharaka
Dear kin,

Can you tell me what you obtained for matrix P? Then I shall try to help you.

i find
$\displaystyle P= \left(\begin{array}{ccc}-2/3&2&-1\\1/3&1&0\\1&0&1\end{array}\right)$

$\displaystyle P^-1= \left(\begin{array}{ccc}-3&6&-3\\1&-1&1\\3&-6&4\end{array}\right)$

4. For any invertible matrix P, define U by $\displaystyle U= P^{-1}XP$. Then $\displaystyle X= PUP^{-1}$ so $\displaystyle X^2= (PUP^{-1})(PUP^{-1})= PU^2P^{-1}$. Now if also $\displaystyle X^2= A$, $\displaystyle A= PU^2P^{-1}$ and $\displaystyle P^{-1}AP= U^2$.

If P is such that $\displaystyle P^{-1}AP= D$, a diagonal matrix, the $\displaystyle D= U^2$ and so U is also a diagonal matrix having entries equal to the square roots of the diagonal elements of D. That is, with $\displaystyle D= \begin{pmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$, U is also $\displaystyle \begin{pmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$ and, as before, $\displaystyle X= PUP^{-1}$.

5. Originally Posted by HallsofIvy
For any invertible matrix P, define U by $\displaystyle U= P^{-1}XP$. Then $\displaystyle X= PUP^{-1}$ so $\displaystyle X^2= (PUP^{-1})(PUP^{-1})= PU^2P^{-1}$. Now if also $\displaystyle X^2= A$, $\displaystyle A= PU^2P^{-1}$ and $\displaystyle P^{-1}AP= U^2$.

If P is such that $\displaystyle P^{-1}AP= D$, a diagonal matrix, the $\displaystyle D= U^2$ and so U is also a diagonal matrix having entries equal to the square roots of the diagonal elements of D. That is, with $\displaystyle D= \begin{pmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$, U is also $\displaystyle \begin{pmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$ and, as before, $\displaystyle X= PUP^{-1}$.
actually $\displaystyle X= PUP^{-1}$ =$\displaystyle A= PDP^{-1}$...!!!!!!!!!!!!!

by the definition of X, X should not = +A/ - A......!!!!