1. ## Matrix

Find an orthogonal matrix P and a diagonal matrix D such that

A = PDP^(-1).

I found P but how do I find D? Is it correct to do D = PAP^(-1)? Thanks in advance.

2. Originally Posted by Air
Find an orthogonal matrix P and a diagonal matrix D such that

A = PDP^(-1).

I found P but how do I find D? Is it correct to do D = PAP^(-1)? Thanks in advance.
yes, you can do that.. but i think, it is $\displaystyle D = P^{-1}AP$

oh well, as long as they are inverses.. Ü

3. Originally Posted by Air
Find an orthogonal matrix P and a diagonal matrix D such that

A = PDP^(-1).

I found P but how do I find D? Is it correct to do D = PAP^(-1)? Thanks in advance.
No you cannot do that I think because matrices are not commutative.

4. Originally Posted by rednest
No you cannot do that I think because matrices are not commutative.

So, how do I work out D?

5. how do you get the inverse of $\displaystyle P$?

6. Originally Posted by kalagota
how do you get the inverse of $\displaystyle P$?
It's just Transpose.

7. yeah.. so apply the corrected equation i gave..

8. Do you know how to find a matrix P such that $\displaystyle P^{-1}AP = D$?

If your familiar with that all you need to do is some basic matrix algebra

first pre multiply both sides by P , $\displaystyle PP^{-1}AP = PD \ \ \Rightarrow IAP = PD$

then post multiply by $\displaystyle P^{-1}$ so $\displaystyle IAPP^{-1} = PDP^{-1}$

so $\displaystyle IAI = PDP^{-1}$

given $\displaystyle A = PDP^{-1}$

Bobak

9. Originally Posted by Air
Find an orthogonal matrix P and a diagonal matrix D such that

A = PDP^(-1).

I found P but how do I find D? Is it correct to do D = PAP^(-1)? Thanks in advance.
Although you could do $\displaystyle D = P^{-1}AP$, there is a much easier way:

In the process of finding $\displaystyle P$, you probably found the eigenvalues and eigenvectors of $\displaystyle A$. We know that if the eigenvalues of $\displaystyle A$ are $\displaystyle \lambda_1,\;\lambda_2,\;\dots,\;\lambda_n$, and each eigenvalue $\displaystyle \lambda_i$ corresponds to an eigenvector $\displaystyle \textbf{p}_i$ such that $\displaystyle \textbf{p}_1,\;\textbf{p}_2,\;\dots,\;\textbf{p}_n$ are linearly independent, then the matrix $\displaystyle P = \left[\begin{array}{c|c|c|c}\textbf{p}_1 & \textbf{p}_2 & \cdots & \textbf{p}_n\end{array}\right]$ will diagonalize $\displaystyle A$, and $\displaystyle A = PDP^{-1}$. Now, it turns out that

$\displaystyle D = \left[\begin{matrix} \lambda_1 & 0 & 0 & \cdots & 0\\ 0 & \lambda_2 & 0 & \cdots & 0\\ 0 & 0 & \lambda_3 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & \lambda_n \end{matrix}\right]$

Therefore, if you diagonalized $\displaystyle A$ with $\displaystyle P$ to get $\displaystyle A = PDP^{-1}$, then $\displaystyle D$ will be a diagonal matrix whose diagonal entries are the eigenvalues of $\displaystyle A$, listed in the same order as you placed the eigenvectors of $\displaystyle A$ into $\displaystyle P$.

Let me give a brief example:

Let $\displaystyle A = \left[\begin{matrix} 1 & -1 & -1 \\ 1 & 3 & 1 \\ -3 & 1 & -1 \end{matrix}\right]$

We may easily determine the eigenvalues of $\displaystyle A$ to be $\displaystyle \lambda_1 = 2,\;\lambda_2 = -2,\;\lambda_3 = 3$, and their corresponding eigenvectors:

$\displaystyle \begin{array}{lcrl} \lambda_1 & = & 2: & \left[\begin{matrix}-1\\0\\1\end{matrix}\right]\medskip\\ \lambda_2 & = & -2: & \left[\begin{matrix}1\\-1\\4\end{matrix}\right]\medskip\\ \lambda_3 & = & 3: & \left[\begin{matrix}-1\\1\\1\end{matrix}\right] \end{array}$

Putting these eigenvectors into the columns of $\displaystyle P$, we get

$\displaystyle P = \left[\begin{matrix} -1 & 1 & -1\\ 0 &-1 & 1\\ 1 & 4 & 1 \end{matrix}\right]$

so $\displaystyle A = PDP^{-1}$ for some diagonal $\displaystyle D$. And if you find $\displaystyle D$, you will find that it is

$\displaystyle D = \left[\begin{matrix} 2 & 0 & 0\\ 0 &-2 & 0\\ 0 & 0 & 3 \end{matrix}\right]$

(try verifying this) and the entries along the main diagonal correspond exactly to the eigenvalues we found for $\displaystyle A$.

But, if you were to change the order of the eigenvectors in $\displaystyle P$, the entries in $\displaystyle D$ will also be in different order.

For example, if we had set

$\displaystyle P = \left[\begin{matrix} 1 & -1 & -1\\ -1 & 0 & 1\\ 4 & 1 & 1 \end{matrix}\right]$,

then we would get

$\displaystyle D = \left[\begin{matrix} -2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3 \end{matrix}\right]$