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 $D = P^{-1}AP$

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 $P$?

6. Originally Posted by kalagota
how do you get the inverse of $P$?
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 $P^{-1}AP = D$?

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

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

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

so $IAI = PDP^{-1}$

given $A = PDP^{-1}$

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 $D = P^{-1}AP$, there is a much easier way:

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

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

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

Let me give a brief example:

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

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

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

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

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

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

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

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

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

For example, if we had set

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

then we would get

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