Matrix

Simplicity · #1 $Old$ June 19th, 2008, 08:02 AM

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.

kalagota · #2 $Old$ June 19th, 2008, 08:14 AM

Quote:

Originally Posted by Air $View Post$

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

rednest · #3 $Old$ June 19th, 2008, 08:15 AM

Quote:

Originally Posted by Air $View Post$

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.

Simplicity · #4 $Old$ June 19th, 2008, 08:31 AM

Quote:

Originally Posted by rednest $View Post$

No you cannot do that I think because matrices are not commutative.

So, how do I work out D?

kalagota · #5 $Old$ June 19th, 2008, 08:33 AM

how do you get the inverse of $P$ ?

Simplicity · #6 $Old$ June 19th, 2008, 08:35 AM

Quote:

Originally Posted by kalagota $View Post$

how do you get the inverse of $P$ ?

It's just Transpose.

kalagota · #7 $Old$ June 19th, 2008, 08:37 AM

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

bobak · #8 $Old$ June 19th, 2008, 08:48 AM

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

Reckoner · #9 $Old$ June 19th, 2008, 11:24 AM

Quote:

Originally Posted by Air $View Post$

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

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