Matrix similarity transformation

[DryRun]

Homework Statement

For a 3x3 matrix A, i know the eigenvalues and their corresponding 3 eigenvectors.

Define a matrix P such that $PAP^{-1}$ is a diagonal matrix.

Homework Equations

Similarity transformation formula: $D=P^{-1}AP$ where D is the diagonal matrix containing the eigenvalues as its trace elements.

The Attempt at a Solution

The 3 eigenvectors form a 3x3 matrix P.
The problem is that I've learned the formula to be: $D=P^{-1}AP$
But what is being asked here is: $PAP^{-1}$

I've re-arranged the matrix, $D=P^{-1}AP$, into: $PDP^{-1}=IAI$ which gives $PDP^{-1}=A$ but i have no idea what to do next.

 

[Karamata]

Well, you can transponse left and right sides of $D=P^{-1}AP$, notice that $D = D^T$

 

[DryRun]

Hi Karamata

Well, you can transponse left and right sides of $D=P^{-1}AP$, notice that $D = D^T$

I don't understand what you mean.
$$D^T=PAP^{-1}?$$What would P be then?

 

[spamiam]

Homework Statement

For a 3x3 matrix A, i know the eigenvalues and their corresponding 3 eigenvectors.

Define a matrix P such that $PAP^{-1}$ is a diagonal matrix.

Homework Equations

Similarity transformation formula: $D=P^{-1}AP$ where D is the diagonal matrix containing the eigenvalues as its trace elements.

The Attempt at a Solution

The 3 eigenvectors form a 3x3 matrix P.
The problem is that I've learned the formula to be: $D=P^{-1}AP$
But what is being asked here is: $PAP^{-1}$

I've re-arranged the matrix, $D=P^{-1}AP$, into: $PDP^{-1}=IAI$ which gives $PDP^{-1}=A$ but i have no idea what to do next.

If you have $D = P^{-1} A P$, then let $Q = P^{-1}$. Do you see why this gives you the result you want?

 

[DryRun]

If you have $D = P^{-1} A P$, then let $Q = P^{-1}$. Do you see why this gives you the result you want?

If $Q = P^{-1}$ then does $Q^{-1} = P$? If yes, then $D=QAQ^{-1}$? Is there any proof for it? I'm not convinced. In that case, what would be the matrix Q?

 

[spamiam]

Proving $(Q^{-1})^{-1} = Q$ is easy. What's the definition of the inverse of a matrix?

Well Q would be the inverse of the matrix P! :P Seriously though, if you have $D = P^{-1} A P$ or equivalently $PD = AP$ then what must the columns of P be? Think about eigenvectors.

 

[DryRun]

OK, i think i understand the idea here.

Let $Q = P^{-1}$, then $Q^{-1} = P$ and the problem changes to $D=Q^{-1}AQ$, which corresponds to the standard form for the diagonal, D. Here, Q is the matrix of all 3 eigenvectors, and D is the diagonal matrix with the eigenvalues as its trace elements. Now, i need to find P, where $P=Q^{-1}$. Is there an easier way to find $Q^{-1}$ other than finding the adjoint matrix of Q and its determinant?

Instead of using this formula to find $Q^{-1}$: $\frac{1}{|Q|}adj(Q)$

I'm thinking it might be easier to derive it from: $D=Q^{-1}AQ$ but i have no idea how.

 

[spamiam]

OK, i think i understand the idea here.

Let $Q = P^{-1}$, then $Q^{-1} = P$ and the problem changes to $D=Q^{-1}AQ$

Don't you mean $D = Q A Q^{-1}$?

, which corresponds to the standard form for the diagonal, D. Here, Q is the matrix of all 3 eigenvectors, and D is the diagonal matrix with the eigenvalues as its trace elements.

Right, except I think $Q^{-1}$ would be the matrix whose columns are the eigenvectors of A. Try writing out what PD = AP tells you in terms of the columns of P:
$$
PD =
\begin{pmatrix}
| & | & | \\
P_1 & P_2 & P_3 \\
| & | & |
\end{pmatrix}
\begin{pmatrix}
\lambda_1 & 0 & 0 \\
0 & \lambda_2 & 0 \\
0 & 0 & \lambda_3
\end{pmatrix}= \cdots
$$
$$
AP = A
\begin{pmatrix}
| & | & | \\
P_1 & P_2 & P_3 \\
| & | & |
\end{pmatrix} = \cdots
$$

 

[DryRun]

I think you misunderstood the problem:

Define a matrix P such that $PAP^{-1}$ is a diagonal matrix.

So, $D=Q^{-1}AQ$ and Q is the matrix of eigenvectors of A.

 

[spamiam]

Ah, sorry I hadn't noticed you had edited your post!

Since you already know the eigenvalues and eigenvectors, just put them in a matrix and invert it using whatever method you prefer. You can use Cramer's rule, the equation for which you've listed below, or write the augemented matrix P|I and then row-reduce the left-hand side to the identity.

I don't see any shortcut using the equation $D = Q A Q^{-1}$. Certainly you know what $Q^{-1}$ must be, but I think you just have to calculate its inverse as usual.

Edit: I think we're confusing each other about what P is and what Q is. I took P to be as in the similarity formula you listed in the OP, $D = P^{-1} A P$, which would then make $Q = P^{-1}$ the solution you're looking for.

 

[DryRun]

Edit: I think we're confusing each other about what P is and what Q is. I took P to be as in the similarity formula you listed in the OP, $D = P^{-1} A P$, which would then make $Q = P^{-1}$ the solution you're looking for.

I know. So, if i find $Q^{-1}$ then i'll know the required matrix P. According to my calculations:
$$Q^{-1}= \begin{bmatrix}-1 & 0 & -2 \\ 1 & 1 & 3 \\ -1 & -1 & -2\end{bmatrix}$$
I tested the answer and got $QQ^{-1}=I$

Thanks for the help.

 

[spamiam]

I know. So, if i find $Q^{-1}$ then i'll know the required matrix P. According to my calculations:
$$Q^{-1}= \begin{bmatrix}-1 & 0 & -2 \\ 1 & 1 & 3 \\ -1 & -1 & -2\end{bmatrix}$$
I tested the answer and got $QQ^{-1}=I$

Thanks for the help.

Great! And just to make sure, you might want to calculate $P A P^{-1}$ and verify that you get the diagonal matrix D. And if it doesn't give you D, then I've succeeded in confusing you and you should see if $P^{-1} A P$ works! :tongue: