# Thread: Deriving a matrix from eigenvalues?

1. ## Deriving a matrix from eigenvalues?

Sorry I haven't learned the Latex system here...

If we have a matrix C it is simple to find the eigenvalues L and eigenvectors <e>, however, if I have a list of eigenvectors <e> and eigenvalues L, is it possible to reconstruct the matrix C ?

2. Originally Posted by HowardF
If we have a matrix C it is simple to find the eigenvalues L and eigenvectors <e>, however, if I have a list of eigenvectors <e> and eigenvalues L, is it possible to reconstruct the matrix C ?
Let D be the matrix whose entries down the main diagonal are the eigenvalues, with all the off-diagonal entries zero. Let P be the matrix whose columns are the associated eigenvectors. Then $C=PDP^{-1}$.

3. IF you are given that the matrix is "n by n" and n independent eigenvalues that works. If not then you will have to know which eigenvalues are of multiplicity greater than the number of eigenvalues corresponding to them and construct the appropriate Jordan Normal matrix.

4. ## deriving matrix from eigenvalues

This did not work for me. I have eigenvalues of 1,-8 with corresponding eigenvectors of [1,-1] and [2,-3] respectively.

C=P^-1DP would be the following

P^-1 = [3,2;-1,-1]
D=[1,0;0,-8]
P=[1,2;-1,-3]

So multiplying those 3 matrices woud yield [19,54;-9,-26] but I know the a11 position would be a 1. Please assist

5. The eigenvectors are scalable. This means they define a direction only not a magnitude. This means that the matrix P and its inverse are not unique. But I don't know if the product of the three matrices will be unique or not.

6. Originally Posted by reap76
This did not work for me. I have eigenvalues of 1,-8 with corresponding eigenvectors of [1,-1] and [2,-3] respectively.

C=P^-1DP would be the following

P^-1 = [3,2;-1,-1]
D=[1,0;0,-8]
P=[1,2;-1,-3]

So multiplying those 3 matrices would yield [19,54;-9,-26] but I know the a11 position would be a 1. Please assist
First, there is a bad mistake in my previous comment above, which I have now corrected. The formula for the matrix should be $PDP^{-1}$, not $P^{-1}DP$.

However, if the eigenvalues are 1 and –8, with corresponding eigenvectors $\begin{bmatrix}1\\-1\end{bmatrix}$ and $\begin{bmatrix}2\\-3\end{bmatrix}$, then the matrix comes out as $\begin{bmatrix}19&18\\-27&-26\end{bmatrix}$, which still has 19 in the (1,1)-position. As far as I can see, you can't get a 1 in that position from the given data.

Originally Posted by samer_guirguis_2000
The eigenvectors are scalable. This means they define a direction only not a magnitude. This means that the matrix P and its inverse are not unique. But I don't know if the product of the three matrices will be unique or not.
That is very true. But if you multiply the columns of P by scalars then the rows of the inverse of P will be multiplied by the reciprocals of those scalars, and $PDP^{-1}$ will turn out to be the same as before.