# Finding all 2 by 2 matrices that have eigenvalues x and y

• Apr 5th 2009, 03:37 PM
precious_pearl13
Finding all 2 by 2 matrices that have eigenvalues x and y
I have been pulling my hair in solving the following problem and I have reached to the deadline. If anyone can give me an idea of how can I solve the problem, please help me...

Problem: Find all the 2 by 2 matrices that have x and y as eigenvalues.

You may replace the values of x and y, say by 4 and 5 to show me an example of how the problem can be solved...

Thank you very much for your time... It means a lot to me... Hope to get some idea of how the problem can be solved....
• Apr 5th 2009, 04:46 PM
Mush
Quote:

Originally Posted by precious_pearl13
I have been pulling my hair in solving the following problem and I have reached to the deadline. If anyone can give me an idea of how can I solve the problem, please help me...

Problem: Find all the 2 by 2 matrices that have x and y as eigenvalues.

You may replace the values of x and y, say by 4 and 5 to show me an example of how the problem can be solved...

Thank you very much for your time... It means a lot to me... Hope to get some idea of how the problem can be solved....

Just reverse engineer the problem.

If x and y are eigenvalues, then the characteristic polynomial is:

$(\lambda - x)(\lambda - y) = 0$

Since there are two eigenvalues you are expecting a matrix or order 2. Meaning your characteristic matrix is:

$\left( \begin{array}{ccc}
a & b \\
c & d \end{array} \right)$

You know the determinant of such a matrix must give yhou the characteristic polynomial, so:

$ad-bd = (\lambda - x)(\lambda - y)$

See where I'm going with this?