Find a 2x2 Matrix with given EigenValues

1. Jul 23, 2015

Jtechguy21

1. The problem statement, all variables and given/known data
Find a 2X2 matrix that has all non-zero entries where 3 is an eigenvalue

2. Relevant equations

3. The attempt at a solution
well since the 2x2 matrix cannot be triangular, it makes things harder for me.
I have no idea where to start. I am not given any eigenvectors either.
It seems like a simple problem, but I've been stuck on it for a while.

If I did not have the zero entries restriction i would have selected 2x2 matrix
A= {{3,0}{0,3}}τ

Can Someone point me in the right direction?

2. Jul 23, 2015

ShayanJ

The eigenvalue equation is for the 2X2 matrix, if written as a system of homogeneous equations, will have a solution if the determinant of the matrix of coefficients is zero. So we have the equation $\lambda^2-(a+d)\lambda+ad-bc=0$ where $\lambda$ is the given eigenvalue and a,b,c and d are the unknown matrix entries. So we have one equation for four unknowns which means this is an under-determined system. So there are an infinite number of 2X2 matrices with a given eigenvalue. Just choose three of the entries arbitrarily and determine the fourth via the above equation.

3. Jul 23, 2015

Jtechguy21

thanks for your help Shyan Using that system I was able to solve my problem :)

4. Jul 24, 2015

HallsofIvy

Staff Emeritus
First, while you said that "3 is an eigenvalue", you did NOT say that it was the only eigenvalue, a double eigenvalue. Even assuming that there are an infinite number of such 2 by 2 matrices. One way to find one is to start with the diagonal matrix $$M= \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}$$. Now, for any invertible matrix, A, $AMA^{-1}$ will have the same eigenvalues so you only need to choose A so that has no 0 entries.

Completely different solutions can be found by starting, not with a diagonal matrix, but with the "Jordan Normal Form", $$M= \begin{bmatrix}3 & 1 \\ 0 & 3\end{bmatrix}$$

5. Jul 24, 2015

pasmith

If $M$ is a multiple of the identity, then it commutes with $A$ so that $AMA^{-1} = MAA^{-1} = M$, which has zero entries.

This is necessary if 3 is to be a repeated eigenvalue, but the question doesn't require that.

6. Jul 25, 2015

HallsofIvy

Staff Emeritus
Good points- thank you.