Find a 2x2 Matrix with given EigenValues

[Jtechguy21]

Homework Statement

Find a 2X2 matrix that has all non-zero entries where 3 is an eigenvalue

Homework Equations

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?

 

[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.

 

[Jtechguy21]

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

 

[HallsofIvy]

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}

 

[pasmith]

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.

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

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

 

[HallsofIvy]

Good points- thank you.