some foundational linear algebra

[gusl]

Let U be a matrix such that det(U) = 1.
Show that for any square matrix A, the eigenvalues of UA have the same modulus as the eigenvalues of A.

I don't know how to get the eigenvectors of UA from the eigenvectors of A.

Comments

[altamira16.livejournal.com]

I want to believe that there is some trickery involving invertible matrices here, but I am not absolutely sure.

[jcreed.livejournal.com]

This does not hold.

In[1]:= A = {{2, 0}, {0, 1}}
In[2]:= U = {{0, -1}, {1, 0}}

In[3]:= Det[U] == 1
Out[3]= True

In[4]:= Eigenvalues[A]
Out[4]= {2, 1}

In[5]:= Eigenvalues[U . A]
Out[5]= {I Sqrt[2], -I Sqrt[2]}

I think you probably want the group-theoretic conjugate UAU^-1. The existence of the inverse is guaranteed by the weaker assumption that det U ≠ 0.

In[6]:= Eigenvalues[U . A . Inverse[U]]
Out[6]= {2, 1}

In general:
Let v be an eigenvector of A with eigenvalue λ. Then Uv is an eigenvector of UAU^-1 with the same eigenvalue: UAU^-1(Uv) = UAv = Uλv = λUv. Since det U ≠ 0, the mapping v → Uv is injective, and U has the same eigenvalues with the same multiplicities.

[gustavolacerda.livejournal.com]

thanks.