The discussion revolves around the relationship between the eigenvalues and eigenvectors of a matrix \( A \) and its transpose \( A^T \), specifically under the condition that \( AA^T = A^TA \). Participants are exploring the implications of this condition on the eigenvectors of both matrices.
Some participants are providing insights into the properties of square matrices and questioning the assumptions made about eigenvalues. There is an acknowledgment of incorrect statements regarding matrix inverses, and a participant expresses uncertainty about the implications of zero being an eigenvalue.
Participants are navigating through the implications of the condition \( AA^T = A^TA \) and are considering the validity of certain mathematical properties, including the potential presence of zero as an eigenvalue of \( A \).
What you have to show is that if x is an eigenvector of A, then x, not ATx, is also an eigenvector of AT.arpon said:Homework Statement
If ##AA^T=A^TA##, then prove that ##A## and ##A^T## have the same eigenvectors.
Homework Equations
The Attempt at a Solution
##Ax=\lambda x##
##A^TAx=\lambda A^Tx##
##AA^Tx=\lambda A^Tx##
##A(A^Tx)=\lambda (A^Tx)##
So, ##A^Tx## is also an eigenvector of ##A##.
arpon said:What should be the next steps?
Mark44 said:A fact that might be pertinent here is that for square matrices A and B, if AB = BA, then each one is the inverse of the other.
No, I meant the first, so my "fact" was incorrect. I guess I had that confused with if AB = I, then A and B are inverses.Ray Vickson said:If
$$A = \pmatrix{0&1\\1&0} \; \text{and} \; B = \pmatrix{1&0\\0&1}$$
then ##AB = BA = A##, but ##B \neq A^{-1}## and ##A \neq B^{-1}##.
However, if you mean that ##AB = (BA)^{-1}##, etc., then that seems OK, in this example at least.
Ray Vickson said:However, if you mean that ##AB = (BA)^{-1}##, etc., then that seems OK, in this example at least.