Are Similar Matrices Always Similar to Their Transpose?

[calstudent]

1. Show that two matrices A,B ∈ Mn(C) are similar if and only if they share a Jordan canonical form.

2. Prove or disprove: A square matrix A ∈ Mn (F) is similar to its transpose A^T. If the statement is false, find a condition which makes it true.
(I'm pretty sure that this is true and can be proven using the above by showing that A and A^T share a Jordan form.)

I have a basic understanding of what Jordan blocks are and what JCF matrices look like, but I don't know what the identifying characteristics of a specific JCF are or how to show that two arbitrary matrices share the same form. I know in both questions that the two matrices have the same characteristic polynomial and spectrum, but I don't know where to go from there.

 

[CompuChip]

So let's start with the direct implication. Suppose that two matrices A and B share the same JNF. What does this mean explicitly? If A' is a JNF of A, then there exists a matrix P such that $A' = P^{-1} A P$ and A' has a specific form, right? So start working this out and try to show that they are similar (what does it mean explicitly if two matrices are similar?)

 

[calstudent]

But the fact that there exists a matrix P such that A'= P^-1AP, is the direct definition of similarity and the proof would be trivial. Does it follow directly from the definition of JNF that such a matrix P always exists?

 

[CompuChip]

No, it's not trivial (very easy, though). If, in what you wrote down, A' is the Jordan normal form, such a matrix P exists and indeed you express there that A is similar to its JNF. So you can assume that. Similarly, suppose that there is a(n invertible) matrix Q such that Q^-1A^TQ is the same Jordan normal form (and Q is the similarity transformation between the transpose of A and that normal form). Now you still have to prove that A and A^T are similar, that is: construct a matrix S such that A = S^-1 A^T S.
In fact this is a more general statement: if A is similar to C, and B is similar to C, then A is similar to B. You can try and prove that instead (it's actually the same proof as you don't have to use that C is in JNF anywhere). (In fact I think that similarity is even an equivalence relation, and the direct implication of your original question also shows the "hardest" part of that )