Proving Similarity of Transposed Matrices

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Homework Help Overview

The discussion revolves around proving that the transposes of similar matrices are also similar. The original poster presents an attempt to demonstrate this property using matrix transposition and properties of similar matrices.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to transpose the equation relating matrices A and B to explore the similarity of their transposes. They express uncertainty about how to manipulate the resulting equation to achieve the desired form.

Discussion Status

Some participants engage with the original poster's reasoning, suggesting a substitution that may simplify the expression. There appears to be a positive exchange, with one participant affirming the approach taken by the original poster.

Contextual Notes

The original poster mentions having limited experience with mathematical proofs, which may influence their confidence in the approach being taken.

phyzmatix
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Homework Statement



Let A and B be similar matrices. Prove that [tex]A^T[/tex] and [tex]B^T[/tex] are also similar.

2. The attempt at a solution

We know that, since A and B are similar matrices, there exists an invertible matrix P such that

[tex]B=P^{-1}AP[/tex]

so, I thought if we transposed both sides, this could lead to a proof, but this gives

[tex]B^T=\left(P^{-1}AP\right)^T[/tex]

[tex]B^T=P^TA^T\left(P^{-1}\right)^T[/tex]

[tex]B^T=P^TA^T\left(P^T\right)^{-1}[/tex]

Which feels close, but how do I get it to the form

[tex]B^T=\left(P^T\right)^{-1}A^TP^T[/tex] ?

Provided of course, that this is an actually viable route to take (I have virtually zero experience with mathematical proofs).

Thanks!
phyz
 
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hi phyz,
you reached [tex]B^T=P^TA^T(P^T)^{-1}[/tex].
if you put [tex]Q = (P^T)^{-1}[/tex] , what form would the above statement take?
 
Ha! Easy as that is it? :biggrin:

Thanks winter85!
 
yep :)
 

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