Symmetric, invertible transformation matrix for showing similarity between A^T and A

[sleek oyster]

\begin{problem} Any $A \in M_n(K)$ is similar to $A^{\tr}$. Show that there exists a \underline{symmetric}, invertible $P \in M_n(K)$ with $P^{-1} \cdot A \cdot P =A^{\tr}$. \end{problem}

[spring gorge]

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[sleek oyster]

How do we do this?

[left sky]

lemme find a better source

[left sky]

Kaplansky "Linear algebra and geometry, a second course"
Theorem 66

@sleek oyster

literally the thing you're looking for

the proof is quite involved however, so take your sweet time

[sleek oyster]

Well yeah the first part

[sleek oyster]

Do you have any hints or something?

[left sky]

the opening of the proof is what I'd pretty much hint at as well

[sleek oyster]

The exercise refers to this section of my lecture notes:

Not sure if that's similar to Kaplansky's proof

[left sky]

oh god I don't speak Reich 😦

[sleek oyster]

Basically it says one can define an invertible C_0 with A_f * C_0 = C_0 * A_f^(tr)

And that C_0 is given there

With C_0 = C_0^(tr)

[left sky]

what is f supposed to be

[sleek oyster]

Just some degree n polynomial

It says that we already showed A_f and A_f^(tr) are similar

Where A_f is the companion matrix to f

[sleek oyster]

Thanks

[mossy nimbus]

@sleek oyster has given 1 rep to @left sky

[sleek oyster]

+close