<s>[INST] Find invertible matrices without Jordan form Here is the statement of the problem: Let $ A $ be the following matrix with entries from $ \mathbb{R} $ $$ \begin{pmatrix} 1 & 2 & -4 & 4 \\ 2 & -1 & 4 & -8 \\ 1 & 0 & 1 & -2 \\ 0 & 1 & -2 & 3\end{pmatrix} $$
Let $ A^t $ be its transpose. Find an invertible matrix $ P \in M_4\left(\mathbb{R}\right) $ such that $ PAP^{-1}=A^t $
By using Jordan form, there are invertible matrices $ B,C $ so that $ BAB^{-1}=J=CA^tC^{-1} $. In other words, $ \left(C^{-1}B\right)A\left(C^{-1}B\right)^{-1}=A^t $, which is what we want. So it suffices to find such matrices. However, it seems one should calculate a lot to get the desired matrices. My question is ... 
Is it possible to answer this problem without using Jordan form?
Let $ B = \begin{pmatrix} 0 & 2 \\ 2 & -2 \end{pmatrix} $, then we have $ A-I_4=\begin{pmatrix} B & -B^2 \\ I_2 & -B \end{pmatrix} $. Perhaps it's a useful identity?
 [/INST] Normally, for "small" computational problems, it is usually best to work them in a straightforward manner, even if it is a bit tedious, because that is usually faster than finding a shortcut. In this problem however, you seem to have found the key to the shortcut. 
By using the useful identity in your question, we have:
\begin{align*}A - I_4 &= \begin{bmatrix}B&-B^2\\I_2&-B\end{bmatrix} \\ &= \begin{bmatrix}B&B\\I_2&I_2\end{bmatrix} \begin{bmatrix}I_2& \\ &-B\end{bmatrix} \\ &= \begin{bmatrix}B&\\ &I_2\end{bmatrix} \begin{bmatrix}I_2&I_2\\I_2&I_2\end{bmatrix} \begin{bmatrix}I_2& \\ &-B\end{bmatrix} \ \ \color{red}{(1)}\end{align*}
Since $B^T = B$, by transposing both sides of equation $\color{red}{(1)}$, we get: 
\begin{align*}A^T-I_4 &= \begin{bmatrix}I_2& \\ &-B\end{bmatrix}^T \begin{bmatrix}I_2&I_2\\I_2&I_2\end{bmatrix}^T \begin{bmatrix}B&\\ &I_2\end{bmatrix}^T \\ &= \begin{bmatrix}I_2& \\ &-B\end{bmatrix} \begin{bmatrix}I_2&I_2\\I_2&I_2\end{bmatrix} \begin{bmatrix}B&\\ &I_2\end{bmatrix} \ \ \color{red}{(2)}\end{align*}
Since $B$ is invertible, we can rearrange equation $\color{red}{(1)}$ as follows: 
\begin{align*}\begin{bmatrix}I_2&I_2\\I_2&I_2\end{bmatrix} &= \begin{bmatrix}B&\\ &I_2\end{bmatrix}^{-1}(A-I_4)\begin{bmatrix}I_2&\\ &-B\end{bmatrix}^{-1} \\ &= \begin{bmatrix}B^{-1}&\\ &I_2\end{bmatrix}(A-I_4)\begin{bmatrix}I_2&\\ &-B^{-1}\end{bmatrix} \ \ \color{red}{(3)}\end{align*}
Then, substituting equation $\color{red}{(3)}$ into equation $\color{red}{(2)}$ yields: 
\begin{align*}A^T-I_4 &= \begin{bmatrix}I_2& \\ &-B\end{bmatrix} \begin{bmatrix}I_2&I_2\\I_2&I_2\end{bmatrix} \begin{bmatrix}B&\\ &I_2\end{bmatrix} \\ &= \begin{bmatrix}I_2& \\ &-B\end{bmatrix}\begin{bmatrix}B^{-1}&\\ &I_2\end{bmatrix}(A-I_4)\begin{bmatrix}I_2&\\ &-B^{-1}\end{bmatrix}\begin{bmatrix}B&\\ &I_2\end{bmatrix} \\ &= \begin{bmatrix}B^{-1}& \\ &-B\end{bmatrix}(A-I_4)\begin{bmatrix}B& \\ &-B^{-1}\end{bmatrix} \\ &= \begin{bmatrix}B^{-1}& \\ &-B\end{bmatrix}A\begin{bmatrix}B& \\ &-B^{-1}\end{bmatrix} - \begin{bmatrix}B^{-1}& \\ &-B\end{bmatrix}I_4\begin{bmatrix}B& \\ &-B^{-1}\end{bmatrix} \\ &= \begin{bmatrix}B^{-1}& \\ &-B\end{bmatrix}A\begin{bmatrix}B& \\ &-B^{-1}\end{bmatrix} - I_4 \ \ \color{red}{(4)}\end{align*}
Adding $I_4$ to both sides of equation $\color{red}{(4)}$ yields $A^T = \begin{bmatrix}B^{-1}& \\ &-B\end{bmatrix}A\begin{bmatrix}B& \\ &-B^{-1}\end{bmatrix}$.
Therefore, we can set $P = \begin{bmatrix}B^{-1}& \\ &-B\end{bmatrix}$ to get $A^T = PAP^{-1}$.
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