How to decompose a matrix into an antisymmetric matrix plus a multiple of the identity I was given a problem to solve earlier that I couldn't figure out.  I don't still have it, but it was basically: Given the invertible matrix $A$, find the invertible matrix $P$, such that $A=P^{-1}CP$.  Where $$C=\begin{bmatrix} c & -d \\ d & c\end{bmatrix}$$ and $A$ I don't exactly remember, but it was something like $$A=\begin{bmatrix} 3 & 1 \\ 1 & 0\end{bmatrix}$$
I was completely stumped.  I see that $A$ and $C$ are similar so they should have the same determinant and trace.  So I should be able to find $C$ from that.  But how would I find $P$?
And I can see that this is similar to eigendecomposition, but is this decomposition at all useful, or did some textbook writer just come up with this problem with no underlying use?

Hint: If two matrices are diagonalizable with the same eigenvalues, then they are similar.
If an $n \times n$ matrix has $n$ distinct eigenvalues, it is necessarily diagonalizable.
The punchline: you should try to find a matrix $C$ that has the same eigenvalues as $A$.

Note that if we have invertible matrices $S,T$ such that
$$
SCS^{-1} = D = TAT^{-1}
$$
then
$$
A = T^{-1}SCS^{-1}T = (S^{-1}T)C(S^{-1}T)
$$
