How to decompose a matrix into an antisymmetric matrix plus a multiple of the identity

253 Views Asked by At

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?

1

There are 1 best solutions below

3
On BEST ANSWER

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 $$ A = T^{-1}SCS^{-1}T = (S^{-1}T)C(S^{-1}T) $$