Is it true that the only matrix that is similar to a scalar matrix is itself A square matrix $A$ is said to be similar to a square matrix $B$, of the same order, if there is an invertible matrix $P$ such that
$A = PBP^{-1}$.
Is this true?
The only matrix that is similar to a scalar matrix is itself.
I.e., if $C$ is a scalar matrix and $T = P CP^{−1}$, then $T = C$.
If yes, how do I prove it?
If $C$ is a scalar matrix, then $C = \mu I_{d}$, then $T = PCP^{-1} = P(\mu I_{d})P^{-1} = \mu I_{d} (P P^{-1}) = \mu I_{d}$
Note : This proof relies on a simple detail and funny exercise, but that should not be underestimated; In the third equality I used that the center of matrices are only the scalar ones, i.e $Z(M_{n}(\mathbb{K})) = \mu I_{d}$
If $C$ is a scalar matrix, then $C = \mu I_{d}$, then $T = PCP^{-1} = P(\mu I_{d})P^{-1} = \mu I_{d} (P P^{-1}) = \mu I_{d}$
Note : This proof relies on a simple detail and funny exercise, but that should not be underestimated; In the third equality I used that the center of matrices are only the scalar ones, i.e $Z(M_{n}(\mathbb{K})) = \mu I_{d}$