Which matrices are similar to themselves only?

Question

Which matrices are similar to themselves only?

If $A$ is similar to itself then $A=P^{-1}AP$ then $PA=AP.$ Then if $\lambda$ is an eigenvalue of $A$ then for $x\not =0$ we have $$PAx=APx\implies APx-\lambda Px=0\implies (A-\lambda I)Px=0.$$

Can I conclude that $A=\lambda I$ from this?

Answer 1

Every square matrix is similar to itself: you can take $P = I$.

No, $(A-\lambda I) P x = 0$ doesn't say $A=\lambda I$, it just says $Px$ is an eigenvector of $A$ for eigenvalue $\lambda$.

Answer 2

Your proof fails for two reasons:

1. $ABx=0$ means only that $Bx \in \operatorname{Ker}A$, not that $A=0$.
2. In your proof, $\lambda$ depends on $x$: different $x$ (= different eigenvectors) in general have different $\lambda$ (different eigenvalues).

Concerning your main question, any matrix is similar to itself: just take $P=I$ (the identity matrix):

$$I^{-1}AI = A$$

In fact, similarity is an equivalence relation. That is,

1. Any matrix is similar to itself: $I^{-1}AI=A$
2. If $A$ is similar to $B$, then $B$ is similar to $A$: if $B=P^{-1}AP$, then $A=PBP^{-1}=(P^{-1})^{-1} B P^{-1}$
3. If $A$ is similar to $B$ via $B = P^{-1}AP$, and $C$ is similar to $B$ via $C=Q^{-1}BQ$, then $A$ is similar to $C$: $C = Q^{-1}P^{-1}APQ = (PQ)^{-1} A PQ$

Answer 3

You're on the right track, but you're expressing your ideas in an imprecise fashion.

If $A$ is only similar to itself, then $A=P^{-1}AP$ for every invertible matrix $P$. Now, choose an invertible matrix $P$ and let $\lambda$ be an eigenvalue for $A$. Then, for some $x\ne0$, we have $Ax=\lambda x$, so $$ \lambda Px=PAx=APx $$ and therefore $Px$ is an eigenvector for $A$. Since $x\ne0$, for every $y\ne0$, there exists an invertible matrix $P$ such that $Px=y$ (why?). Therefore every nonzero vector is an eigenvector for $A$ relative to $\lambda$ and so the eigenspace has dimension $n$ (the order of the matrix $A$). Hence $A$ is diagonalizable; since there's a unique eigenvalue, we conclude $A=\lambda I$.

Note that this relies on the fact that an eigenvalue exists to begin with, so it holds for matrices over the complex numbers.