I would like to understand why the following is true for $A \in M_{nxn}(K)$: $$\text{the equivalence class of A in the equivalence relation of two matrices being similar consists only of A} \iff \exists a\in K \ (A=aI)$$ Two matrices are similar when $B=C^{-1}AC$. I am also aware that if $A$ and $B$ are similar matrices, then: 1) $\text{det}A=\text{det}B$, 2) $\text{tr}(A)=\text{tr}(B)$, 3) $\text{r}(A)=\text{r}(B)$.

  • $\begingroup$ It is actually not true: not every matrix is similar to a multiple of the identity. If anything, then an interesting class of matrices consists of matrices that are similar to a diagonal matrix, but a diagonal matrix can have different elements on its diagonal. $\endgroup$ – uniquesolution Mar 3 '17 at 9:36
  • $\begingroup$ @uniquesolution Perhaps the wording of my question was somewhat unclear. I didn't mean to claim that every matrix was similar to $aI$ but rather that if we have a matrix that is only similar to itself than it is of the form $aI$. $\endgroup$ – Zelazny Mar 3 '17 at 9:41
  • $\begingroup$ Oh, Ok, so all you need to do is to verify that if $C^{-1}AC=A$ for all matrices $C$, then $C$ is a multiple of the identity. Can you do that? $\endgroup$ – uniquesolution Mar 3 '17 at 9:44
  • $\begingroup$ Have you learned Jordan canonical form? $\endgroup$ – Li Li Mar 3 '17 at 9:53
  • $\begingroup$ @LiLi no, the material I'm going over hasn't covered this topic so far. $\endgroup$ – Zelazny Mar 3 '17 at 9:55

Let $C^{-1}AC=A$ for all invertible matrices $C \in M_{nxn}(K)$.

Hence $AC=CA$ for all invertible matrices $C \in M_{nxn}(K)$.

If $D \in M_{nxn}(K)$, take $t \in K$ such that $C:=D-tI$ is invertible ( hence take $t$ such that $t$ is not an eigenvalue of $D$).

Then it is easy to see that $AD=DA$.

Consequence: $AD=DA$ for all matrices $D \in M_{nxn}(K)$.

Can you now show that , for some $a \in K$ we have $A=aI$ ?

Hint: consider the matrices $E_{ij}$, where $E_{ij}$ is the matrx whose $(i,j)$ - entry $=1$ and all other entries are $=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.