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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) $\text{r}(A)=\text{r}(B)$.

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  • $\begingroup$ 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
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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 $=0$.

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