User:IssaRice/Linear algebra/A matrix is only similar to itself if and only if it is a scalar multiple of the identity matrix

"matrix is only similar to itself" means that the linear map expressed in any single basis has the same matrix

The identity matrix has the remarkable property that it is only similar to itself: if A is matrix similar to I, then A=I. Why? We have $A=QI{Q}^{-1}$ for some invertible Q by definition of matrix similarity. but the rhs simplifies to I.

Are there any other matrices with this property? If $\lambda \in \mathbf{R}$, then for $\lambda I$ we have $Q(\lambda I)Q^{-1}=\lambda (QI{Q}^{-1})=\lambda I$ so any scalar multiple of the identity matrix also has this property.

Are there any others? It turns out there aren't! We want to show that if a matrix A does not have the form $\lambda I$, then there is a distinct matrix B that it is similar to. More precisely, suppose $A\ne \lambda I$ for any $\lambda \in \mathbf{R}$. Then there exists a matrix B and an invertible matrix Q such that $A=QB{Q}^{-1}$.

We split this proof into two cases:

1. Suppose A is not a diagonal matrix. Then there is an off-diagonal entry that is non-zero, say $a_{jk}$ (row j, column k, j!=k). Let E be the elementary matrix that multiplies row j by 2. Then $E^{-1}$, when applied from the right, divides column j by 2. Now if we consider $EAE{-}^{}1$, its entry j,k will be $2a_{jk}\ne a_{jk}$. So $EAE{-}^{}1\ne A$, even though the two matrices are similar.
2. Suppose A is a diagonal matrix where not all entries on the diagonal are equal. Call two of those non-diagonal entries j and k, so that $a_{jj}\ne a_{kk}$. Let E be the elementary matrix that transposes row j and row k. Then $E^{-1}$, when applied from the right, transposes column j with column k. Thus $EA{E}^{-1}$ will be the matrix with $a_{jj}$ and $a_{kk}$ swapped, but all other entries staying the same. So we see that $EAE{-}^{}1\ne A$, even though the two matrices are similar.