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If a matrix representation of a linear transformation is similar to a diagonal matrix, why does this imply that the Jordan normal form must also be diagonal?

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    $\begingroup$ The Jordan normal form is unique, up to rearrangement, and a diagonal matrix is in Jordan normal form. $\endgroup$ – Paul Sinclair Aug 22 '15 at 4:32
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The comment by Paul Sinclair gives the fundamental reason why this is so: the Jordan normal form is a form that generalises the diagonal form such that all complex matrices are similar to some Jordan normal form, which is unique up to permutation of the blocks; diagonalisable matrices already were similar to some diagonal matrix, so that must be their Jordan normal form.

But explicitly, let $A$ have (i.e., be similar to) a Jordan normal form $J$ that has at least one Jordan block of size larger than$~1$ (so that it is not diagonal). By permutation of the blocks we may assume the first Jordan block has size${}>1$, and say it has $\lambda$ as diagonal entries; then the second standard basis vector $e_2$ satisfies $(J-\lambda I)e_2=e_1\neq0$ but $(J-\lambda I)^2e_2=0$. Since $J$ is similar to $A$, the vector $v$ corresponding to$~e_2$ under the change of basis satisfies $(A-\lambda I)v\neq0$ but $(A-\lambda I)^2v=0$ which shows that $A$ is not diagonalisable (since for diagonalisable $A$ one has that $\ker(A-\lambda I)^2$ is equal to the eigenspace $\ker(A-\lambda I)$ for$~\lambda$).

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Suppose the minimum polynomial of matrix $A$ is $$ m(\lambda)=(\lambda-\lambda_1)^{p_1}\cdots(\lambda-\lambda_k)^{p_k} $$ where $\lambda_1,\cdots,\lambda_k$ are distinct eigenvalues of $A$.

Then if $A$ is diagonalizable, $m(\lambda)$ must have linear order for all distinct eigenvalues, i.e. $p_i=1$. The maximum size of Jordan block for $\lambda_i$ is $p_i$. So if all $p_i=1$, then all Jordan blocks are of size $1$. Thus if $A$ is diagonalizable, the Jordan normal form of $A$ must also be diagonal.

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  • $\begingroup$ Why does diagonalizability of $A$ imply that $m(\lambda )$ has linear order in all distinct eigenvalues? $\endgroup$ – mathemather Apr 23 at 9:43
  • $\begingroup$ This is a well known theorem in linear algebra $\endgroup$ – Math Wizard Apr 23 at 14:12
  • $\begingroup$ yeah it is very easy to see also we just have to write down $(D- e_1)...(d-e_k)$ where D is diagonalization of A. $\endgroup$ – mathemather Apr 24 at 8:05

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