Is a matrix diagonalizable if and only if it has an eigendecomposition?

If not, can you give an example of a diagonalizable matrix which doesn't have an eigendecomposition, or vice-versa?


1 Answer 1


Yes. If $M$ is a $n\times n$ matrix and if there are vector $v_1,\ldots,v_n$ each of which is an eigenvector of $M$ and such that $\{v_1,\ldots,v_n\}$ is a basis of $M$, if $P$ is the $n\times n$ matrix whose columns are the $v_k$'s, then $P^{-1}MP$ is a diagonal matrix.

And if $M$ is diagonalisable and $P$ is an invertible matrix such that $P^{-1}MP$ is a diagonal matrix, then the columns of $P$ are a basis of eigenvectors of $M$.

  • $\begingroup$ Can you give a source for this? I would imagine it depends on the field the matrix is over $\endgroup$ Apr 15, 2020 at 18:04
  • $\begingroup$ For what exactly? $\endgroup$ Apr 15, 2020 at 18:06
  • $\begingroup$ In the first part, how do we know that P is invertable? In the second part, how do we know that M only scales the columns of P? $\endgroup$
    – JacKeown
    Oct 13, 2021 at 18:46
  • $\begingroup$ @JacKeown Because its columns are a basis, and therefore linearly independent. So, $\det P\ne0$. $\endgroup$ Oct 13, 2021 at 18:47
  • $\begingroup$ @JoséCarlosSantos Ah yes. I see now. $\endgroup$
    – JacKeown
    Oct 13, 2021 at 18:52

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