I remember there exist a decomposition such that it transforms any given matrix to a block diagonal matrix, and each block having single eigenvalue. I couldn't find the name, please help me.

I tried to do it myself. Using Shur decomposition, we can transform any matrix to upper triangular matrix, such that diagonal elements are eigenvalues, and we can order them from smallest to largest. But I am not sure how to transform it to block diagonal matrix.

Please note that I am not talking about Jordan Canonical form, since in JCF different blocks can contain same eigenvalue.

$A\Longrightarrow \begin{bmatrix} A_1& 0&\ldots&0 \\0 & A_2 &\ldots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\ldots&A_n\end{bmatrix}$, such that all eigenvalues of $A_i$ equal to $\lambda_i$. And $\lambda_i\neq\lambda_j$ for all $i\neq j$.

Assume $A\in\mathbb{R}^{n\times n}$.

  • $\begingroup$ In Jordan form, if you group together blocks for the same eigenvalue (as any reasonable approach would do, though the Wikipedia definition does not seem to require it), you can consider all of them together to form a single block. Maybe one with zeros in places they are not required to be, but still a block that appears to match your requirements. $\endgroup$ Apr 13, 2020 at 9:26
  • $\begingroup$ @MarcvanLeeuwen I am looking to something that I can use to coding. Unfortunately, it is not possible to find JCF of any given matrix. $\endgroup$
    – Lee
    Apr 13, 2020 at 9:30
  • $\begingroup$ Coding? If you are not working over the complex numbers, you won't necessarily have all the eigenvalues. If you are working with (approximated) complex coefficients, the finding a JNF is possible (though probably not much fun). $\endgroup$ Apr 13, 2020 at 9:32
  • $\begingroup$ To add to Marc's comments: it is true that trying to compute the Jordan decomposition in inexact arithmetic is not usually tenable, and you'll have to settle for the Murnaghan-Winter (block Schur) decomposition. (See e.g. Golub and Van Loan.) If you can symbolics (e.g. in a computer algebra system), then Jordan might be possible. $\endgroup$ Apr 13, 2020 at 14:03

1 Answer 1


Assuming you are working over an algebraically closed field (like the complex numbers), any basis adapted to the direct sum decomposition of the whole space into generalised eigenspaces would match the requirement. There is a lot of choice there, you can improve the individual blocks to be upper triangular, like in a Schur form (but without the unitary requirement for the change of basis matrix), and even a Jordan form if you try harder.


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