I have just proven that a Matrix $A \in M_{2}(\mathbb C)$ is either diagonalizable or similar to a matrix $B=\begin {pmatrix}\lambda &1 \\0& \lambda \end {pmatrix}$ $\lambda \in \mathbb C$.

I used the fact that if A is non-diagonalizable, its Eigenvalues are not pairwise distinct and since A only has 2 Eigenvalues, they have to be the same.

I feel that this feature might be pretty useful in order to show whether or not a matrix is diagonalizable, assuming this is possible in a general setting. So I considered a general case, namely,

if $A \in M_{n}(\mathbb C)$ then A is either diagonalizable or similar to a matrix

$C=\begin {pmatrix}\lambda &1&...&1 \\0& \ddots&\ddots &\vdots \\ \vdots&\ddots&\ddots &1 \\0&\dots&0 &\lambda \end {pmatrix}$ $\lambda \in \mathbb C$

So I tried to prove it for the $A \in M_{3}(\mathbb C)$ case first. There I noticed, if I assume A is non-diagonalizable, its Eigenvalues are either all the same (which would result in being similar to C) or 2 of them are equal and the third one is distinct. In order to prove the proposition I need to prove that the latter case cannot occur or that even if it occurs it is still similar to C. But now I am stuck.

I thought that if I can prove it for the $A \in M_{3}(\mathbb C)$ case I can see a pattern which coincides with the $M_{2}(\mathbb C)$ one and prove it for $ M_{n}(\mathbb C)$ via induction.

But I somehow feel that it is kinda pedestrian or not even true in the general case.

Any thoughts and hints are appreciated. Thank you in advance.

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    $\begingroup$ en.wikipedia.org/wiki/Jordan_normal_form $\endgroup$ – Will Jagy Apr 13 '14 at 20:02
  • $\begingroup$ Did you forget to restrict this to matrices with only one eigenvalue? $\endgroup$ – Git Gud Apr 13 '14 at 20:04
  • $\begingroup$ Expanding on Will's comment, this is not possible in general. The correct generalization is the Jordan Normal Form. $\endgroup$ – Alex Becker Apr 13 '14 at 20:05
  • $\begingroup$ I haven't heard of this one yet. Thank you very much. $\endgroup$ – user114193 Apr 13 '14 at 20:44

The Jordan normal form shows the possible canonical matrices you can get equivalent to if you have a repeated eigenvalue. In particular, for $n > 2$ there are multiple possible Jordan normal forms and they are not equivalent under change of basis by definition, so it's impossible to always get to the upper triangular matrix you are trying to reach through change of basis. (Otherwise all possible Jordan normal forms would be equivalent, a contradiction.)

  • $\begingroup$ So the proposition is rather: A is diagonalizable or similar to the Jordan normal form? $\endgroup$ – user114193 Apr 13 '14 at 20:47
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    $\begingroup$ Yes, exactly. And the Jordan normal form could be a diagonal matrix, in which case your matrix is diagonalizable. But a matrix is always similar to its Jordan normal form, and no two different Jordan normal forms are similar if the eigenvalues are placed in increasing order. $\endgroup$ – user2566092 Apr 13 '14 at 20:55

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