Is there any quick way to check whether a matrix is diagonalizable or not?

In exam if a question is asked like "Which of the following matrix is diagonalizable?" and four options are given then how can one check it quickly? I hope my question makes sense.

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    Find the eigenvalues, if the multiplicity of each of them is one then it is diagonalizable (by Hamilton-Cayley theorem). If there are eigenvalues with multiplicity more than one check for the dimension of the Eigenspace related to those eigenvalues, if the dimension is less than the multiplicity then it is not diagonalizable. – Kat Nov 06 '16 at 04:43
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    @Ron But it will take time. There are 4 matrices in option. – zafran Nov 06 '16 at 04:46
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    Another tool: the spectral theorem. (All symmetric matrices are diagonalizable over the reals. More generally, all normal matrices are diagonalizable over the complex numbers.) – symplectomorphic Nov 06 '16 at 05:09

2 Answers2


Firstly make sure you are aware of the conditions of Diagonalizable matrix.

In a multiple choice setting as you described the worst case scenario would be for you to diagonalize each one and see if it's eigenvalues meet the necessary conditions.

However, as mentioned here:

A matrix is diagonalizable if and only if for each eigenvalue the dimension of the eigenspace is equal to the multiplicity of the eigenvalue.

Meaning, if you find matrices with distinct eigenvalues (multiplicity = 1) you should quickly identify those as diagonizable.

It also depends on how tricky your exam is. For instance if one of the choices is not square you can count it out immediately. On the other hand, they could give you several cases where you have eigenvalues of multiplicity greater than 1 forcing you to double check if the dimension of the eigenspace is equal to their multiplicity.

Again, depending on the complexity of the matrices given, there is no way to really spot-check this unless you're REALLY good at doing this all in your head.

Robert Lee
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One nice characterization is this: A matrix or linear map is diagonalizable over the field F if and only if its minimal polynomial is a product of distinct linear factors over F.

So first, you can find the characteristic polynomial (https://en.wikipedia.org/wiki/Characteristic_polynomial). If the characteristic polynomial itself is a product of linear factors over $F$, then you are lucky, no extra work needed, the matrix is diagonalizable.

If not, then use the fact that minimal polynomial divides the characteristic polynomial, to find the minimal polynomial. (This may not be easy, depending on degree of characteristic polynomial)

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