 Quick way to check if a matrix is diagonalizable. Is there any qui Juan Hewlett 2021-12-15 Answered
Quick way to check if a matrix is diagonalizable.
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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Firstly make sure you are aware of the conditions of Diagonalize 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 $$\displaystyle{\left(\text{multiplicity}={1}\right.}$$) 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.
Not exactly what you’re looking for? Mary Goodson
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. 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)