Harshal posted an Question
July 26, 2020 • 17:33 pm 30 points
  • IIT JAM
  • Mathematics (MA)

Given the matrix

3 Answer(s) Answer Now
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  • comment-profile-img>
    Deepak singh 1

    in the above example AM is not equal to GM also .. result- A matrix is diagonizable iff it has n eigenvalues and AM= GM of each eigenvalues.. this result holds in both real and complex..

  • comment-profile-img>
    Deepak singh 1

    see, this symmetric complex matrix is no diagonizable.. so, A symmtric matrix may not be diagonizable.

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  • comment-profile-img>
    Deepak singh 1 Best Answer

    see attached, i have mentioned how to find AM and GM

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  • Shashi ranjan sinha

    Similar question has been asked in CSIR NET Dec 19 in part C

  • Shashi ranjan sinha best-answer

    see the attachment

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    The result that I have wrote for such a matrix M is very helpful.

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    can you please explain what am & gm is...

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    i am getting a bit confusion there

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    see

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    Yes 2nd is also correct

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    let me explain u am and gm

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    However, GM of the Eigen value say "d" is the dimension of the null space of the matrix A- dI, where I is identity matrix

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    Am of Eigen value is its multiplicity as a root of the characteristics polynomial of the matrix

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    Harshal, if you still have any doubt....I will surely try to clear it...you can ask without hesitation

  • comment-profile-img>
    Deepak singh 1 best-answer

    see

    cropped2129266006.jpg
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    what is AM and GM?

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    and it is msq type que

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    AM - algebraic multiplicity GM = geometric multiplicity

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    this result is true only for real symmetric matrix, not over complex field

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    use the result for finding eigen value ..which is -i and 3 i

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    then since rank is 4, there will be 4 eigenvalues and rank of this matrix is 4 , it is diagonalizable also (AM=GM) ..

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    read about AM any GM ..to know more..

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    any doubt then ask..

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    u said rank = 4, so it must have 4 Eigen values..... but Eigen values has nothing to do with rank of a matrix.... for example rotational matrix has rank 2 but it has no Eigen values

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    rank of matrix = no. of eigenvalue

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    there will be 4 eigenvalues , no doubt ..

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    I didn't understand ur reason for diagonalizability

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    read this

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    read it carefully...these statements are true for real symmetric only

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    dear shashi , diagonalizable iff AM=GM holds for both complex and real..

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