Diagonalizable and Jordan Canonical Form | CSIR NET 2011 to 2023

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  • Опубликовано: 9 сен 2024
  • This lecture explains the PYQs on Diagonalizable and Jordan Canonical Form

Комментарии • 65

  • @hajeeranasreenqueenofscience
    @hajeeranasreenqueenofscience 8 месяцев назад +11

    A²=A idempotant sir and A²=I is involuntary.

  • @SADDAMHUSSAIN-mw3cv
    @SADDAMHUSSAIN-mw3cv 9 месяцев назад +10

    1:06:03. But Correct options are (a) and (d) Only.

  • @mcbc01
    @mcbc01 10 месяцев назад +3

    This is series will be very help full for upcoming csir net

  • @rajithak2606
    @rajithak2606 2 месяца назад

    Thanks a lot sir for providing the tips for every concept of u taken sir it helps me a lot to understand it in better level and think in a objective way and I always thankful to you sir🙏

  • @charitylyngdoh8912
    @charitylyngdoh8912 21 день назад

    Thank you so much sir 🙏

  • @mystory5438
    @mystory5438 6 месяцев назад

    Thank you so much for your lecture sir, . It is very useful for us.

  • @pranavninghot3777
    @pranavninghot3777 6 месяцев назад +2

    A^2 = A is idempotent matrix
    A^2 = I is involuntary matrix

  • @ishadevi2071
    @ishadevi2071 Месяц назад

    Very helpful 😊

  • @mak511
    @mak511 9 месяцев назад +2

    Sir in this video many mistakes in the solution...
    overall most helpful thanks for this .

  • @nidhiyadav1577
    @nidhiyadav1577 10 месяцев назад

    Good evening sir,
    Thank you so much sir for making helpful lectures.
    THANK YOU SIR.

  • @yogeshjadhavpatil4120
    @yogeshjadhavpatil4120 9 месяцев назад

    Aafke video se bahut help milati hai

  • @jittaboinaramu1537
    @jittaboinaramu1537 10 месяцев назад +6

    Sir , in some problems at A²=I is involuntary matrix but you told that it is idempotent matrix ..

  • @professorragavan9783
    @professorragavan9783 Месяц назад

    Thank you sir

  • @sat4maths
    @sat4maths 5 месяцев назад

    Hi sir, All your short trick are very nice and useful to solve the problems quickly. But few problems ,there some mistakes that rank of Matrix is we count no.if non zero rows right..... In some problems you took no of zero rows.....

  • @masterkapil6214
    @masterkapil6214 10 месяцев назад

    Thnku sir aap hmare liye bhut acha content tyar krte ho

  • @SADDAMHUSSAIN-mw3cv
    @SADDAMHUSSAIN-mw3cv 9 месяцев назад +2

    59:33 sir here you taken an example of a matrix of zero eigen values.

  • @yashrajtripathi4832
    @yashrajtripathi4832 2 месяца назад

    Thank you sir 😊

  • @kritikagrover4980
    @kritikagrover4980 10 месяцев назад

    Thank you so much sir for all your efforts🙏🏻🙏🏻

  • @roshaningale5907
    @roshaningale5907 3 месяца назад +1

    10:57 dec 2014 questions AM(1)= 2,GM=1

    • @Achukutty25
      @Achukutty25 3 месяца назад

      Yes, rank is 1, so GM is 2

  • @user-ec3dq2bf1c
    @user-ec3dq2bf1c Месяц назад

    Dear Sir If A=I then A is always Diagonalizable Bcoz am(1) = Gm(1)@1.11.05

  • @anitakaushik-be8mo
    @anitakaushik-be8mo 10 месяцев назад +2

    In que june 2017 ..option b is not correct sir ..for idempotent matrix it is not necessary to be an diagonal matrix ..it can be all entries with (1/n) type ..so how it is correct by your side

  • @shrutik5302
    @shrutik5302 5 месяцев назад

    Thanks sir 🌼

  • @artidevi8628
    @artidevi8628 9 месяцев назад

    thank you so much sir

  • @906mukesh8
    @906mukesh8 8 месяцев назад

    Thanks you sir

  • @indhui1911
    @indhui1911 10 месяцев назад +3

    A^2 =A matrix is Idempotent **

  • @906mukesh8
    @906mukesh8 10 месяцев назад

    Thank you so much

  • @ronneysaini7763
    @ronneysaini7763 10 месяцев назад

    Thanks Sir ji❤

  • @shwetarathi304
    @shwetarathi304 10 месяцев назад

    Thankyou sir

  • @anupamsamanta6690
    @anupamsamanta6690 10 месяцев назад +1

    Sir Identify matrix always diagonalizable

  • @shimorikichishorts8302
    @shimorikichishorts8302 2 месяца назад

    How can we know rank of a matrix from characteristic polynomial 40:10

  • @archanakumarimeena3058
    @archanakumarimeena3058 3 месяца назад

    24:42 par A2=0 to A^3=0 hi.hoga zero matrix se A ka.multipliy karenge to zero honhoga A^3=A^2.A

  • @mainakbanerjee9282
    @mainakbanerjee9282 20 дней назад

    7:44 are the eigen values correct?

  • @ummehabeeba9026
    @ummehabeeba9026 10 месяцев назад +1

    Sir can you solve modern algebra pyq?

  • @saikatmalla7390
    @saikatmalla7390 3 месяца назад

    sir please upload ring theory previous year question lecture

  • @super40educationalinstitut29
    @super40educationalinstitut29 9 месяцев назад

    Very good sir.. mja aa gaya

  • @physicsmagic786
    @physicsmagic786 5 месяцев назад

    11 : 25 e. Values are 1,1,2 which are not distinct so how it can be diagonalisable.

  • @mirshabir3672
    @mirshabir3672 9 месяцев назад +1

    Net Dec. 2018 A is a real matrix and Characteristic polynomial (X-1)^3......Sir, can you explain the question again ............

  • @HeWhoKnows-
    @HeWhoKnows- 10 месяцев назад

    Sir can you please do a short video on Rate of convergence of Müller's method.
    I've searched in my library and non of the books has detailed proofs and just a statement that it convergence to 1.84

  • @GameFeverHQ
    @GameFeverHQ 9 месяцев назад +1

    There are so many mistakes sir,like you said identity matrix is not diagonalisable . overall helpful

  • @kalyandasguruji
    @kalyandasguruji 2 месяца назад

    1:10 sir why option d is wrong .... In this option they didn't ask the minimal polynomial...so why we check tgis with minimal polynomial....matrix M satisfy this option d....and also one is not an eigen value in d option ...
    Please clarify this option

  • @rintumondal5658
    @rintumondal5658 3 месяца назад

    1:02:07

  • @amitgupta-sk4hw
    @amitgupta-sk4hw 10 месяцев назад

  • @hajeeranasreenqueenofscience
    @hajeeranasreenqueenofscience 8 месяцев назад

    Sir 1:10 dec18 ...A=I ev's (1,1,1) not distinct so not diagonalizable Right

  • @ishitaagrawal987
    @ishitaagrawal987 9 месяцев назад +1

    sir identity matrix is always diagonalizable then how are you discarding options saying it is not.

    • @bantumath3527
      @bantumath3527 9 месяцев назад

      Im also thinking that I is not diagonalizable,but it is not true
      But how sir prove AM(a=1) is not same as GM(a=1)

    • @ishitaagrawal987
      @ishitaagrawal987 9 месяцев назад

      Am is always equal to Gm for identity matrix

    • @bantumath3527
      @bantumath3527 9 месяцев назад

      @@ishitaagrawal987 no
      Am(1)=3 and gm(1)= nullity(A-0.I)
      Which is 3 but sir take rank of A ,that why it is not equal
      Now i know it is equal

  • @mathematical2508
    @mathematical2508 10 месяцев назад

    Sir aapne diagonalizable ka jo first point likhvaya h vo iff nahi h
    Diagonalizable ke liye jaruri nahi h ki eigen value distinct aayengi
    Distinct imply krta h diagonalizable honga

  • @mathmadeeasy2601
    @mathmadeeasy2601 3 месяца назад

    54:00 can't be use matrix [1/2 1/2 1/2 1/2]

    • @Urbarman
      @Urbarman 2 месяца назад

      Why ?? Although it satisfy A^2= A

  • @SADDAMHUSSAIN-mw3cv
    @SADDAMHUSSAIN-mw3cv 9 месяцев назад +1

    54:48 only options (a) ,(c), and (d). are correct.

    • @waseemayoub6227
      @waseemayoub6227 8 месяцев назад

      We can take nilpotent matrix to discard option (a).
      Why is b not true...if eigen values are distinct then it is necessarily diagonalisable.
      Correct me if I am wrong sir.

  • @yogeshjadhavpatil4120
    @yogeshjadhavpatil4120 9 месяцев назад

    Thank you sir

  • @Manish-uq5le
    @Manish-uq5le 7 месяцев назад

    Thanks you sir ❤

  • @shrawankumarpatel2148
    @shrawankumarpatel2148 10 месяцев назад

    Thank you so much sir

  • @sureshm92
    @sureshm92 10 месяцев назад

    Thank you sir...

  • @ummehabeeba9026
    @ummehabeeba9026 10 месяцев назад

    Thank you sir