Diagonalizable and Jordan Canonical Form | CSIR NET 2011 to 2023

[@mcbc01]

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

[@hajeeranasreenqueenofscience]

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

[@suniltiwari9215]

True

[@yogeshjadhavpatil4120]

Aafke video se bahut help milati hai

[@pranavninghot3777]

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

[@DrHarishGarg]

Thanks

[@mystory5438]

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

[@masterkapil6214]

Thnku sir aap hmare liye bhut acha content tyar krte ho

[@rajithak2606]

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🙏

[@mak511]

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

[@nidhiyadav1577]

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

[@SADDAMHUSSAIN-mw3cv]

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

[@ishadevi2071]

Very helpful 😊

[@super40educationalinstitut29]

Very good sir.. mja aa gaya

[@sat4maths]

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.....

[@RanjangamingYT7477]

55:36 jast example [1/2,1/2][1/2,1/2] this is idempotent but diagonal entries not 0 and 1 Therefore b option discard

[@kritikagrover4980]

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

[@roshaningale5907]

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

[@Achukutty25]

Yes, rank is 1, so GM is 2

[@BalajiPadhy-l8f]

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

[@jittaboinaramu1537]

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

[@amitgupta-sk4hw]

Correct buddy

[@hajeeranasreenqueenofscience]

In hurry sir has said like that...

[@Manish-uq5le]

Thanks you sir ❤

[@906mukesh8]

Thanks you sir

[@SADDAMHUSSAIN-mw3cv]

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

[@ronneysaini7763]

Thanks Sir ji❤

[@shrawankumarpatel2148]

Thank you so much sir

[@sureshm92]

Thank you sir...

[@indhui1911]

A^2 =A matrix is Idempotent **

[@maharaja8085]

Both

[@anitakaushik-be8mo]

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

[@906mukesh8]

Thank you so much

[@shwetarathi304]

Thankyou sir

[@lm-sajid1985]

sir,at 47:48 if we take the trace of only A then option c is true but if we are having A= -I then the sum of trace of A and A(square) is 0..then option b must be true ?

[@aaqibwani7406]

Dear Dr,,,,at 55.06 just take an example of [[1/2,1/2],[1/2,1/2]] to discard option 2.

[@ummehabeeba9026]

Sir can you solve modern algebra pyq?

[@HeWhoKnows-]

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

[@mirshabir3672]

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

[@hajeeranasreenqueenofscience]

A=I ,evs( 1,1,1) not distinct so not diagonalizable.

[@anupamsamanta6690]

Sir Identify matrix always diagonalizable

[@GameFeverHQ]

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

[@shimorikichishorts8302]

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

[@archanakumarimeena3058]

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

[@saikatmalla7390]

sir please upload ring theory previous year question lecture

[@physicsmagic786]

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

[@hajeeranasreenqueenofscience]

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

[@Radharanilove-y5d]

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

[@RanjangamingYT7477]

I think only C option true hoga. B,d option ka rank 2 ha question me rank of Matrix 1 ha

[@mathematical2508]

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

[@ishitaagrawal987]

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

[@bantumath3527]

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]

Am is always equal to Gm for identity matrix

[@bantumath3527]

@@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

[@mainakbanerjee9282]

7:44 are the eigen values correct?

[@manishasharma9637]

I think not correct

[@mathmadeeasy2601]

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

[@Urbarman]

Why ?? Although it satisfy A^2= A

[@RanjangamingYT7477]

1:6:20 B option not true . Only A true

[@SADDAMHUSSAIN-mw3cv]

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

[@waseemayoub6227]

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.

[@amitgupta-sk4hw]

❤

[@rintumondal5658]

1:02:07

[@charitylyngdoh8912]

Thank you so much sir 🙏

[@artidevi8628]

thank you so much sir

[@ummehabeeba9026]

Thank you sir

[@shrutik5302]

Thanks sir 🌼

[@yashrajtripathi4832]

Thank you sir 😊

[@yogeshjadhavpatil4120]

Thank you sir

[@professorragavan9783]

Thank you sir