Target CSIR NET June 2024 Exam with PYQ Discussion Series | Linear Algebra by Sumit Sir | CSIR NET

Prepare to ace your CSIR NET June 2024 exam with our focused live class on Linear Algebra! Sumit Sir will discuss and solve previous years' questions (PYQs) from the CSIR NET June 2023 exam, providing you with the expertise and confidence to tackle this crucial subject.
Download Slides of this lecture - forms.gle/BgCi...
What You'll Learn:
Detailed step-by-step solutions to PYQs from the CSIR NET June 2023 Linear Algebra section
Fundamental and advanced concepts in Linear Algebra
Effective problem-solving techniques and strategies
Real-time Q&A session with Sumit Sir to address your doubts and questions
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[@akashsingh-hl6lk]

Apki class may time ka pata hi nahi chalta hai lugta hai ki abhi to start Kiya tha ...abhi hi khatam Thank you so much sir 🙏❤❤❤❤❤😊😊😊😊😊😊😊❤❤❤

[@samikshajakhar]

Nice explanation sir...very helpful❤

[@divyasingh-mb5ek]

Thanq sir some doubts are clear

[@sandeepsirmaths7500]

Hero bhai ❤

[@bhavishyaacademy9264]

Nice explanation sir ❤

[@anandnishant77]

Nice

[@bhavishyaacademy9264]

Bhut sare doubt clear ho gaye

[@shweta9227]

Thankyou sir 😊

[@velmurugandon1241]

Option c is also correct.T has 1 as an eigen value so Tv= v for some nonzero vector v

[@gauravtyagitv7538]

How can you claim that one is an eigen value?

[@velmurugandon1241]

You are correct one need not be an eigen value

[@dipasahu3507]

But 1 is the root of minimal polynomial. It is diagonalizable so it have 3 distinct root then why 1 is not eigen value????

[@gauravtyagitv7538]

It is one sided implication i.e. n distinct eigenvalues imply diagonalizability not conversely

[@user-vn7yx8ek7n]

Sir sep ka paper bhi kra dijiye