# Understand the problem

Is it possible to have a non-identity diagonalizable, invertible, complex matrix s.t characteristics polynomials of and are the same?

##### Source of the problem

TIFR GS 2019, Part B Problem 4

##### Topic

Linear algebra

##### Difficulty Level

Hard

##### Suggested Book

Linear Algebra; Hoffman and Kunze

# Start with hints

Do you really need a hint? Try it first!

Think in terms of eigenvalues. Let and are eigenvalues of then eigenvalues of are and .

You need So as exists. Can you think from here? What about and ?

,,

Now, what are the cases?

Discard as and diagonalizable. Discard the case of both eigenvalues equal as is not diagonalizable for repeated eigenvalues. The only possibility is as the eigenvalue set for . So . Hence, the statement is false.

# Watch the video (Coming Soon)

# Connected Program at Cheenta

#### College Mathematics Program

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.