### CSIR JUNE 2011 PART C QUESTION 77 SOLUTION (Diagonalizability of Nilpotent matrices)

Let $N$ be a $3 \times 3$ non-zero matrix with the property $N^3 = 0$. Which of the following is/are true?
1. $N$ is not similar to a diagonal matrix,
2. $N$ is similar to a diagonal matrix,
3. $N$ has one non-zero eigenvector,
4. $N$ has three linearly independent eigenvectors.

I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you.
Solution: An $n \times n$ matrix $N$ is said to be nilpotent if $N^k = 0$ for some $k \in \Bbb N$. A matrix $M$ is said to be diagonalizable over a field $\Bbb F$ if there exists an invertible matrix $P$ with entries from the field $\Bbb F$ such that $$P A P^{-1} = \text{diag}(\lambda_1,\lambda_2,\dots,\lambda_n)$$ where $\text{diag}(\lambda_1,\dots,\lambda_n)$ is the diagonal matrix with the diagonal entries eigenvalues of $A$.
Observation1: All the eigenvalues of a nilpotent matrix are zero.
Proof: Let $N$ be a nilpotent matrix and $\lambda$ be an eigenvalue of $N$. Then $\lambda^k$ is an eigenvalue of $N^k = 0$. Hence $\lambda^k = 0$ and in turn $\lambda = 0$.
Result1: If $A$ is a non-diagonal matrix whose eigenvalues are equal. Then $A$ is not diagonalizable.
Proof: Suppose $A$ is diagonalizable, then there exists an invertible matrix $P$ such that $P A P^{-1} = \text{diag}(\lambda,\lambda,\dots,\lambda)$. This implies that $A = P^{-1}\text{diag}(\lambda,\lambda,\dots,\lambda)P$. But $\text{diag}(\lambda,\dots,\lambda)$ is a scalar matrix and hence it commutes with all the matrices. So $A = P^{-1}\text{diag}(\lambda,\dots,\lambda)P = \text{diag}(\lambda,\dots,\lambda)$ Contradiction to the fact that $A$ is a non-diagonal matrix.
Result2: An $n \times n$ matrix $M$ is diagonalizable if and only if it has $n$ linearly independent eigenvectors.
option 1:(True) Follows from observation 1 and result 1.
option 2:(False) Because option 1 is true.
option 3:(False) We have shown that $0$ is an eigen value of $N$. Hence there exists an eigen vector $v \ne 0$ such that $N v = 0$. Let $\alpha$ be any scalar then $N (\alpha v) = \alpha Nv = 0$. So any scalar multiple of eigenvector is again an eigenvector. So there are more than one eigen vectors.
option 4:(False) We have shown that $N$ is not diagonalizable in option 1. Hence by result 2 it cannot have $n$ linearly independent eigen vectors.

### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...