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.

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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.


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