This is part of tifr 2017 detailed math solution series.

**State True or False**:

The set of nilpotent matrices of \( M_3 (\mathbb{R} ) \) spans \( M_3 (\mathbb{R} ) \) considered as an \( \mathbb {R} \) – vector space ( a matrix A is said to be nilpotent if there exists \( n \in \mathbb{N} \) such that \( A^n = 0 \) ).

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**Discussion TIFR 2017 math solution: ***False *

If such a basis exist. Then it will contain 9 matrices (because dimension of \( M_3 ( \mathbb{R} ) \) is 9 ).

Suppose \( \{ N_1, … , N_9 \} \) be the 9 nilpotent matrices which span \( M_3 (\mathbb{R} ) \). Now consider the usual basis of \( M_3 (\mathbb{R} ) \) :

$$ M_1 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} , M_2 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} , … , M_9 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

There is a linear transformation **L **(via a \( 9 \times 9 \) change of basis matrix) that sends \( \{ N_1, … , N_9 \} \) to \( \{ M_1, … , M_9 \} \).