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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 $$\{ M_1, … , M_9 \}$$.